adasalai.org POWERS OF IMAGINARY UNIT = i i 2001 Division algorithm : n = 4(q) + r

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1 .COMPLEX NUMBERS Pai POWERS OF IMAGINARY UNIT i =, i = i, i 4 = Division algorithm : n = 4(q) + r i n = (i) 4q+r = (i) 4q (i) r = (i 4 ) q (i) r if r = 0, i n = ; if r =, i n = i; if r =, i n = ; if r =, i n = i ab = a b is valid onl if at least one of a, b is non-negative. For R, > 0 EXERCISE. Simplif the following:.i i = 4(486) + ; 950 = 4(487) + i i 950 = (i 4 ) 486 (i) + (i 4 ) 487 (i) = i = i.i 948 i = 4(487) + 0 ; = 4(467) + i 948 i 869 = i 948 = i 869 (i4 ) 487 (i) 0 = = (i 4 ) 467 i i ( i ) i i = ( i i ) = + i ( i = ). n= i n n= i n = i + i + i + i 4 + i 5 + i 6 + i 7 + +i 8 + i 9 + i 0 + i + i = i + i + i + i 4 + i 4 i + (i ) + (i ) i + (i ) 4 + (i ) + (i ) 5 + (i ) 5 i + (i ) 6 = i i + + i i + + i i + = 0 4.i 59 + i = 4(4) + i 59 + = i 59 (i4 ) 4 i + (i 4 ) 4 i = i + = i + i i i i = i + i = i i = 0 5.i. i. i. i 4.. i 000 = i ( ) ai.org ai.org ai.org ai.org ai.org ai.org ai.org ai.org ai.org.p.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.p.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.p.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.p = i TrbTnpsc = i = i 000. i 00 = (i 4 ) 50 (i 4 ) 500 i ( 00 4 = 4(500) + ) =.. i = i 6. 0 n= i n+50 0 n= i n+50 = i 5 + i 5 + i i 60 = i 50 (i + i + i + i 4 + i 5 + i 6 + i 7 + i 8 + i 9 + i 0 ) = (i 4 ) i (i i + + i i + + i ) = (i ) = i EXAMPLE. Simplif the following: (i) i 7 (ii)i 79 (iii)i 94 + i 08 (iv) 0 n= i n (v)i. i. i. i 4.. i 40 COMPLEX NUMBERS Rectangular Form Of a Comple Number: A comple number is of the form z = + i,, R. is called the real part and is called the imaginar part of the comple number. Two comple numbers z = a + ib, z = c + id are said to be equal if and onl if Re(z ) = Re(z ) and Im(z ) = Im(z ). ie. a = c and b = d Scalar multiplication of comple numbers: If z = + i and k R then kz = k( + i) = k + ki Addition of comple numbers: If z = a + ib, z = c + id, then z + z = (a + c) + i(b + d) Subtraction of comple numbers: If z = a + ib, z = c + id, then z z = (a c) + i(b d) Multiplication of comple numbers: If z = a + ib, z = c + id, then z z = (ac bd) + i(ad + bc) Multiplication of a comple z b i successivel gives a 90 0 counter clockwise rotation successivel about the origin. EXERCISE.. Evaluate the following if z = 5 i and w = + i (i)z + w (ii)z iw (iii)z + w (iv)zw (v)z + zw + w (vi)(z + w).pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.p.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.p.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.p.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org V.GNANAMURUGAN, M.Sc.,M.Ed., P.G.T, GHSS, S.S.KOTTAI, SIVAGANGAI DT , Page

2 (i)z + w = 5 i + i = 4 + i Pai (ii)z iw = 5 i i( + i) = 5 i + i i = 5 i + i + = 8 i i = (iii)z + w = (5 i ) + ( + i) = 0 4i + 9i = 7 + 5i (iv) z z = (ac bd) + i(ad + bc); a = 5, b =, c =, d = zw = (5 i)( + i) = [(5)( ) ( )()] + i[(5)() + ( )( )] = [ 5 + 6] + i[5 + ] = + 7i (v)z = (5 i) = 5 0i + 4i = 5 0i 4 = 0i zw = (5 i )( + i) = [ + 7i] from (iv) zw = + 4i w = ( + i) = 6i + 9i = 6i 9 = 8 6i z + zw + w = 0i + + 4i 8 6i = 5 + 8i (vi)(z + w) = (4 + i) from(i) = 6 + 8i + i = 6 + 8i = 5 + 8i. Given the comple number z = + i, represent the comple numbers in Argand diagram. (i)z, iz and z + iz (ii)z, iz and z iz (i)z = + i = (,) iz = i( + i) = i + i = + i = (,) z + iz = + i + i( + i) = + i + i = + 5i = (,5) ai.org ai.org ai.org ai.org ai.org ai.org ai.org ai.org.p.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.p.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.p (i)z = + i = (,) iz = i( + i) = i i = i = (, ) z iz = + i i( + i) = + i + i = 5 + i = (5,) ai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.p.pai.org.pai.org.pai.org.pai.org.p.pai.org.pai.org.pai.org.pai.org. Find the values of the real numbers and, if the comple numbers ( i) ( i) + i + 5 and + ( + i) + + i are equal. z = ( i) ( i) + i + 5 = i + i + i + 5 = ( + 5) + i( + + ) z = + ( + i) + + i = + i + + i = ( + ) + i( + ) Given : z = z ( + 5) + i( + + ) = ( + ) + i( + ) Equating the real and imaginar parts on both sides,.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org + 5 = = 0 = () () + () = =.Pai.Org.Pai.Org.Pai.Org.Pai.Org.P.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.P + + = = 0 = 0 ().Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org Sub. = in () = = EXAMPLE. Find the values of the real numbers and, if the comple numbers ( + i) + ( i) + i and + ( + i) + + i are equal. [Ans: = and = ] PROPERTIES OF COMPLEX NUMBERS Properties of comple numbers under addition: I. Closure propert: z, z C, (z + z ) C II. Commutative propert: z, z C, z + z = z + z III. Associative propert: z, z, z C, (z + z ) + z = z + (z + z ) IV. Additive identit: z C, z + 0 = 0 + z = z 0 = 0 + 0i is an additive.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org TrbTnpsc.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org V.GNANAMURUGAN, M.Sc.,M.Ed., P.G.T, GHSS, S.S.KOTTAI, SIVAGANGAI DT , Page

3 identit. Pai V. Additive inverse: z C, z C z + ( z) = ( z) + z = 0 C, so z is an additive identit. Properties of comple numbers under multiplication: I. Closure propert: z, z C, (z. z ) C II. Commutative propert: z, z C, z. z = z. z III. Associative propert: z, z, z C, (z. z ). z = z. (z. z ) IV. Multiplicative identit: z C, z. =. z = z = + 0i is a multiplicative identit. ai.org ai.org.p.pai.org.pai.org.pai.org.pai.org V. Multiplicative inverse: z C z z C z. z = z. z = + i0 C, so z is ai.org multiplicative identit..pai.org.pai.org If z = + i then its multiplicative inverse is z = ( + ) + i ( Distributive propert: z, z, z C, (z + z ). z = z z + z z o z, z, z C, z. (z + z ) = z z + z z Comple numbers obe the laws of indices: ai.org.pai.org.pai.org (i)z m z n = z m+n (ii) zm.p.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org z n = zm n.p.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org (iii)(z m ) n = z m z n (iv)(z z ) m = z m m z EXERCISE..If z = i, z = 4i and z = 5, show that (i)(z + z ) + z = z + (z + z ) (ii)(z z )z = z (z z ) (i)(z + z ) + z = ( i 4i) + 5 = 7i + 5 = 6 7i () z + (z + z ) = i + ( 4i + 5) = i 4i + 5 = 6 7i () From () & (), (z + z ) + z = z + (z + z ) (ii)(z z )z = [( i)( 4i)]5 = ( 4i + i )5 = ( 4i )5 = 60 0i () z (z z ) = ( i)[( 4i)(5)] = ( i)[ 0i] = 0i + 60i = 60 0i () From () & (), (z z )z = z (z z ).If z =, z = 7i and z = 5 + 4i, show that (i)z. (z + z ) = z z + z z (ii)(z + z ). z = z z + z z (i)z. (z + z ) = ( 7i i) = (5 i) = 5 9i () z z + z z = ()( 7i) + ()(5 + 4i) = i i = 5 9i () ai.org ai.org ai.org ai.org ai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org + ) (or).pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.p From () & (), TrbTnpsc z. (z + z ) = z z + z z (ii)(z + z ). z = ( 7i)(5 + 4i) = 5 + i 5i 8i = 5 + i 5i + 8 = 4 i () z z + z z = ()(5 + 4i) + ( 7i)(5 + 4i) = 5 + i 5i 8i = 5 + i 5i + 8 = 4 i () From () & (), (z + z ). z = z z + z z.if z = + 5i, z = 4i and z = + i, find the additive and multiplicative inverse of z, z and z Additive inverse of z is z. So, Additive inverse of z = + 5i is z = 5i, Additive inverse of z = 4i is- z = + 4i Additive inverse of z = + i is z = i.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.p.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org Multiplicative inverse of z = + i is z = ( +) + i ( + ) Multiplicative inverse of z = + 5i is z = ( 5 +5) + i ( +5 ) = 5 i 9 9 Multiplicative inverse of z = 4i isz 4 = ( ) + i (.Pai.Org.Pai.Org.Pai.Org.Pai.Org V.GNANAMURUGAN, M.Sc.,M.Ed., P.G.T, GHSS, S.S.KOTTAI, SIVAGANGAI DT , Page ( ) +( 4) = i ( ) +( 4) ) Multiplicative inverse of z = + i is z = ( +) + i ( + ) = i CONJUGATE OF A COMPLEX NUMBER The conjugate of the comple number z = + i is z = i. zz = + The conjugate is useful in division of comple numbers. PROPERTIES OF COMPLEX CONJUGATES: z + z = z + z Im(z) = z z i z z = z z (z z. z = z. z n ) = (z ) n z is real if and onl if z = z ( z z ) =, z z z 0 z is purel imaginar if and onl if z = z Re(z) = z+z z = z.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.p.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org EXERCISE.4.Write the following in the rectangular form:.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.p.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org

4 (i)(5 + 9i) + ( 4i) (ii) 0 5i (iii)i + Pai 6+i i (i)(5 + 9i) + ( 4i) = 5 + 9i + 4i = 5 9i + + 4i = 7 5i (ii) 0 5i = 0 5i 6 i = 60 0i 0i+0i 6+i 6+i 6 i 6 + = 60 50i 0 = 50 50i (iii)i + = i + +i i i +i ai.org ai.org = i = i + +i + = i i.if z = + i, find the following in rectangular form: = + 5i+i = 4 i (i)re ( ) (ii)re(iz ) (iii)im(z + 4z 4i) z (i) = z ai.org i = i = i +i i i ) = Re ( z ) = Re (.P.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org (ii)iz = i( ) + i = i( i) = i i = + i Re(iz ) = Re( + i) = (iii)z + 4z 4i = ( + i) i 4i = + i + 4( i) 4i = + i + 4 4i 4i = 7 i 4i = 7 + i( 4) Im(z + 4z 4i) = Im(7 + i( 4)) = 4. If z = i and z = 4 + i, find the inverse of z z and z ai.org ai.org.pai.org.pai.org.pai.org.pai.org z z = ( i )( 4 + i) = 8 + 6i + 4i i = 8 + 0i + = 5 + 0i Inverse of z z = ( +) + i ( + ) = ( 5 ( 5) +0 ) + i ( 0 ( 5) +0 ) = 5 0 i = i ai.org.pai.org.pai.org z = i = i 4 i = 8 6i+4i+i = 8 i z 4+i 4+i 4 i ( 4) + 5 Inverse of z = ( z +) + i ( + ) ai.org ai.org.pai.org.pai.org = ( 5 ( 5) +( =.Pai.Org.Pai.Org.P.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org ) ) + i ( i = i 5 5 ( 5) +( 5.Pai.Org.Pai.Org.P.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org z.pai.org.pai.org = 5 5 i.pai.org.pai.org ) ).Pai.Org.Pai.Org 4. The comple numbers u, v and w are related b = +. If v = 4i and u v w ai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.p w = 4 + i, find TrbTnpsc u in the rectangular form. = + = + u v w u 4i 4+i = ( +4i ) + ( 4i +4i.Pai.Org.Pai.Org.Pai.Org.Pai.Org.P = +4i + 4 i = 5.Pai.Org.Pai.Org u = 7+i ( + 4i + 4 i) 4 i ) 4+i 4 i 5 = 5 7 i = 5(7 i) = 5(7 i) = 7 i 7+i 7+i 7 i Pai.Org.Pai.Org u = 5 = 7 i 5.Prove the following properties: (i)z is real if and onl if z = z (ii)re(z) = z+z (iii)im(z) = z z.pai.org.pai.org.pai.org.pai.org (i)z is real z = + 0i z = + 0i = 0i = z is real if and onl if z = z (ii) Let z = + i ; Re(z) = () z = i z + z = + i + i = z+z = ().Pai.Org.Pai.Org.Pai.Org.Pai.Org From () & (), Re(z) = z+z (iii) Let z = + i ; Im(z) = () z = i z z = + i + i = i z z = ().Pai.Org.Pai.Org.Pai.Org.Pai.Org From () & (), Im(z) = z z.pai.org.pai.org.pai.org.pai.org V.GNANAMURUGAN, M.Sc.,M.Ed., P.G.T, GHSS, S.S.KOTTAI, SIVAGANGAI DT , Page 4 i 6.Find the least value of the positive integer n for which ( + i) n (i)real (ii)purel imaginar ( + i) = + 9i i = 8i.Pai.Org.Pai.Org.Pai.Org.Pai.Org ( + i) 6 = (( + i) ) = (8i) = 64 (i)when n = 6, ( + i) n is real.pai.org.pai.org.pai.org.pai.org.p.pai.org.pai.org.pai.org.pai.org (ii)when n =, ( + i) n is purel imaginar 7.Show that (i)( + i ) 0 ( i ) 0 is purel imaginar..pai.org.pai.org.pai.org.pai.org i.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.p.pai.org.pai.org.pai.org.pai.org i.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org

5 (ii) ( 9 7i 9+i ) + ( 0 5i Pai 7 6i ) is real. (i) z = ( + i ) 0 ( i ) 0 z = ( + i ) 0 ( i ) 0 = ( + i ) 0 ( i ) 0 ai.org )0 = ( + i ( )0 0 0 i = ( i ) ( + i ) = {( + i ) 0 ( i ) 0 } = z ai.org ( + i ) 0 ( i ) 0 is purel imaginar. (ii) 9 7i = 9 7i 9+i 9+i 0 5i = 0 5i 7+6i = 40+0i 5i 0i = 70+85i 7 6i 7 6i 7+6i ai.org 9 i = 7 9i 6i+7i = 64 8i = 8( i) = i 9 i = + i z = ( 9 7i 9+i ) + ( 0 5i 7 6i ) = ( i) ( + i) z = ( i) ( + i) = ( i) ( + i) = ( + i) ( i) = z ai.org ( 9 7i 9+i ) + ( 0 5i 7 6i ) is real. EXAMPLE. 8 Show that (i)( + i ) 0 + ( i ) 0 is real and ai.org (ii) ( 9+9ii 5 i )5 ( 8+i +i )5 is purel imaginar. EXAMPLE. Write +4i in the + i form, hence find its real and imaginar 5 i parts. +4i = ai.org = +4i 5+i = 5+6i+0i 48 = +56i 5 i 5 i 5+i Real part = 69 ai.org.p.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org EXAMPLE. 4 Simplif ( +i 56 and imaginar part= 69 i ) ( i +i ).Pai.Org.Pai.Org +i = +i +i = +i+i+i = +i = i i i +i + i = +i (+i i ) = i ( +i ai.org.pai.org.pai.org = i + 56 i i ) ( i +i ) = (i) ( i) = i i = i z+ EXAMPLE. 5 If, find the comple number z. ai.org = +4i z 5i.Pai.Org.Pai.Org.P.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.P.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org z+ = +4i z 5i.P.P (z + ) = ( + 4i)(z 5i) z + 6 = z 5i + 4zi 0i z z 4zi = 5i Pai.Org.Pai.Org z 4zi = 4 5i z( 4i) = 4 5i z = 4 5i 4i.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org = 4 5i +4i = 4+56i 5i 0i = 4+5i+0 = 4+5i = 7(+i) 4i +4i EXAMPLE. 6 If z = i and z = 6 + 4i, find z z. z = i = i 6 4i = 8 i i 8 = 0 4i = (5 i) = 5 i z 6+4i 6+4i 6 4i Pai.Org.Pai.Org.Pai.Org.Pai.Org EXAMPLE. 7 Find z, if z = ( + i)( i). z = ( + i)( i) = i + i i = + i + = 5 + i z = ( +) + i ( + ) = ( 5 5 +) + i ( 5 + ) = 5 i 6 6 MODULUS OF A COMPLEX NUMBER.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org TrbTnpsc = + i V.GNANAMURUGAN, M.Sc.,M.Ed., P.G.T, GHSS, S.S.KOTTAI, SIVAGANGAI DT , Page 5.Pai.Org.Pai.Org.P.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.P.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org If z = + i, then the modulus of z is z = + zz = z PROPERTIES OF MODULUS OF A COMPLEX NUMBER: Z = Z z + z z + z ( Triangle inequalit ) z z z z z z = z z z = z, z z z 0 z n = z n, n is an integer. Re(z) z Im(z) z z z z + z z + z The distance between the two points z and z in the comple plane is z z (or) ( ) + ( ) Formula for finding the square root of the comple number z = a + ib is.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org

6 a + ib = ± ( z +a + i b ai.org If b is negative b If b is positive b b b b z a ), =, and have different signs. =, and have same signs. Eercise.5.Find the modulus of the following comple numbers: (ii) i (iii)( i) 0 (iv)i( 4i)(4 i) ai.org (i) i +4i ai.org (i) i = +4i + i +i i i = 0 + = 4 = +4i z = + (ii) i + i = +i i ( i)( i)+(+i)( i) = i i+i + i+i i (+i)( i) + = i i + i+ = 4 4i = i = + = 8 = (iii) ( i) 0 = i 0 = ( + ) 0 = 0 = 5 = ai.org (iv) i( 4i)(4 i) = i 4i 4 i = = =.5.5 = 50 EXAMPLE. 9 If z = + 4i, z = 5 i and z = 6 + 8i, find z, z, z, z + z, z z and z + z. EXAMPLE. 0 Find the following : (i) +i (ii) ( ( + i) + i)(4i ) ai.org ai.org +i (iii) i(+i) (+i).for ant two comple numbers z and z, such that z = z = and z z then show that z +z is a real number. +z z Given z = z = z = z = z z = z = z ai.org ai.org lll l z = z ai.org.p.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.p.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org Pai.P.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.P TrbTnpsc Let w = z +z w = ( z +z z ) = +z z +z z.p +z = +z = z + z +z z +z z + zz.p V.GNANAMURUGAN, M.Sc.,M.Ed., P.G.T, GHSS, S.S.KOTTAI, SIVAGANGAI DT , Page 6 = z+z zz = z +z +zz zz +z z = w Since w = w, w is purel real..which one of the points 0 8i, + 6i is closest to + i. Distance between the two points z and z = ( ) + ( ) z = + i, z = 0 8i Distance between the two points z and z = ( 0) + ( + 8) = = 6 = 8 = 9 z = + i, z = + 6i Distance between the two points z and z = ( ) + ( 6) = = 5 = 5 5 = < 9, + 6i is the closest point to + i EXAMPLE. Which one of the points i, + i and is farthest from the origin? 4.If z =, show that 7 z + 6 8i. Let z = z and z = 6 8i W.K.T, z z z + z z + z.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org z + 6 8i z + 6 8i z + 6 8i z + 6 8i 7 z + 6 8i EXAMPLE. If z =, show that z + + 4i If z =, show that z 4. Let z = z and z = W.K.T, z z z + z z + z z z z + z + z 4 z 4.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.P.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org

7 6.If z =, show that the greatest and least value of Pai z are + and z respectivel..p z z = = z z z z ai.org.pai.org.pai.org z z z z a r r a r z z z z z () z z From () z z z z z + z 0 z + z z + z 0 [ z ( + )][ z ( )] 0 z ( + ) ; z ( ) ai.org ai.org ai.org.pai.org.pai.org.pai.org.pai.org z ( ) ().Pai.Org.Pai.Org From () z z z z 0 [ z ( + )][ z ( )] 0 ( ) z ( + ) ai.org.pai.org.pai.org z ( + ) () From () & (), z + Hence the greatest and least value of z are + and respectivel. 7.If z, z and z are three comple numbers such that z =, z =, z = and z + z + z =, show that 9z z + 4z z + z z = 6. z = z = z z = z = zz = z z z = z = 4 z z = 4 z = 4 z z = z = 9 z z = 9 z = 9 z z + z + z = z z z z + z + z = z +4z z +9z z ai.org ai.org ai.org ai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org z z z.pai.org.pai.org.p.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.p.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.p TrbTnpsc z + z + z = z +4z z +9z z z + z + z = z z z z z +4z z +9z z z z z z + z + z = 9z z +4z z +z z.pai.org.pai.org z z z z + z + z = 9z z +4z z +z z z z z = 9z z +4z z +z z.pai.org.pai.org.. z = z z z = z z 9z z + 4z z + z z = 6 EXAMPLE. If z, z and z are three comple numbers such that z = z = z = z + z + z =, find the value of z + z + z..pai.org.pai.org EXAMPLE. 5 If z, z and z are three comple numbers such that z = z = z = r > 0 and z + z + z 0. Prove that z z +z z +z z = r z +z +z z = r z = r z z = r z = r zz = z.pai.org.pai.org z = r z = r z z = r z = r V.GNANAMURUGAN, M.Sc.,M.Ed., P.G.T, GHSS, S.S.KOTTAI, SIVAGANGAI DT , Page 7 z z = r z = r z z = r z = r.pai.org.pai.org z + z + z = r.p.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org z + z + z = r z.pai.org.pai.org z z + r + r = z z z r ( z +z z +z z ) z z z z +z z +z z = r z z +z z +z z z z z z.pai.org.pai.org.pai.org.pai.org z + z + z = r z z +z z +z z r.r.r.pai.org.pai.org.pai.org.pai.org.p.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org = r z z +z z +z z z z z z + z + z = z z +z z +z z r z z +z z +z z = r z +z +z.pai.org.pai.org.p.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org 8.If the area of the triangle formed b the vertices z, iz and z + iz is 50 square units, find the value of z. Let z = a + ib = (a, b) z = a + b Vertices : z = (a, b) iz = i(a + ib) = ia + i b = b + ia = ( b, a) z + iz = a + ib b + ia = (a b) + i(a + b) = (a b, a + b).pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org

8 Area of the triangle = 50 sq. units Pai b a b a a b a a + b b = 50 (a ab b + ab b ) ( b + a ab + a + ab) = 50 a b + b a = 00 a b = 00 (a + b ) = ±00 a + b = 00 a + b = 0 z = 0 ( a + b is not negative) NOTE: This can also done b using distance formula, find the length of base and ai.org ai.org ai.org.p.pai.org.pai.org.pai.org.pai.org height and use Area of the = bh.pai.org.pai.org EXAMPLE. 4 Show that the points, of an equilateral triangle. ai.org Let z =, z = + i and z = i The length of the sides of the triangle are ai.org.pai.org.pai.org + i and z z = ( + i ) = i = ( ) + (.Pai.Org.Pai.Org z z = ( + i ) ( ai.org.p.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org i ) = ).Pai.Org.Pai.Org + i +.P.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org i are the vertices.pai.org.pai.org = = 4 =.Pai.Org.Pai.Org + i = ( ) = z z = ( i ) = i = ( ) + ( = 9 + = 4 4 Since all the three sides are equal, the given points form an equilateral triangle. EXAMPLE. 8 Given the comple number = + i, represent the comple numbers z, iz and z + iz in one Argand diagram. Show that these comple numbers form the vertices of an isosceles right triangle. Given : z = + i = A(,) iz = i( + i) = + i = B(,) z + iz = + i + i = + 5i = C(,5) The length of the sides of the triangle are ai.org ai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org AB = ( ) + ( ) = ( + ) + ( ) = 5 + = 6 ai.org.pai.org.pai.org.pai.org.pai.org ).Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.P BC = ( TrbTnpsc ) + ( ) = ( ) + ( 5) = = CA = ( ) + ( ) = ( ) + (5 ) = = AB = 6, BC =, CA = Since (i)bc = CA = and (ii)bc + CA = + = 6 = AB, So the given vertices form an isosceles right triangle. 9.Show that the equation z + z = 0 has five solutions. z + z = 0 z = z z = z z = z z z = 0 z = 0 z = 0 is one of the solution z ( z ) = 0 { z = 0 zz = 0 z = z.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org Put z = in z z + z = 0 z + ( ) = 0 z z4 + 4 = 0, which provides four solution. Hence z + z = 0 has five solutions. EXAMPLE. 6 Show that the equation z = z has four solutions. 0.Find the square root of (i)4 + i (ii) 6 + 8i (iii) 5 i..pai.org.pai.org.pai.org.pai.org.pai.org.pai.org z + a a + ib = ± ( + i b a z ) b (i)4 + i a = 4, b =, z = 4 + = 5 = i = ± ( i 5 4) = ± ( + i ).Pai.Org.Pai.Org (ii) 6 + 8i a = 6, b = 8, z = = 00 = 0.Pai.Org.Pai.Org 6 + 8i (iii) 5 i 4.Pai.Org.Pai.Org.P.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org = ± ( i ) = ±( + i 8) = ±( + i ) V.GNANAMURUGAN, M.Sc.,M.Ed., P.G.T, GHSS, S.S.KOTTAI, SIVAGANGAI DT , Page 8.Pai.Org.Pai.Org.P.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.P.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org

9 a = 5, b =, z = 5 + = 69 = 5 i ai.org = ± ( 5 + i +5 ) = ±( 4 i 9) = ±( i) EXAMPLE. 7 Find the square root of 6 + 8i GEOMETRY AND LOCUS OF COMPLEX NUMBERS A circle is defined as the locus of a point which moves in a plane such that its distance from a fied point in that plane is alwas a constant. The fied point is the centre and the constant distant is the radius of the circle. Equation of comple form of a circle is z z 0 = r. z z 0 < r represents the points interior of the circle. z z 0 > r represents the points eterior of the circle. + = r represents a circle with centre at the origin and the radius r units. r = + If z = a+ib ac+bd then Re(Z) = ai.org ai.org ai.org c+id c +d If z = a+ib bc ad then Im(Z) = c+id c +d EXERCISE.6.If z = + i is a comple number such that z 4i =. Show that the locus of z is ai.org real ais. Given: z = + i ai.org z 4i z+4i.p.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org = z 4i = z + 4i.Pai.Org.Pai.Org.P.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org z+4i.pai.org.pai.org + i 4i = + i + 4i + i( 4) = + i( + 4) + ( 4) = + ( 4) Squaring on both sides, + ( 4) = + ( 4) = = 0 = 0 The locus of z is real ais. ai.org ai.org ai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org Pai.P.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.P. If z = + i is TrbTnpsc a comple number such that Im ( z+ ) = 0, show that the locus.p of z is + + = 0. Given: z = + i z+ = (+i)+ = (+)+i iz+ i(+i)+ ( )+i.pai.org.pai.org.pai.org.pai.org Here a = +, b =, c =, d = Im ( z+ ) = 0 ( ) (+) iz+ + + = 0.Pai.Org.Pai.Org V.GNANAMURUGAN, M.Sc.,M.Ed., P.G.T, GHSS, S.S.KOTTAI, SIVAGANGAI DT , Page 9 iz+ ( ) + = 0 = 0.Pai.Org.Pai.Org Im(Z) = bc ad c +d.obtain the Cartesian form of the locus of z = + i in each of the following cases: (i)[re(iz)] = (ii)im[( i)z + ] = 0 (iii) z + i = z (iv)z = z.pai.org.pai.org.pai.org.pai.org Given: z = + i (i)iz = i( + i) = i [Re(iz)] = [Re( i )] =.Pai.Org.Pai.Org.Pai.Org.Pai.Org = (ii)( i)z + = ( i)( + i) + = + i i + + = ( + + ) + i( ) Im[( i)z + ] = 0.Pai.Org.Pai.Org.Pai.Org.Pai.Org Im[( + + ) + i( )] = 0 = 0 (or) = 0.P.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org (iii) z + i = z + i + i = + i + i( + ) = ( ) + i () + ( + ) = ( ) + () Squaring on both sides,.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org () + ( + ) = ( ) + () = = 0 + = 0.Pai.Org.Pai.Org.Pai.Org.Pai.Org.P.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org (iv)z = z z = z zz = ( + i)( i) = + = 4.Show that the following equations represent a circle, and find its centre and.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org

10 radius: (i) z i = (ii) z + 4i = (iii) zpai 6 + i = 8 Equation of the circle is z z 0 = r (i) z i = z ( + i) = Centre = + i = (,) ; Radius = units (ii) z + 4i = z + i = z ( + i) = Centre = + i = (,) ; Radius = unit (iii) z 6 + i = 8 z + 4i = 8 z ( 4i) = 8 Centre = 4i = (, 4) ; Radius = 8 units ai.org ai.org.p.pai.org.pai.org.pai.org.pai.org.p.pai.org.pai.org.pai.org.pai.org.p.pai.org.pai.org.pai.org.pai.org EXAMPLE. 9 Show that z 5 + i = 4 represent a circle, and find its centre and radius. 5. Obtain the Cartesian form of the locus of z = + i in each of the following cases: (i) z 4 = 6 (ii) z 4 z = 6 Given: z = + i (i) z 4 = 6 + i 4 = 6 ( 4) + i = 6 ( 4) + = 6 Squaring on both sides, ( 4) + = = = 0 is the locus of z. (ii) z 4 z = 6 + i 4 + i = 6 ( 4) + i ( ) + i = 6 ai.org ai.org ai.org ai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org ( ( 4) + ) ( ) + = 6 ( 4) + [( ) + ] = = 6 6 = = 0 is the locus of z. EXAMPLE. Obtain the Cartesian form of the locus of z in each of the following cases: (i) z = z i (ii) z i = EXAMPLE. 0 Show that z + i < represents interior points of a circle. Find its centre and radius. ai.org ai.org ai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.p Consider z + TrbTnpsc i = z ( + i) =. Centre = + i = (,) ; Radius = units z + i < represents all points inside the circle. POLAR AND EULER FORM OF A COMPLEX NUMBER Polar form of z = + i is z = r(cos θ + i sin θ) (or) z = rcisθ Modulus r = +.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.P.Pai.Org.Pai.Org α = tan ( ) θ = α ; if(+, +) θ lies in the Ist quadrant θ = π α ; if(, +) θ lies in the nd quadrant arg z = θ ; θ = π + α ; if(, ) θ lies in the rd quadrant { θ = α ; if(+, ) θ lies in the 4th quadrant In general arg z = Arg z + nπ, n z. Properties of arguments: arg(z z ) = arg(z ) + arg(z ).Pai.Org.Pai.Org.Pai.Org.Pai.Org arg ( z ) = arg(z z ) arg(z ) arg(z n ) = n argz Some of the principal argument and argument: z i i.pai.org.pai.org.pai.org.pai.org Arg z 0 π.pai.org.pai.org.p.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org argz nπ nπ + π nπ + π nπ π Euler s form of the comple number: z = re iθ ; e iθ = cos θ + i sin θ Properties of polar form: z = r(cos θ + i sin θ) z = (cos θ i sin θ) r If z = r (cos θ + i sin θ ), z = r (cos θ + i sin θ ) then z z = r r [cos(θ + θ ) + i sin(θ + θ )] If z = r (cos θ + i sin θ ), z = r (cos θ + i sin θ ) then z = r [cos(θ z r θ ) + i sin(θ θ )].Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org V.GNANAMURUGAN, M.Sc.,M.Ed., P.G.T, GHSS, S.S.KOTTAI, SIVAGANGAI DT , Page 0 π.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.p.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org π.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org

11 Some of the useful polar forms : = cis(0) ; = cis(π) ; i = cis ( π ) ; i = cis ( π ) EXAMPLE. Find the modulus and principal argument of the following comple numbers: (i) + i (ii) + i (iii) i (iv) i (i) + i ai.org Modulus r = + = + = 4 = ai.org α = tan = tan = π 6 Since the comple number + i lies in the first quadrant, principal argument θ = α θ = π 6 ai.org (ii) + i Modulus r = + = ( ) + = 4 = ai.org α = tan ( ) = tan = π 6 Since the comple number + i lies in the second quadrant, principal argument θ = π α θ = π π 6 = 5π 6 ai.org (iii) i Modulus r = + = ( ) + ( ) = 4 = ai.org α = tan ( ) = tan = π 6 Since the comple number i lies in the third quadrant, principal argument θ = π + α θ = π + π 6 = 5π 6 ai.org (iv) i.p.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.p.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org Modulus r = + = + ( ) = 4 = ai.org.p.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org α = tan ( ) = tan = π 6 Since the comple number i lies in the fourth quadrant, principal argument θ = α θ = π 6 ai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org Pai.Pai.Org.Pai.Org.Pai.Org.Pai.Org.P EXAMPLE. 4 TrbTnpsc Find the principal argument Arg z, when z =.P argz = arg ( ) = arg( ) arg( + i ) () +i arg( ) = tan = tan 0 = 0.Pai.Org.Pai.Org.Pai.Org.Pai.Org V.GNANAMURUGAN, M.Sc.,M.Ed., P.G.T, GHSS, S.S.KOTTAI, SIVAGANGAI DT , Page +i Since the comple number lies in the second quadrant, principal argument θ = π α θ = π 0 = π arg( + i ) = tan = tan = π Since the comple number + i lies in the first quadrant, principal argument θ = α θ = π () arg ( ) = π π = π +i EXERCISE.7.Write in polar form of the following comple numbers:.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org (i) + i (ii) i (iii) i (iv).pai.org.pai.org.pai.org.pai.org (i) + i = rcis(θ) () i cos π +i sinπ r = + = () + ( ) = 4 + = 6 = 4.Pai.Org.Pai.Org.Pai.Org.Pai.Org α = tan = tan = π.p.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org Principal argument lies in the first quadrant θ = α = π () + i = 4cis ( π ) = 4cis (kπ + π ), k z (ii) i = rcis(θ) ().Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org r = + = () + ( ) = 9 + = 4 =.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org α = tan = tan = tan = π 6 Principal argument lies in the 4th quadrant θ = α = π 6 () i = cis ( π ) = cis (kπ π ), k z 6 6 (iii) i = rcis(θ) ().Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org r = + = ( ) + ( ) = = 4 =.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.P.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org

12 α = tan = tan = π 4 Principal argument lies in the rd quadrant θ = π + α = π + π = π 4 4 () i = cis ( π ) = cis (kπ π ), k z 4 4 ai.org (iv) i cos π +i sinπ i = + i = rcis(θ) () r = + = ( ) + () = ai.org α = tan = tan = π 4 Principal argument lies in the nd quadrant θ = π α θ = π π 4 = π 4 ai.org () i = cis ( π 4 ) i cos π +i sinπ ai.org = cis(π 4 ) = cis ( π π 5π ) = cis cis( π ) (5π) = cis (kπ + ), k z 4 EXAMPLE. Represent the comple number (i) i (ii) + i in polar form..find the rectangular form of the comple numbers: ai.org (i) (cos π 6 + i sin π 6 ) (cos π + i sin π ) (ii) cos π 6 i sinπ 6 (i) (cos π 6 + i sin π 6 ) (cos π + i sin π ) ai.org (ii) = cis ( π 6 + π ) = cis ( π ) = cis (π) 4 = cos π + i sin π 4 4 ai.org = + i cos π 6 i sinπ 6 (cos π +i sinπ ) ai.org.p.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org (cos π +i sinπ ) = cis ( π 6 π ) = cis ( π 6 ) = cis ( π ) = [cos ( π ) + i sin ( π )] = (cos π i sin π ) ai.org.pai.org.pai.org.pai.org.pai.org.p.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org Pai.P.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org =.P = i (0 i) TrbTnpsc EXAMPLE. 5 Find the product (cos π + i sin π ). 6 (cos 5π 6 rectangular form..pai.org.pai.org EXAMPLE. 6 Find the quotient (cos 9π 4 +i sin9π 4 ) 4(cos( π )+i sin( π + i sin 5π 6 ) in )) in rectangular form..if ( + i )( + i ).. ( n + i n ) = a + ib, show that (i) ( + )( +.Pai.Org.Pai.Org ).. ( n + n ) = a + b (ii) tan ( b r V.GNANAMURUGAN, M.Sc.,M.Ed., P.G.T, GHSS, S.S.KOTTAI, SIVAGANGAI DT , Page n r= (i)( + i )( + i ).. ( n + i n ) = a + ib Taking modulus on both sides, ( + i )( + i ).. ( n + i n ) = a + ib + i + i.. n + i n = a + ib.pai.org.pai.org.pai.org.pai.org ar ) ( + ) ( + ).. ( n + n ) = a + b = tan ( b ) + kπ, kεz6 a Squaring on both sides, ( + )( + ).. ( n + n ) = a + b (i)( + i )( + i ).. ( n + i n ) = a + ib Taking argument on both sides, arg[( + i )( + i ).. ( n + i n )] = arg(a + ib).pai.org.pai.org tan ( ) + tan ( ) +. +tan ( n n ) = tan ( b a ) 4.If +z z n r=.pai.org.pai.org.p.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org tan ( b r ar ) = tan ( b ) + kπ, kεz6 a.pai.org.pai.org = cos θ + i sin θ, show that z = i tan θ. +z z = cos θ + i sin θ Add on both sides, +z + = ( + cos θ) + i sin θ.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org z +z+ z z z.pai.org.pai.org b = cos θ + i sin θ cos θ = cos θ [cos θ + i sin θ] z.p = cos θ [cos θ + i sin θ] cos θ[cos θ+i sin θ] = z.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.p.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org

13 ai.org z = z = [ z =.P z = cos θ[cos θ+i sin θ] cos θ[cos θ+i sin θ] cos θ i sin θ cos θ(cos θ+sin θ) cos θ sin θ i ] cos θ cos θ.pai.org.pai.org cos θ i sin θ cos θ i sin θ z = + i tan θ z = i tan θ 5.If cosα + cosβ + cosγ = sinα + sinβ + sinγ = 0, then show that (i) cosα + cosβ + cosγ = cos(α + β + γ) (ii)sinα + sinβ + sinγ = sin (α + β + γ) Let a = cisα, b = cisβ, c = cisγ W.K.T, If a + b + c = 0 then a + b + c = abc a + b + c = (cosα + cosβ + cosγ) + i(sinα + sinβ + sinγ) = 0 a + b + c = abc ai.org ai.org ai.org.pai.org.pai.org.pai.org.pai.org (cisα) + (cisβ) + (cisγ) = cisα. cisβ. cisγ cis(α) + cis(β) + cis(γ) = cis(α + β + γ) (cosα + cosβ + cosγ) + i(sinα + sinβ + sinγ) = (cos(α + β +.Pai.Org.Pai.Org γ)) + i(sin (α + β + γ)) Equating the real and imaginar parts on both sides, cosα + cosβ + cosγ = cos(α + β + γ) sinα + sinβ + sinγ = sin (α + β + γ) 6.If z = + i and arg ( z i ) = π, then show that z = 0. Given: Z = + i arg ( z i ) = π z+ 4 ai.org ai.org ai.org.pai.org.pai.org.pai.org.pai.org arg(z i) arg(z + ) = π 4.Pai.Org.Pai.Org.P arg( + i i) arg( + i + ) = π 4 ai.org arg( + i( )) arg( + + i) = π 4 tan tan + = π 4 ai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org Pai.P.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org TrbTnpsc tan + π + ( = + ) 4 (( ))( + ) ( + ) = tan π + 4 ( + ) + ( + ) + + ( + ) = = = = 0 EXAMPLE. 7 If z = + i and arg ( z z+ + =. De MOIVRE S THEOREM AND ITS APPLICATIONS De Moivre s theorem: Given an comple number cos θ + i sin θ and an integer n, (cos θ + i sin θ) n = cos nθ + i sin nθ Some results: (cos θ i sin θ) n = cos nθ i sin nθ (cos θ + i sin θ) n = cos nθ i sin nθ (cos θ i sin θ) n = cos nθ + i sin nθ sin θ + i cos θ = i(cos θ i sin θ) Formula for finding n th root of a comple number: z n = r ncis ( θ+kπ ), k = 0,,., (n ).P.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.P.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org V.GNANAMURUGAN, M.Sc.,M.Ed., P.G.T, GHSS, S.S.KOTTAI, SIVAGANGAI DT , Page n.p.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.p.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org ω = + ω + ω = 0 The sum of all the n th roots of unit is 0.(ie., + ω + ω +. +ω n = 0) The product of all the n th roots of unit is ( ) n. (ie., + ω + ω +. +ω n = ( ) n ) All the n roots of n th roots of unit are in geometrical progression. All the n roots of n th roots of unit lie on the circumference of a circle whose.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org

14 centre is at the origin and radius equal to and these Pai roots divide the circle into n equal parts and form a polgon of n sides. ai.org EXERCISE.8.If ω is a cube root of unit, then show that a+bω+cω a+bω+cω + a+bω+cω b+cω+aω c+aω+bω ai.org b+cω+aω + a+bω+cω c+aω+bω =. = ( a+bω+cω ω ) + b+cω+aω ω (a+bω+cω ω c+aω+bω ω ) = ω(a+bω+cω ) + ω (a+bω+cω ) aω +cω +bω aω +bω 4 +cω ω = ; + ω + ω = 0 ai.org = ω(a+bω+cω ) + ω (a+bω+cω ) a+bω+cω a+bω+cω = ω + ω =.Show that ( + i 5 ) + ( i 5 ) = ai.org + i.p = rcis(θ) () r = + = ( ai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org ).Pai.Org.Pai.Org α = tan = tan = π 6 + ( ) = = =.P.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.P.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org Principal argument lies in the first quadrant θ = α = π 6 () + i = cis (π) 6 ( + i 5 ) = [cis ( π 6 )]5 = cis ( 5π ) = cos 6 (5π) + i sin 6 (5π) 6 Similarl, ( i 5 ) = cos ( 5π ) i sin 6 (5π) 6 ( + i 5 ) + ( i 5 ) = cos ( 5π ) + i sin 6 (5π) + cos 6 (5π) i sin 6 (5π) 6 ai.org ai.org ai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org = cos ( 5π ) = cos (5 80) = cos 50 = cos 0 = = 6 6 EXAMPLE. Simplif (i)( + i) 8 (ii)( + i) ai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.p TrbTnpsc.Find the value of ( +sin π 0 +i cos π 0 +sin π 0 i cos π 0 V.GNANAMURUGAN, M.Sc.,M.Ed., P.G.T, GHSS, S.S.KOTTAI, SIVAGANGAI DT , Page 4 0 ) Let z = sin π 0 + i cos π 0 z = sin π 0 i cos π 0 ( +sin π 0 +i cos π 0 0 +sin π 0 i cos π ) 0.Pai.Org.Pai.Org = ( +z + z 0 ) = ( +z +z z)0 = z 0 z 0 = (sin π 0 + i cos π 0 )0 = [i (cos π 0 i sin π 0 )]0.Pai.Org.Pai.Org.Pai.Org.Pai.Org = i 0 [cos (0 π 0 ) i sin (0 π 0 )] = [cos π i sin π] = ( 0) = EXAMPLE. 9 Simplif (sin π 6 + i cos π 6 )8 +cos θ+i sin θ EXAMPLE. 0 Simplif (.Pai.Org.Pai.Org +cos θ i sin θ )0 4.If cos α = + and cos β = +, show that (i) + = cos(α β) (ii) = i sin(α + β) (iii) m n n = i sin(mα nβ) (iv) m n + = cos(mα + nβ) m m n + = cos α = = cos α + i sin α + = cos β = cos β + i sin β (i) cos α+i sin α = = cos(α β) + i sin(α β) cos β+i sin β.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org = ( ) = cos(α β) i sin(α β) = cos(α β) + (ii) = (cos α + i sin α)(cos β + i sin β) = cos(α + β) + i sin(α + β) = () = cos(α + β) i sin(α + β) = i sin(α + β).pai.org.pai.org.p.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.p.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org (iii) m = (cos α + i sin α) m = cos mα + i sin mα.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.p.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org

15 n = (cos β + i sin β) n = cos nβ + i sin nβ m cos mα+i sin mα = n cos nβ+i sin nβ n = ( m m m n ai.org = cos(mα nβ) + i sin(mα nβ) n) = cos(mα nβ) i sin(mα nβ) n = i sin(mα nβ) m (iv) m = (cos α + i sin α) m = cos mα + i sin mα n = (cos β + i sin β) n = cos nβ + i sin nβ m n = (cos mα + i sin mα)(cos nβ + i sin nβ) = cos(mα + nβ) + i sin(mα + nβ) = cos(mα + nβ) i sin(mα + nβ) n ai.org ai.org m m n + m n = cos(mα + nβ) EXAMPLE. 8 If z = cos θ + i sin θ, show that z n + = cos nθ and zn z n = sin nθ. zn 5.Solve the equation z + 7 = 0 ai.org z + 7 = 0 z = 7 z = ( 7) = (7 ) z = ( ) = [cis(π)] = [cis(kπ + π)], k = 0,, = cis(k + ) π, k = 0,, = cis ( π ), cis(π), cis (5π) ai.org ai.org 6. If ω is a cube root of unit, show that the roots of the equation (z ) + 8 = 0 are, ω, ω. (z ) + 8 = 0 (z ) = 8 (z ) = ( ) (z ) = ( ) ai.org ( z ) = ai.org z.p.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.p.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org = () z = () z = [ (or) ω (or) ω ] ai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org Pai.P.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.P z = () TrbTnpsc z = z = { z = ω z = ω z = ω z = ω Thus the roots of the equation (z ) + 8 = 0 are, ω, ω. EXAMPLE. 4 Solve the equation z + 8i = 0, where z C..Pai.Org.Pai.Org Note: z + 8i = 0 z = (8i) z = (i) find polar form for i and use de Moivre theorem EXAMPLE. 5 Find all cube roots of + i. Note: ( + i) ; find polar form for + i and use de Moivre theorem 8 7.Find the value of (cos kπ kπ + i sin.pai.org.pai.org V.GNANAMURUGAN, M.Sc.,M.Ed., P.G.T, GHSS, S.S.KOTTAI, SIVAGANGAI DT , Page 5 k= 9 Let = () 9 = [cis(0)] 9 = [cis(kπ + 0)] 9, k = 0,,,,8 = cis ( kπ ), k = 0,,,,8 9 Thus the roots are cis(0), cis ( π ), cis 9 (4π), cis 9 (6π), cis 9 (8π ), cis (0π ), 9 9 cis ( π ), cis (4π), cis (6π ) W.K.T, sum of all the roots Is equal to zero..pai.org.pai.org.pai.org.pai.org 9 ) cis(0) + cis ( π ) + cis 9 (4π) + cis 9 (6π) + cis 9 (8π ) + cis (0π ) cis ( π ) + cis (4π) + cis (6π ) = 0 + cis ( π ) + cis 9 (4π) + cis 9 (6π) + cis 9 (8π ) + cis (0π ) cis ( π ) + cis (4π) + cis (6π ) = 0 cis ( π ) + cis 9 (4π) + cis 9 (6π) + cis 9 (8π ) + cis (0π ) cis ( π ) + cis (4π) + cis (6π ) = 8 cis ( kπ =.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org k= 8 k=.pai.org.pai.org.p.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org 9 ) (cos kπ kπ + i sin 9 9 ) =.Pai.Org.Pai.Org.P 8. If ω is a cube root of unit, show that (i)( ω + ω ) 6 + ( + ω ω ) 6 = 8.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.P.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org

16 (ii)( + ω)( + ω )( + ω 4 )( + ω 8 ). ( + ω ) = Pai.P W.K.T, ω = ; + ω + ω = 0 (i)( ω + ω ) 6 + ( + ω ω ) 6 = ( ω ω) 6 + ( ω ω ) 6 = ( ω) 6 + ( ω ) 6 = 6 ω ω = 64(ω ) + 64(ω ) 4 = = 8 (ii)( + ω)( + ω )( + ω 4 )( + ω 8 ). ( + ω ) = ( + ω)( + ω )( + ω )( + ω ). ( + ω ) Clearl the above epression contains terms, = ( + ω)( + ω )( + ω 4 )( + ω 8 ).. terms = [( + ω)( + ω )][( + ω)( + ω )].6 terms = [( + ω)( + ω )] 6 = ( + ω + ω + ω ) 6 = (0 + ) 6 = 9.If z = i, find the rotation of z b θ radians in the counter clockwise direction about the origin when (i)θ = π (ii)θ = π (iii)θ = π z = i = rcis(θ) () r = + = () + ( ) = = 4 = α = tan = tan = π 4 Principal argument lies in the 4th quadrant θ = α = π 4 ai.org ai.org ai.org ai.org ai.org ai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org () z = i = cis ( π 4 ).Pai.Org.Pai.Org (i)when θ = π z = cis ( π 4 ) cis (π ) ai.org ai.org = cis ( π 4 + π ).Pai.Org.Pai.Org = cis ( π+4π ) = cis ( π ) (ii)when θ = π z = cis ( π 4 ) cis (π ) ai.org.pai.org.pai.org = cis ( π 4 + π ).Pai.Org.Pai.Org.P.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.P.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.P TrbTnpsc = cis ( π+8π ) = cis ( 5π ) (ii)when θ = π z = cis ( π ) cis 4 (π).pai.org.pai.org.pai.org.pai.org V.GNANAMURUGAN, M.Sc.,M.Ed., P.G.T, GHSS, S.S.KOTTAI, SIVAGANGAI DT , Page 6 = cis ( π 4 + π ) = cis ( π+6π ) 4 = cis ( 5π ) 4 EXAMPLE. 6 Suppose z, z and z are the vertices of an equilateral triangle inscribed in the circle z =. If z = + i, then find z and z. z = represents the circle with centre (0,0) and radius. Let A, B and C be the vertices of the given triangle. Since the vertices z, z and z form an equilateral triangle inscribed in the circle z =, the sides of this triangle.pai.org.pai.org AB, BC and CA subtend π radian at the origin. We obtain z and z b the rotation of z b π and 4π respectivel..pai.org.pai.org Given: z = + i z = z cis ( π ) = ( + i ) [cos (π ) + i sin (π )].Pai.Org.Pai.Org.Pai.Org.Pai.Org = ( + i )[cos(80 60) + i sin(80 60)] = ( + i )( cos 60 + i sin 60) = ( + i ) ( + i = ) z = z cis ( π ) = [cos (π ) + i sin (π )].Pai.Org.Pai.Org = [cos(80 60) + i sin(80 60)] = ( cos 60 + i sin 60) = ( + i ) = i 4 0.Prove that the values of are ± ( ± i).pai.org.pai.org 4 Let =.Pai.Org.Pai.Org.P.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.P.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.P.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org

17 = ( ) 4 4 = 4 + = 0 ( ) () = 0 ( + ) ( ) = 0 ( + + )( + ) = 0 ( + + ) = 0 ; ( + ) = 0 ai.org ai.org = ± ( ) 4()() ai.org () = ± 4 = ± = ± i ai.org = ± i = ai.org.p.pai.org.pai.org.pai.org.pai.org = b ± b 4ac a ; ( )± ( ) 4()() () ; ± 4 ; ± ; ± i ; ± i.pai.org.pai.org.pai.org.pai.org ( ± i) ; ( ± i) = ± ( ± i) EXAMPLE. Find the cube roots of unit..pai.org.pai.org Let z = z = () = [cis(0)] ai.org ai.org.p.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org z = [cis(kπ + 0)], k = 0,, z = cis ( kπ ), k = 0,, z = cis(0), cis ( π ), cis (4π).Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org z = cis(0), cis (π π ), cis (π + π ) z = cos 0 + i sin 0, cos (π π ) + i sin (π π ), cos (π + π ) + i sin (π + π ) ai.org z =, cos π + i sin π, cos π i sin π z =, + i, +i + i z =, ai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org, i.pai.org.pai.org Pai.P.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.P TrbTnpsc The cube roots of unit, ω, ω are, +i EXAMPLE. Find the fourth roots of unit. Let z 4 = z = () 4 = [cis(0)] 4.Pai.Org.Pai.Org.Pai.Org.Pai.Org V.GNANAMURUGAN, M.Sc.,M.Ed., P.G.T, GHSS, S.S.KOTTAI, SIVAGANGAI DT , Page 7, i z = [cis(kπ + 0)] 4, k = 0,,, z = cis ( kπ ), k = 0,,, 4 z = cis(0), cis ( π ), cis 4 (4π), cis 4 (6π) 4 z = cis(0), cis ( π ), cis(π), cis (π ) z = cos 0 + i sin 0, cos ( π ) + i sin (π ), cos(π) + i sin(π), cos (π ) + i sin (π ) z =, i,, i The fourth roots of unit, ω, ω, ω are, i,, i..prove that the multiplicative inverse of a non- zero comple number z = + i.pai.org.pai.org.p.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org is z = ( +) + i ( + ) PROOF: Let z = u + iv be the inverse of z = + i W.K.T, zz = ( + i)(u + iv) = (u v) + i(v + u) = + i0 Equating the Real and Imaginar parts on both sides, u v = () ; v + u = 0 () = = + ; u = 0 = ; v= 0 = u = u = ; v = v = + +.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.P.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org.Pai.Org z = u + iv z = ( +) + i ( + ).For an two comple numbers z and z prove that z + = z + z PROOF: Let z = a + ib and z = c + id, a, b, c, d R z + z = (a + ib) + (c + id) = (a + c) + i(b + d) = (a + c) i(b + d) = (a ib) + (c id).pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.p.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org.pai.org