Question: Write the polar form of $$\frac{(1+i)^{13}}{(1-i)^7}$$
Well its obviously impractical to expand it and try and solve it. Multiplying the denominator by $(1+i)^7$ will simplify the denominator, and a single term in the numerator.
Answer I got: $$(\frac{1}{\sqrt2}(cos(\frac{\pi}{4}) + sin(\frac{\pi}{4})i)^{20}$$
Is this correct?
On
You can also convert numerator and denominator into polar form immediately to write
$$ \frac{ [ \ \sqrt{2} \ cis(\frac{\pi}{4}) \ ]^{13} \ }{[ \ \sqrt{2} \ cis(-\frac{\pi}{4}) \ ]^7} \ \ . $$
DeMoivre's Theorem for powers gives us
$$ = \ \frac{ (\sqrt{2})^{13} \ cis(\frac{13\pi}{4}) }{(\sqrt{2})^7 \ cis(-\frac{7\pi}{4})} \ \ . $$
Division of complex numbers in polar form then produces
$$ = \ \frac{ (\sqrt{2})^{13} \ }{(\sqrt{2})^7} \ cis( \ \left[\frac{13\pi}{4} \right] \ - \ \left[-\frac{7\pi}{4} \right] \ ) \ \ . $$
You would simplify things from there. (Since the answer's already been posted, I'll finish this off:
$$ = \ 2^{6/2} \ cis \left( \frac{20 \pi}{4} \right) \ = \ 2^3 \ cis(5 \pi) \ = \ 8 \ cis \ \pi \ \ \text{or} \ \ -8 \ \ . ) $$
No, that's not correct. You must have made a couple of errors in your expansions. \begin{align} \frac{(1+i)^{13}}{(1-i)^7} &= \frac{(1+i)^{13}(1+i)^7}{(1-i)^7(1+i)^7} \\ &= \frac{1}{2^7}(1+i)^{20} \\ &= \frac{1}{2^7}\left(\sqrt{2}\left(\cos\frac{\pi}{4} + i \sin\frac{\pi}{4}\right)\right)^{20} \\ &= \frac{2^{10}}{2^7}\left(e^{i\pi/4}\right)^{20} \\ &= 8e^{5\pi i} \\ &= -8. \end{align} The polar form is $8(\cos\pi + i\sin\pi)$, or $(8,\pi)$.