Ask Me Help Desk - Can you help answer how this problem is solved?

Two ways to do this:

1) Brute foirce, using a polynomial expansion:

$ (a-b)^5 = a^5 - 5a^4b + 10a^3b^2 - 10 a^2b^3 +5ab^4 - b^5 $

Apply this to (2-2i), and you get

$ (2-2i)^5 = 2^5(1-i)^5 = 32(1-5i+10i^2-10i^3+5i^4-i^5) = 32(1-5i-10+10i+5 -i) = 32(-4+4i) =-128+128i $

2) Use D'Moivre's theorem:

$ (R \cdot cis(\theta))^n = R^n cis(n \theta) $
where
$ cis(\theta) = \cos(\theta) + i \sin(\theta) $

Convert 2-2i to polar cordinates for the cis form: (2-2i) is radius 2 $\sqrt 2$ and angle $\theta = -\pi/4$

$ (2-2i)^5 = (2 \sqrt 2)^5 cis(- \frac 5 4 \pi) = 2^{15/2}cis( \frac 3 4 \pi) $

Finally, convert back to rectangular coordinates:

$ -2^7 + 2^7 i = -128 + 128i $