Answer to Question #301381 in Complex Analysis for pianowoman

Question

1a) Find the real numbers a and b such that

i5(3 - 2i)2 - (a - bi) = a - bi.

b. Let z1 =  √3 + i and z2 = √3 - i. Show that

z₁⁶¹ + z₂⁶¹ = 2⁶¹√3.

(Remember that: cos(- θ) = cos(θ), sin(θ) = -sin(θ) = -sin(θ), cos(2kπ + θ) = cos(θ) and sin(2kπ + θ) = sin(θ) )

Answer 1

1a)

$i^5(3 - 2i)^2 - (a - bi) = a - bi.$

$i(9-12i-4)=2a-2bi$

$12+5i=2a-2bi$

$a=6, b=-5/2$

b.

$z_1=\sqrt{3}+i=2(\cos (\pi/6)+i\sin (\pi/6))$

$z_2=\sqrt{3}-i=2(\cos (-\pi/6)+i\sin (-\pi/6))$

$z_1^{61}=(2(\cos (\pi/6)+i\sin (\pi/6)))^{61}$

$=2^{61}(\cos (61\pi/6)+i\sin (61\pi/6))$

$=2^{61}(\cos (\pi/6)+i\sin (\pi/6))$

$z_2^{61}=(2(\cos (-\pi/6)+i\sin (-\pi/6)))^{61}$

$=2^{61}(\cos (-61\pi/6)+i\sin (-61\pi/6))$

$=2^{61}(\cos (-\pi/6)+i\sin (-\pi/6))$

$=2^{61}(\cos (\pi/6)-i\sin (\pi/6))$

$z_1^{61}+z_2^{61}=2^{61}(\cos (\pi/6)+i\sin (\pi/6))$

$+2^{61}(\cos (\pi/6)-i\sin (\pi/6))$

$=2^{61}(2\cos (\pi/6))=2^{61}\sqrt{3}$