Chapter 1 Complex numbers

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1 Complex numbers MC Qld- Chapter Complex numbers Exercise A Operations on and representations of complex numbers a u ( i) 8i b u + v ( i) + ( + i) + i c u + v ( i) + ( + i) i + + i + 8i d u v ( i) ( + i) ( i) e u v ( i)( + i) + i i 8i + 8i f u v i i + i i i i+ 8i i i 0 ( + 8 i) 0 ( + 8i) i g u ( i) ( i) 9 i + i i a v + i ( ) + (). b v + w + i + ( + i) + i + + 8i 9i 9 c v i d w+ v e ( + i+ ( + i) ( + i + i) ( + i) i w i () + ( ) 7. f vv ( + i)( i) + i i i g w w ( + i) ( + i) ( + 8i)( 8i) i + i i 8 h v v i + i i i i ( + i) 0. 0.i i w i + i i i i i 7 ( i) i a z a + bi Show that z z z LHS z a bi a+ bi a bi a bi a + b z a bi RHS z ( a + b ) z z z b zz LHS RHS z LHS RHS So zz z a Let z + i r () + () r 8 r zz (a + bi) (a bi) a b i a + b z ( a + b ) a + b a bi a + b LHS

2 MC Qld- Complex numbers θ tan π θ ( st quadrant) z cis z. cis b Let z i r () + ( ) r θ tan ( ) ( th quadrant) π θ π r cis π r. cis c z + i r ( ) + () r θ tan ( ) ( nd quadrant) θ π r cis π r. cis π d z + i r ( ) + () r + r 8 r 7 r.9 θ tan ( st quadrant) θ 0.8 z.9 cis(0.8) e z i r ( ) + ( ) r + r 8 r 7 r.9 θ tan ( th quadrant) θ 0.8 z.9 cis( 0.8) f z + i r () + () r + 9 r r θ tan ( st quadrant) θ 0. z cis(0.) g z i r ( ) + ( ) r r θ tan ( rd quadrant) θ ( π + 0.) θ.0 z cis(.0) h z + i r ( ) + () r r θ tan θ (π 0.) θ.0 z cis(.0) i z + i r () + () r r θ tan θ 0.9 z cis(0.9) a z cis ( nd quadrant) ( st quadrant) z cos + isin z 0. + i 0.8 z +.i b z cis z cos + isin z i 0. z.98 +.i z. +.i c z 0 cis π z 0cos + isin z i z i π d z 8 cis π π z 8cos + isin z i z..i e z i cis(π) z i (cos(π) + i sin(π)) i ( + 0) i

3 Complex numbers MC Qld- f z cis() z (cos() + i sin()) 0. + i i g z cis() z (cos() + i sin()) z 0. + i 0.9 z. +.7i h z cis( ) z (cos( ) + i sin( )) z 0. + i 0.8 z.08.8i See answers above 7 See answers above 8 a u v cis 8 cis π π ( 8) cis + cis π 8 π + cis π b u v 8 8 cis π cis π 8 π cis π c v u 8 8 cis π π cis 8 π π cis π d u cis cis π cis + cis e v 8 π cis cis(π) (cos(π) + i sin(π)) 9 a Let z + i r + r. tan(θ) θ tan θ 0. In polar form z. cis(0.) z b + i z. cis(0.) z. 0. z. cis(0.) c + i. cis(0.). (cos(0.) + i sin(0.)). + 0.i 0 a Let z + i r + r. tan(θ) θ tan () θ. z. cis(.) z z.. z. cis(0.) z. cos(0.) +. sin(0.)i z i i i i z i ( ) i i b i i (0 + i) c Let z r cis cos + isin i i + +

4 MC Qld- Complex numbers tan(θ) θ π π z z π z cis π z π π π ( th quadrant) π <θ π cos + isin i Operation on z Geometric relationship to z a z Reflection in x-axis b iz Anticlockwise rotation of 90 c z Rotation of 80 d iz Clockwise rotation of 90 Investigation Complex numbers and matrices a z + i A I + H b A + ( ) ( ) + ( ) + ( ) This corresponds to the complex number a + bi where 7 ai + BH ai + BH a b a b b a a b 7 So b a 7 Therefore a 7 and b, giving the complex number 7 + i z ( + i)( + i) 9 + i + i + i 7 + i So A corresponds to the complex number z c A ( ) 9 + A corresponds to the complex number c + di where ci + dh 0 0 ci + dh c d c d d c c d So d c c d d c Therefore c and d So A corresponds to the complex number i z ( + i) i + i i

5 Complex numbers MC Qld- a b i 9 i i i So the complex number corresponding to A is the inverse of z, z. a b 0 0 b a a b ai + bh which corresponds to the complex number a + bi a b T b a a b b a a 0 0 b 0 0 ai bh which corresponds to the complex number a bi, the complex conjugate of a + bi. So the transpose corresponds to the conjugate. a b b a a b a b b c + a b a b a b + + b a a + b a + b a 0 b 0 a b 0 a b a b I + a + b a + b which corresponds to the complex number a b i a + b a + b (a + bi) a bi a+ bi a bi a bi a b i a bi a + b a b a + b a + b which is the inverse of a + bi So the inverse of the matrix corresponds to the inverse of its corresponding complex number. a c det b b a a a b ( b) a + b a+ bi ( a + b ) a + b So the determinant corresponds to the modulus squared. Investigation e iθ Show that e iθ cos(θ)+ i sin(θ) LΗS e iθ n θ θ n θ ( ) +...!! ( n)! + n+ θ θ n θ i θ ( ) +...!! (n + )! i But n θ θ n θ cos(θ) ( ) +...!! ( n)! n+ θ θ n θ sin(θ) θ ( ) +...!! (n + )! So LHS (cos(θ)) + i(sin(θ)) cos(θ) + i sin(θ) RHS, as required iθ iθ e + e Show that cos(θ) iθ iθ RHS e + e ( ) ( iθ i θ e + e ) [(cos(θ) + i sin(θ)) + (cos( θ) + i sin( θ))] (cos(θ) + i sin(θ) + cos(θ) i sin(θ)) ( cos(θ)) cos(θ) LHS, as required To express sin(θ) in terms of e iθ and e iθ we first need to have an i sin(θ) term, so we multiply and divide by i. We also multiply and divide by, because we expect to have had two i sin(θ) terms added together, similarly to question. sin(θ) (i sin(θ)) i To have only a sin(θ) term remaining, the cos(θ) terms must cancel out so sin(θ) (cos(θ) + i sin(θ) cos(θ) + i sin(θ)) i [(cos(θ) + i sin(θ)) (cos(θ) i sin(θ))] i [(cos(θ) + i sin(θ)) (cos( θ) + i sin( θ))] i i (eiθ e iθ ) iθ iθ e e i θ Show that cos(iθ) e + e LHS cos(iθ) ii ( θ) ii ( θ) e + e e θ θ + e θ θ e + e RHS, as required log ( x ) xlog e () e θ x e e using the expansion for e x x xlog e () ( xlog e()) ( xlog e()) e + xlog e() !! x In e x x In e x In ( x In ) ( x In ) e + x In+ + +!!

6 MC Qld- Complex numbers By using the value of log e () from a table and by taking many terms from the expansion, an accurate value for x can be computed to a certain number of decimal places, depending on the amount of terms used. Investigation e iθ and de Moivre s theorem a Prove that v w rr cis( θ + θ ), where v r cis(θ ) and w r cis(θ ) LHS v w [r (cos(θ ) + i sin(θ ))] [r (cos(θ ) + i sin(θ ))] iθ iθ ( re )( r e ) rr i i e θ + θ ( ) rr i e θ + θ rr cis( θ + θ ) RHS, as required b Prove that v w rr cis( θ θ), where v r cis(θ ) and w r cis(θ ) LHS v w rcis( θ) r cis( θ ) iθ re iθ re r iθ iθ e r r i( θ θ ) e r r θ θ r cis( ) RHS, as required c Prove that v n n r cis nθ, where v r cis(θ ) LHS v n (r cis(θ )) n i ( n re θ ) n ni ( ) r e θ n in ( ) r e θ n r cis( nθ ) RHS, as required a Show that cos(θ) cos (θ) sin (θ) LHS cos(θ) ix ix ( ) ( ) ( i θ i θ e + e e + e ), since cos(x) ( iθ iθ e + e ) ( iθ ) ( iθ ) e + e (cos( θ ) + i sin( θ )) + (cos( θ ) + i sin( θ )) (cos( ) sin( )) (cos( ) sin( )) θ + i θ + θ i θ (cos ( θ ) + i sin( θ )cos( θ ) + i sin ( θ ) + cos ( θ) isin( θ)cos( θ) + i sin( θ)) (cos ( ) sin θ ( θ )) cos (θ) sin (θ) RHS, as required b Show sin(θ) sin(θ) cos(θ) LHS sin(θ) ( ) ( ) ( i θ i θ e e ), i Since sin(x) ix ix ( e e ) i ( iθ iθ e e ) i ( iθ ) ( iθ e e ) i (cos( θ ) + i sin( θ )) (cos( θ ) + i sin( θ )) i (cos( θ ) + i sin( θ )) (cos( θ ) i sin( θ )) i (cos ( θ ) + i sin( θ )cos( θ ) + i sin ( θ ) i cos ( θ) + isin( θ)cos( θ) i sin ( θ)) (i sin( θ)cos( θ )) i sin(θ) cos(θ) RHS, as required a i i i i 0 + i cos + isin e iπ i So i i ( e ) i π e π e b e π i e e π i e[ cos( π) + isin( π) ] e( + 0i) e c log ( i) e First express i in polar form Let z i r cis(θ) r ( ) + ( ) + tan(θ) tan(θ) θ tan, (th quadrant) π So z π π So log e( i) loge iπ e e log ( ) π iπ e + e e log() log( )

7 Complex numbers MC Qld- 7 π log e() i log e( e) iπ loge Exercise B Factorisation of polynomials in C a z + z i z (i) (z + i)(z i) b z + 7 z 7i z ( 7 i) ( z+ 7 i)( z 7 i) c z + 8z + (z + 8z + ) + (z + ) + 9 (z + ) 9i (z + ) (i) (z + + i)(z + i) d z z z z z + 7 z + 7i z 7 i z 7 7 z + i z i e z z + 7 z z+ + 7 z + z i (z ) (i) (z + i)(z i) f 9z + z 8 9z z z z + [(z ) i ] [(z ) (i) ] (z + i)(z i) a z 8 Let w z z 8 w 8 (w + 9)(w 9) (z + 9)(z 9) (z 9i )(z + )(z ) (z (i) )(z + )(z ) (z + i)(z i) (z + )(z ) b z z Let w z z z w w (w )(w + ) (z )(z + ) ( z ( ) )( z i ) ( z+ )( z )( z+ i)( z i) c z + 0z + Let w z z + 0z + w + 0w + (w + )(w + ) (z + )(z + ) (z i )(z i ) (z (i) )(z (i) ) (z + i)(z i)(z + i)(z i) d z + z 0 Let w z z + z 0 w + w 0 (w + )(w ) (z + )(z ) ( z + i )( z ( ) ) ( z + ( i) )( z ( ) ) ( z+ i)( z i)( z+ )( z ) a f(z) z z + z + 8 f( ) ( ) ( ) + ( ) So (z + ) is a factor of f(z) Let f(z) (z + )(z + pz + q) So z z + z + 8 (z + )(z + pz + q) z z + z + 8 z + pz + qz + z + pz +q z z + z + 8 z + (p + )z + (p + q)z + q p + q 8 p q So f(z) (z + )(z z + ) (z + )(z z ) (z + )[(z ) + ] (z + )[(z ) i ] ( z+ )[( z ) ( i) ] ( z+ )( z + i)( z i) The three factors of f(z) are z +, z + i and z i b f(z) z + z + z f() () + () + () so (z ) is a factor of f(z) Let f(z) (z )(z + pz + q) So z + z + z (z )(z + pz + q) z + z + z z + pz + qz z pz q z + z + z z + (p )z + (q p)z q p q p q So f(z) (z )(z + z + ) (z )(z + z + + ) (z )[(z + ) + ] (z )[(z + ) i ]

8 MC Qld- 8 Complex numbers ( z )[( z+ ) ( i) ] ( z )( z+ + i)( z+ i) The three factors of f(z) are z, z+ + i and z+ i c f(z) z z z + 0 f( ) ( ) ( ) So (z + ) is a factor of f(z) Let f(z) (z + )(z + pz +q) So z z z + 0 (z + )(z + pz + q) z z z + 0 z + pz + qz + z +pz + q z z z + 0 z + (p + )z + (p + q)z + q p + q 0 p q So f(z) (z + )(z z + ) 9 9 ( z+ ) z z ( z+ ) z + ( z+ ) z + ( z+ ) z i ( z+ ) z i ( z+ ) z + i z i The three factors of f(z) are z +, z + i and z i d f(z) z + z z f( ) ( ) + ( ) ( ) So (z + ) is a factor of f(z) Let f(z) (z + )(z + pz + q) So z + z z (z + )(z + pz + q) z + z z z + pz + qz + z + pz +q z + z z z + (p + )z + (p + q)z + q p + q p So f(z) (z + )(z + z ) (z + )(z )(z + ) The three factors of f(z) are z +, z and z + e f(z) z z z f( ) ( ) ( ) ( ) + 0 So (z + ) is a factor of f(z) f() () () () So (z ) is a factor of f(z) Let f(z) (z + )(z )(z + pz + q) So z z z (z + )(z )(z + pz + q) z z z (z z + z )(z + pz + q) z z z (z z )(z + pz + q) z z z z + pz + qz z pz qz z pz z + 0z z z z + (p )z + (q p )z + ( q p)z q p 0 q p q So f(z) (z + )(z )(z + z + ) (z + )(z )(z + z + + ) (z + )(z )[(z + ) + ] (z + )(z )[(z + ) i ] (z + )(z )[(z + ) (i) ] (z + )(z )(z + + i)(z + i) The four factors of f(z) are z +, z, z + +i and z + i. f f(z) z Let w z Then z w (w + )(w ) So f(z) (z + )(z ) Let f(z) g(z)h(z) such that g(z) z + and h(z) z g(z) z + g( ) ( ) So (z + ) is a factor of g(z) Let g(z) (z + )(z +pz + q) So z + (z + )(z +pz + q) z + z + pz + qz + z + pz + q z + 0z + 0z + z + (p + )z + (p + q)z +q p + 0 q p So g(z) (z + )( z z +) ( z+ ) z z+ + ( z+ ) z + ( z+ ) z i ( z+ ) z i ( z+ ) z + i z i h(z) z h() () 0 So (z ) is a factor of h(z) Let h(z) (z )(z + rz + s) So z (z )(z + rz + s) z z + rz + sz z rz s z + 0z + 0z z + (r )z + (s r)z s r 0 s r s So h(z) (z )(z + z +) ( z ) z + z+ + ( z ) z+ + ( z ) z+ i ( z ) z+ i

9 Complex numbers MC Qld- 9 ( z ) z+ + i z+ i Since f(z) g(z)h(z), we can now write f(z) ( z+ ) z + i z i ( z ) z+ + i z+ i ( z )( z ) + z + i z i z i z i + + The six factors of f(z) are z +, z, z + i, z i, z+ + i and z+ i a + i is a zero of P(z) z + z 0z + Let z + i, then z i is another zero Let z be the third zero. Then P(z) (z z )(z z )(z z ) (z i)(z + i)(z z ) [(z ) + ](z z ) (z z + + )(z z ) (z z + )(z z ) so z + z 0z + (z z + )(z z ) Comparing the left- and right-hand sides: z z Therefore the two zeros required are i and b + i is a zero of P(z) z + 9z +z + Let z + i, then z i is another zero Let z be the third zero. Then P(z) (z z )(z z )(z z ) (z + i)(z + + i)(z z ) [(z + ) + ](z z ) (z + z + + )(z z ) (z + z + )(z z ) So z +9z + z + (z + z + )(z z ) Comparing the left- and right-hand sides: z z Therefore the two zeros required are i and c i is a root of P(z) z 0z + z Let z i, then z + i is another root let z be the third root, then P(z) (z z )(z z )(z z ) (z + i) (z i) (z z ) [(z ) + ](z z ) (z 8z + + )(z z ) (z 8z + 7)(z z ) So z 0z + z (z 8z + 7)(z z ) Comparing the left- and right-hand sides: 7 z z Therefore the two roots required are + i and z z + z z + 7 If z i is a factor, then z +i is a factor Let P(z) z z + z z + 7, then P(z) has factors (z i) and (z + i) Let P(z) (z i) (z + i)(z + pz + q) (z + 9)(z + pz + q) z + pz + qz + 8z + 9pz +9q z + pz + (q + 8)z + 9pz +9q So z z + z z + 7 z + pz + (q + 8)z + 9pz + 9q p 9q 7 q So P(z) (z i)(z + i)(z z + ) (z i)(z + i)(z z + ) (z i)(z + i)((z ) + ) (z i)(z + i) (( z ) i ) (z i)(z + i) z i z + i Therefore the remaining factors are z + i, z i, z + i and. P(z) z + ( + i)z + i P( i) ( i) + ( + i)( i) + i i i i + i 0, as required P( ) ( ) + ( + i)( ) + i 9 9 i + i 0, as required So i and are the zeros pf P(z) E 7 P(z) z ( + i)z + ( + i)z P(z) (z ) Q(z), where Q(z) is a polynomial Let Q(z) z + pz + q. Then P(z) (z )(z + pz + q) z + pz + qz z pz q z + (p )z + (q p)z q So z ( + i)z + ( + i)z z + (p )z + (q p)z q P ( + i) q P i q P i So Q(z) z iz + Q(i) (i) i(i) D 8 P(z) is a polynomial of degree so Let P(z) (z z )(z z )(z z )(z z ) ai and bi are roots, so let z ai and z bi If ai is a root so is ai, and if bi is a root so is bi, so let z ai and z bi The term that does not contain z is given by z z z z ai bi ai bi a b i a b E 9 P(z) z + z z + a, P( i) 0 P( i) ( i) + ( i) ( i) + a () () (i) + ()(i) (i) + ( i + i ) + i + a i + i + i + i + a 8 7i + 7i + a a 8 0 So a 8 C 0 a Let P(z) z + z + az + 8 If (z + ) is a factor of P(z), P( ) 0 So P( ) ( ) + ( ) + a( ) a a a

10 MC Qld- 0 Complex numbers b Let P(z) z + az + z If (z + i) is a factor of P(z), P( i) 0 So P( i) ( i) + a( i) + ( i) 0 i a i 0 a a c Let P(z) z + z +8z + a If (z + i) is a factor of P(z), P( + i) 0 So P( + i) ( + i) + ( + i) + 8( + i) + a 0 [( ) + ( ) (i) + ( )(i) + (i) ] + ( i + i ) 8 + i + a 0 ( + i + 8i) + ( i ) 8 + i + a 0 ( i) + ( i) 8 + i + a 0 i 9 i 8 + i + a 0 + a 0 a d Let P(z) z z i + az i If i is a root of P(z), P(i) 0 P(i) (i) (i) i + a(i) i 0 8i + 8i + ia i 0 ia i a a Let P(z) z + az + 8z + b If and are roots, P( ) 0 and P() 0 P( ) ( ) + a( ) + 8( ) + b 7 + 9a + b 9a + b 0 So 9a + b [] P() () + a() + 8() + b 8 + a + + b a + b + 0 So a + b [] Subtracting [] from [] gives 9a a + a 7 a Substituting back into [] gives 9() + b + b b 8 b Let P(z) z + az + bz 7z + If and are zeros of P(z), P() 0 and P() 0 P() () + a() + b() 7() + + a + b 8 + a + b So a + b 0 a + b [] P() () + a() + b() 7() + + a + b 7 + a + b + 0 So a + b [] Subtracting [] from [] gives a a + a 9 a Substituting back into [] gives ( ) + b + b b c Let P(z) z + aiz + bz i If i and i are roots of P(z), P(i) 0 and P(i) 0 P(i) (i) + ai(i) + b(i) i 8i ia + ib i ia + ib 0i 0 So ia + ib 0i a b 0 [] P(i) (i) + ai(i) + b(i) i 7i 9ia + ib i 9ia + ib 9i 0

11 Complex numbers MC Qld- So 9ia + ib 9i a b [] Subtracting [] from [] gives a a + 0 a Substituting into [] gives ( ) b 0 b 0 b a Let P(z) z iz + aiz + b If (z + ) is a factor of P(z), P( ) 0 P( ) ( ) i( ) + ai( ) + b i ai + b (b ) (a + )i 0 + 0i Therefore, b 0 a + 0 b a b Let P(z) az z + biz + i If (z i) is a factor, P(i) 0 P(i) a(i) (i) + bi(i) + i ai + b + i ( b) (a )i 0 + 0i Therefore, b 0 a 0 b a c Let P(z) z + aiz + iz + ( + i)b If (z + i) is a factor, P( i) 0 P( i) ( i) + ai( i) + i( i)+(+i)b 8i ai + + b + bi ( + b) + (8 + b a) 0 + 0i Therefore + b b a 0 b 8 a 0 a a Since complex roots occur in conjugate pairs, an odd number of roots must have at least one real root. Let P(z) (z z )(z z )(z z ) and i are zeros, so let z and z i. Since i is a zero, i must also be a zero, so let z i So P(z) (z )(z i)( z + i) (z )(z + ) z + z z z z + z a P(z) z + z + z + 08 ai is a solution to P(z), so P(ai) 0 P(ai) (ai) + (ai) + (ai) + 08 a i a + ai + 08 (08 a ) + (a a )i 0 + 0i So 08 a 0 a a 0 a 0 a(a ) 0 a ± a 0 a a ± a must be such that both real and imaginary parts equal zero, so we discard the a 0 solution. Therefore a ±. b P(z) z + iz z i If ai is a solution to P(z), P(ai) 0 P(ai) (ai) + i(ai) ai i a i a i ai i (a + a + a + )i 0i So a + a + a + 0 Let f(a) a + a + a + f( ) ( ) + ( ) + ( ) So a + is a factor of f(a) Let f(a) (a + )(a + pa + q) a + pa + qa + a + pa + q a + (p + )a + (p + q)a + q So a + a + a + a + (p + )a + (p + q)a + q p + q p So f(a) (a + )(a + a + ) (a + )(a + )(a + ) 0 a, a, a Let P(z) z + i P(i) (i) + i i + i 0 So (z i) is a factor of P(z) Let P(z) (z i)[z + (a + bi)z + (c + di)], where a, b, c, d R P(z) (z i)[z + (a + bi)z + (c + di)] z + (a + bi)z + (c + di)z iz i(a + bi)z i(c + di) z + (a + bi)z + (c + di)z iz + (b ai)z + (d ci) z + (a + bi i)z + (c + di + b ai)z + (d ci) z + [a + (b )i]z + [(c + b) + (d a)i]z + (d ci) So z + z + [a + (b )i]z + [(c + b) + (d a)i]z + (d ci) a + (b ) 0 + 0i d ci 0 + i so a 0 b 0 d 0 c b c Therefore P(z) (z i)[z + (0 + i)z + ( + ci)] (z i)(z + iz ) 7 a P(z) z ( + i)z + (i )z + (7 + i)z i Show P() 0 P() () ( + i)() + (i )() + (7 + i)() i + i + i i i 0 + 0i 0, as required b P(z) (z ) Q(z) Let Q(z) [z + (a + bi)z + (c + di)z + (e + fi)], where a, b, c, d, e, f R So P(z) (z )[z + (a + bi)z + (c + di)z + (e + fi)] z + (a + bi)z + (c + di)z + (e + fi)z z (a + bi)z (c + di)z (e + fi) z + (a + bi )z + [(c + di) (a + bi)]z + [(e + fi) (c + di)]z (e + fi) z + [(a ) + bi]z + [(c a) + (d b)i]z + [(e c) + (f d)i]z (e + fi) So z ( + i)z + (i )z + (7 + i)z i z + [(a i) + bi]z + [(c a) + (d b)i]z + [(e c) + (f d)i] (e + fi) Equation coefficients gives ( + i) (a ) + bi i (a ) + bi So a b a 0 (i ) (c a) + (d b)i + i (c a) + (d b)i So c a d b c 0 d + c d 0 i (e + fi) + i e + fi So e f Therefore Q(z) z + (a + bi)z + (c + di)z + (e + fi) z + (0 i)z + ( + 0i)z + ( + i) z z i z + + i c Q(z) (z a) + b, where a c and b R Let a c + di, where c, d R Q(z) (z a) + b [z (c + di)] + b z + z (c + di) + z(c + di) (c + di) + b z (c + di)z + (c + cdi d )z [c + c di + c(di) + (di) ] + b

12 MC Qld- Complex numbers z (c + di)z + [(c d ) + cdi]z (c + c di cd d i) + b z (c + di)z + [(c d ) + cdi]z [(c cd b) + (c d d )i] So z iz z + + i z (c + di)z + [(c d ) + cdi]z [(c cd b) + (c d d )i] i (c + di) + i [(c cd b) (c d d )i] i c + di So (c cd b) So c 0 d (0 (0)() b) So a 0 + i b a i b 8 Let P(z) z + z + 8z + 0z + Since ( z+ i) is a factor of P(z), ( z i) must also be a factor since P(z) has real coefficients Let P(z) ( z+ i)( z i)( z + pz+ q) (z + )(z + pz + q) z + pz + qz + z + pz + q z + pz + (q + )z + pz + q So z + z + 8z + 0z + z + pz + (q + )z + pz + q p q q So P(z) ( z+ i)( z i)( z + p+ ) ( z+ i)( z i)[( z + p+ ) + ] ( z+ i)( z i)[( z+ ) + ] ( z+ i)( z i)[( z+ ) i ] ( z+ i)( z i)[( z+ ) ( i) ] ( z+ i)( z i)( z+ + i)( z+ i) 9 P(z) 9z + (9i )z + ( i)z + i If P( i) 0, (z + i) is a factor of P(z) Let P(z) (z + i)[9z + (a + bi)z + (c + di)], where a, b, c, d R P(z) (z + i)[9z + (a + bi)z + (c + di)] 9z + (a + bi)z + (c + di)z + 9iz + i(a + bi)z + i(c + di) 9z + (a + bi)z + (c +di)z + 9iz + ( b + ai)z + ( d + ci) 9z + [a + (b + 9)i]z + [(c b) + (d + a)i]z + ( d + ci) So 9z + (9i )z + ( i)z + i 9z + (a + (b + a)i)z + [(c b) + (d + a)i]z + ( d + ci) a + (b + a)i 9i d + ci i a + (b + a)i +9i d + ci 0 + i So a b So d 0 c b 0 d 0 Therefore p(z) (z +i)[9z + ( + 0i)z + ( + 0i)] (z + i)(9z z + ) (z + i) 9z z+ + 9 ( z+ i) z + ( z+ i) (z ) i (z + i)(z + i)(z i) 0 a + z a z az + z a z + az + a 0 Let P(z) z + az + a We want to solve for a such that P(z) 0. If i is a zero of P(z), then P ( i) 0. So P ( i) ( i) + a( i) + a 0 9 a + a 0 a a Investigation The exact values of cos and sin z x +yi, z z (x + yi) x + x (yi) + 0x (yi) + 0x + (yi) + x(yi) + (yi) x + x yi 0x y 0x y i + xy + y i So x + x yi 0x y 0x y i + xy + y i x + x yi 0x y 0x y i + xy + y i (x 0x y + xy ) + (x y 0x y + y )i Equating the real parts gives x 0x y + xy z x + y z So z kπ kπ z cis(θ) Therefore z So x + y x + y y x x 0x y + xy x 0x y + x(y ) x 0x ( x ) + x( x ) x 0x + 0x + x( x + x ) x 0x + 0x + x 0x + x x 0x + x x 0x + x 0 The two exact positive solutions for x are and z kπ, k N kπ kπ x + yi cos + isin k x + yi cos + isin So x cos, from part We reject the x solution because this would be when k 0, giving x cos(0). 7 Pythagoras theorem gives cos + sin + sin + + sin

13 Complex numbers MC Qld- + sin sin sin π + π 0 + sin 0 + sin + The positive square root was taken because the angle π is in the first quadrant, so sin must be positive. Exercise C Solving equations in C a x + x + 0 (x + x + ) + 0 (x + ) + 0 (x + ) x + ± x ± i b x 8x + 0 (x 8x + ) + 0 (x ) (x ) 9 x ± 9 x ± i c x x (x x + 9) (x 7) (x 7) 00 x 7 ± 00 x 7 ± 0i d x x + 0 x x x x x x ± x ± i e x x + 0 x x + 0 x x + 0 x x + 0 x x+ + 0 x + 0 x x x ± ± i x ± i a z z z Let P(z) z z z P( ) ( ) ( ) ( ) So z is a solution So (z + ) is a factor Let P(z) (z + )(z + pz + q) z + pz + q + z + pz + q z + (p + )z + (p + q)z + q So z z z + 0 z + (p + )z + (p + q) + q p + q 0 p q So P(z) (z + )(z z + ) 0 So z + 0 z z + 0 z a b c ( ) ± ( ) z z ± 9 0 z ± z i ± b z z + z 0 Let P(z) z z + z 0 P() () () + () + 0 So z is a solution So (z ) is a factor Let P(z) (z )(z + pz + q) z + pz + qz z pz q z + (p )z + (q p)z q So z z + z z + (p )z + (q p)z q p q p q So P(z) (z )(z z + ) 0 So z 0 z z + 0 z a b c ( ) ± ( ) z z ± 8 z ± 7 z 7 ± i c z 7z + 0z 8 0 Let P(z) z 7z + 0z 8 0 P() () 7() + 0() So z is a solution So (z ) is a factor Let P(z) (z )(z + pz + q) z + pz + qz z pz q z + (p )z + (q p)z q

14 MC Qld- Complex numbers So z 7z + 0z 8 z + (p )z + (q p)z q p 7 q 8 p q So P(z) (z )(z z + ) 0 So z 0 z z + 0 z a, b, c ( ) ± ( ) z z ± 9 z ± z ± i d z z + z 0 Let P(z) z z + z 0 P() () () + () So z is a solution So (z ) is a factor Let P(z) (z )(z + pz + q) z + pz + qz z p q z + (p )z + (q p)z q So z z + z z + (p )z + (q p)z q p q p q So P(z) (z )(z z + ) 0 So z 0 z z + 0 z a, b, c ( ) ± ( ) z z ± z ± z i ± e z 0z + z 0 0 Let P(z) z 0z + z 0 0 P() () 0() + () So z is a solution So (z ) is a factor Let P(z) (z )(z + pz + q) z + pz + qz 8z pz q z + (p 8)z + (q p)z q So z 0z + z 0 z + (p 8)z + (q p)z q p 8 0 q 0 p q 0 So P(z) (z )(z z + 0) 9 (z ) z z (z ) z + (z )[(z ) i ] (z )(z + i)(z i) 0 So z 0 z + i 0 z i 0 z z i z + i z i i f(z) z, g(z) z z +, h(z) z z + z Show that f(z) g(z) h(z) f(z) g(z) (z )(z z + ) z z + z z + z z z + z h(z), as required h(z) 0 So f(z) g(z) 0 (z )(z z + ) 0 So z 0 z z + 0 z a b c ( ) ± ( ) z z ± z ± z i a x + x + 0 y + y + 0, with y x (y + )(y + 9) 0 (x + )(x + 9) 0 x + 0 x x x 9 x ± x ± 9 x ±i x ± i b z z 0 y y 0, with y z (y )(y + ) 0 (z )(z + ) 0 z 0 z + 0 z z z ± z ± z ± z ± i c 9z + z 0 9y + y 0, with y z (9y )(y + ) 0 (9z )(z + ) 0 9z 0 z + 0 z 9 z z ± 9 z ± z ± z ± i d x + x y + y + 9 0, with y x (y + )(y + ) 0 (y + ) 0 (x + ) 0 x + 0 x x ±

15 Complex numbers MC Qld- (z ) + 0 (z ) i x ± x ± z ± z ± i z ± i B a + i i Let z + i. Take a and b x + + ( ) x + x + x x ± x ± x ± y y y ± y ± y ± Therefore the two roots z, z are z + i and z i So + i is ( + i) or ( + i ) b + 0i Let z + 0i. Take a and b 0 x x + 7 x + x 7 x x ± y y y ± Therefore the two roots z, z are z + i and z i So + 0i is + i or i c + i Let z + i. Take a and b x + + x + x + x 8 8 x ± 9 x ± 9 x ± y 8 y 8 y 9 y ± 9 7 y ± y 7 ± Therefore the two roots z and z are z 9 7 i and z i So + i is (9 + 7 i ) or (9 + 7 i ) 7 Let z i z i. Take a 0 b x x x ± x ± x ± y 0 y y ±

16 MC Qld- Complex numbers y ± y ± Therefore the two roots z and z are z z So i i is ( + i ) and ( + i ) + i and 8 One root of a ( + i) is a 8 Let z a 8 and z be the other root The root z is obtained by rotating z through an angle of π c So z a π 8 7π a C 8 (Note that we subtract π c rather than adding it to give an angle such that π < Arg(z) π) 9 a z i If w i r cis(θ) r tan(θ) ( ) + ( ) tan π ( th quadrant) π So w + kπ Let z π + kπ π z + kπ π + π k k 0 z cis π π k z + π The two square roots of i in polar form are cis π and b z + i If w + i r cis(θ) r + tan(θ) tan θ θ tan () ( st quadrant) π So w + kπ Let z + kπ k 0 z z + kπ () + kπ 8 + kπ 8 cis 8 π k z + π 8 9 π 8 9 π π 8 7 π 8 The two square roots of + i in polar form are 8 and 7 π 8 c z + i If w + i r cis(θ) r ( ) + ( ) tan(θ) tan(θ) θ π + tan ( ) ( nd quadrant) π π π So w 8 + kπ Let z 8 + kπ z 8 + kπ kπ + 9 k 0 z 9 π k z + 9

17 Complex numbers MC Qld- 7 + π 9 8π cis 9 π k z π 9 π 9 π π 9 π 9 The three cube roots of + i are 9, 8π 9 and π 9 d z i If w i r cis(θ) r 0 + sin(θ), cos(θ) 0 π θ So w + kπ Let z + kπ z + kπ kπ + k 0 z π k z + + π π π k z + + 8π 9π π π π π e z i If w i r cis(θ) r ( ) + ( ) tan(θ) tan(θ) θ π + tan () ( rd quadrant) π π + π π So w + kπ Let z k 0 z π + kπ π + kπ π kπ + z π π π k z + 8 π + π π π π k z + π + π π π π π The three cube roots of i are π, π and π 0 Let z ( i) So z i Let z w w i r cis(θ) r sin(θ) 0 + ( ), π θ cos(θ) 0 π So w + kπ

18 MC Qld- 8 Complex numbers So z π + kπ π z + kπ π kπ + π k 0 z π π cos + isin i i π π k z + π π π cos + isin i π π k z + 7π 7π 7π cos + isin π π cos isin i i Sum i+ i i i + i 0 i+ i i So ( i) is i, i or i and they sum to zero. a Let z If w r cis(θ) r sin(θ) 0 cos(θ) θ 0 So w cis(kπ) Let z cis(kπ) z [ cis( kπ )] kπ b k 0 z cis(0) (cos(0) + i sin(0)) k z cos + isin cos + isin i + + i π k z π π cos + isin cos isin i i The three cube roots of are, + i and i a z z cis(kπ) z (cis( kπ )) kπ kπ k 0 z cis(0) k z i k z cis(π) k z π i The four solutions in Cartesian form are, i, and i b z z cis(kπ) z (cis( kπ )) kπ kπ

19 Complex numbers MC Qld- 9 k 0 z cis(0) k z i k z cis( π ) k z π i The four solutions in Cartesian form are,, i and i c z z cis(kπ) z (cis( kπ )) kπ kπ k 0 z cis(0) k z z cos + isin + i + i k z cos + isin cos + isin i + + i k z cis(π) π k z π π cos + isin cos isin i i π k z π π cos + isin cos isin i i The six solutions in Cartesian form are, + i, + i,, i and. i d z 7 z 7 cis(kπ) z (7cis( kπ )) 7 kπ kπ k 0 z cis(0) k z cos + isin k z z + i + i cos + isin cos + isin i + + i k z cis ( π ) k z π π π cos + isin cos isin k z i i π π π cos + isin

20 MC Qld- 0 Complex numbers cos isin i i The six solutions in Cartesian form are, z z cis(kπ) z (cis( kπ )), + i, + i, i and i kπ k 0 z cis(0) k z π k z π k z π π π 8π k z 8π π π The five solutions in polar form are cis(0),, π, π and π Investigation Current and voltage iωt V Reiω L+ R+ Ice iωc I cos(0t) A I 0 cos(ω t) A So I 0 s and ω 0 R 80 Ω L 0. H C 0 F V i 0t Re i e i 0 0 0it Re 9.i e (.9 0 ) i i Re 9.i (cos(0 t) isin(0 t)) + (.9 0 ) i i Re (cos(0 ) sin(0 )) (.9 0 ) i t i t (cos(0 t)) + i 9. sin(0 t) (.9 0 ) 00 cos(0 t) 9. sin(0 t) (.9 0 ) 00 cos(0t) + 70 sin(0t)

21 Complex numbers MC Qld- Exercise D Graphs of complex relations: rays, lines and circles a b f g g h c h 7 a d i b c e j d f a a C b E B D a e f b b g c c h d d e 8 a e f b

22 MC Qld- Complex numbers c d e f g h 9 a b c d e f 0 B D a b c d e f g h Chapter review a u ( i) 8 i b u + v ( i) + ( + i) + i c u v ( i) ( + i) 8 + 0i i i + i u d v i i + i i 8 0i i+ i i 7 i i e u + v ( i) + (+i) 9i + + 0i + i f u + v + i, from part (e) a z b w + ( ) + c z i d z + w ( + i) + ( + i) + i e zw ( + i)( i) i i 8i 0i f w ( + i) i + i i i i i i z + i r +

23 Complex numbers MC Qld- 8 tan(θ) tan(θ) θ tan () ( st quadrant) π z cos + sin + i + i a u v π π π + 9π + π π π π π u b v π cis π π cis 9 π π cis π c v d π 8 cis(π) u π π π π 8 a Let z + i and w i + z + i r tan(θ) + + θ θ π z So w z tan ( ) ( st quadrant) π cis(π) cos(π)+ isin(π) b Let z i and w i z i r + 8 tan(θ) tan(θ) θ tan ( ) ( th quadrant) π π z 8 So w z z π 8 π 8 cis 8 8

24 MC Qld- Complex numbers i cos sin i 7 x + x + (x + x + 9) 9 + (x + ) 8 ( x + ) ( 8) (x + + 8)(x + 8) ( x + + ) ( x + ) A 8 Let P(z) z 8 z + z If i is a zero, then i must also be a zero. From the multiple choice the other zero must be either or, so we test them P() () () + () P( ) ( ) ( ) + ( ) So the third zero must be. A 9 a P(z) z z + 89 (z z + ) + 89 (z 8) + (z 8) i (z 8) (i) (z 8 + i)(z 8 i) b P(z) z + z 7z + P( ) ( ) + ( ) 7( ) So (z + ) is a factor of p(z) Let P(z) (z + )(z + pz + q) z + pz + qz + 8z + pz + q z + (p + 8)z + (q + p)z + q So z + z 7 + z + (p + 8)z + (q + p)z + q p + 8 q p q So P(z) (z + )(z z + ) (z + )(z z + ) (z + ) z z+ + (z + ) z + (z + ) z i (z + ) z i (z + ) z + iz i 0 P(z) z + z + az + 9z 9, P(i) 0 P(i) (i) + (i) + a(i) + 9(i) 9 8 7i 9a + 7i 9 8 7i 9a+ 7i 9 7 9a 0 7 a 9 a 8 Let z i So z i, Take a and b x x + + ( ) x + 00 x + 0 x x ± ± ± 0 y y y ± ± ± 0 So i is D z + i If w r + i r cis(θ) ( ) + tan(θ) θ tan π So w cis + kπ Let z cis + kπ z + kπ 0 0 i ± ( st quadrant) π kπ + 8 k 0 z π 8 k z π π π 8

25 Complex numbers MC Qld- π 8 k z π π π 8 π 8 C π π 8 π 8 z 8 will have 8 solutions which will be evenly spaced on an Argand diagram, so they will be spaced 0 apart 8 E a x x + 0 a, b, c ± ( ) x ± 0 ± ± i ± i b z 8 + i. Take a 8 and b x x x x x x ± ± ± y 8 y 9 y ± 9 y ± ± Therefore z i ± + ± ( + i ) c z z 8 z 8 cis(kπ) z ( 8 cis(kπ) ) 8 kπ kπ k 0 z cis(0) k z cos + isin cos + isin i + i π k z π π cos + isin cos isin i + i The three cube roots of 8 are, i and + i The graph is a ray with angle π clockwise from the horizontal and shifted in the positive direction of the Im(z) axis so E vertical line at Re(z) 7 So B 7 Ray with angle π anticlockwise from the horizontal and shifted units in the negative direction of the Re(z) axis and units in the positive direction of the Im(z) axis. So C 8 Circle of radius with centre at 0 + i. So A ( ) 9 Circle of radius with centre at i. So D 0 Sketch the graph of π z:arg( z i) on an Argand diagram.

26 MC Qld- Complex numbers Sketch the graph of {z: Im (z i) } on the complex plane. Illustrate {z: z + i } on the complex plane. Modelling and problem solving a z w π π π π cis π π cos + sin i b z cos + isin c + i + i w cos + isin i + z w + i + i + i i + i i i+ i i i ( + ) + ( ) i d i Equating the results from parts (a) and (c) gives π π cos + sin i + + i Equating the real parts gives π + cos ii Equating the imaginary parts of part (d) i gives π sin π sin π iii tan π cos + + ( ) ( + ) (8 ) c z, w cos zw π + π + π π π π cos sin i + π π cos + sin i z cos + isin i + + i + i w cos + isin i + + i zw ( + i) ( + i) + i+ i+ i ( ) + ( + )i Therefore ( ) π π cos + sin i ( ) + ( + ) i

27 Complex numbers MC Qld- 7 Equating the real and imaginary paths gives π cos π sin + π cos π sin + π tan π sin π cos ( + ) , as required a z + z i z (8i) (z + 8i)( z 8i) b z + y i y (8i), with y z (y + 8i)(y 8i) (z + 8i)(z 8i) c i Show that ( + i) 8i ( + i) + i + i + 8i 8i, as required ii Show that ( i) 8i ( i) + i + i 8i 8i, as required d z + (z + 8i)(z 8i) [z ( i) ][z ( + i) ] (z + i)(z + i)(z + + i)(z i) e z + (z + i)(z + i)(z + + i)(z i) [(z + i)(z + + i)][(z + i)(z i)] [z + ( + i) z + ( i) z + ( i)( + i)] [z + ( i) z + ( + i) z + ( + i)( i)] [z + ( + i + i)z + + i i i ] [z + ( i + i) z + + i i i ] (z + z + 8)(z z + 8)