Polar Form?

Question

Write 3-2i in polar form.


Write (2-2i)^8 in the standard form a+bi.

Answer 1

3-2i
has modulus sqrt13 and argument arctan(-2/3) = θ
3 - 2i = sqrt(13) * exp(i*θ)
where θ = arctan(-2/3) = -0.588
3 - 2i = 3.606 * exp(-i*0.588)

2 - 2i = 2*sqrt(2) * exp(-i*π/4)
(2 - 2i)^8 = 2^8 * 2^4 * exp(-i*2π)
= 2^12

Answer 2

3 - 2i is in quad 4
r² = 3² + (-2)² = 13
r = √13

tan Θ = -2/3
Θ = 326.31°

3 - 2i = (√13, 326.31°)

similarly, but using radians,
2 - 2i = (2√2, 7π/4)
then using DeMoivre's Theorem,
(2√2, 7π/4)^8 =
(2√2)^8[cos(8•7π/4) + i sin(8•7π/4)] =
(2^(1.5))^8 [ cos 14π + i sin 14π)] =
2^12 [ 1 + 0i] =
4096