Polar form of a complex number 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?
 A: 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 $(8,\pi)$.
A: 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:
$$ = \ 2^{6/2} \ cis \left( \frac{20 \pi}{4} \right) \ = \ 2^3 \ cis(5 \pi) \ = \ 8 \ cis \ \pi \ \ \text{or} \ \ -8 \ \ .  ) $$ 
