I need this as example, complex analysis...?

Question

It's about powers of complex numbers and I needed the solution for this problem as an example. Answers should be in 1. Trigo Form and 2. Rectangular Form. The final answer should be in its "simplest" form. Thanks! :p

(2*sqrt.2+3i)^8

OMG no answers yet...

Answer 1

z = r angle θ
r² = 8 + 9
r = √17
r^(8) = 83521
θ = tan^(-1) (3 / (2√2)
θ = tan^(-1) (3√2 / 2)
θ = 64.8°
z = √17 (cos θ + i sin θ)
z^8 = 83521 (cos 8θ + i sin 8θ)
8θ = 518°
z^8 = 83521 angle 518°
z^8 = - 77439 + i 31288

Answer 2

a = 2 * sqrt(2) ; b = 3

1. Convert to polar form

r = sqrt (a^2+b^2) = sqrt(4*2+9) = sqrt (17)
or r = 17^(1/2)
displacement angle A = inv tan (3/2sqrt(2)
A = 46.69deg
polar form = r
2. Apply multiplication rule in the polar form

r^8 < 8*(46.69)

[17^(1/2)]^8 = 17^4

8*(46.69) = 373.50 deg

polar form of the answer is 17^4 < 373.50

3. Convert r trig form is 17^4(cos 373.50+i sin 373.50)
or 83521 (cos 13.5 + i sin 13.5) or
83251 cis 13.5

4. Convert r
a = 83251 cos 13.5 = 81213.31
b = 83251 sin 13.5 = 19434.56

81213.31 + 19434.56 i

Answer 3

well, it is
[sqrt(17) *(2 sqrt(2)/sqrt(17) + 3 i/sqrt(17))]^8

Now let Theta be cos^(-1) [2 sqrt(2)/sqrt(17)]
then the expression is
[sqrt(17) ( cos(theta) + i sin(theta))^8
= [sqrt(17) (e^(i * theta))]^8
= 17^4 e^(i * 8 theta)
= 17^4 [cos(8 theta) + i cos(8 theta)]

Hope this helps.