Originally Posted by

**yeongil** Not sure what you're asking. I think it's easy to see that

$\displaystyle 1 + i = \sqrt{2}\left(\cos \frac{\pi}{4} + i\sin \frac{\pi}{4}\right)$,

and that

$\displaystyle 1 - i = \sqrt{2}\left(\cos \frac{7\pi}{4} + i\sin \frac{7\pi}{4}\right)$.

DeMoivre's Theorem states that if

$\displaystyle z = r(\cos \theta + i \sin \theta )$

then

$\displaystyle z^n = r^n(\cos n\theta + i \sin n\theta )$.

So

$\displaystyle \begin{aligned}

(1 + i)^5 &= (\sqrt{2})^5 \left(\cos \frac{5\pi}{4} + i\sin \frac{5\pi}{4}\right) \\

&= 4\sqrt{2} \left(\cos \frac{5\pi}{4} + i\sin \frac{5\pi}{4}\right)

\end{aligned}$

and

$\displaystyle \begin{aligned}

(1 - i)^7 &= (\sqrt{2})^7 \left(\cos \frac{49\pi}{4} + i\sin \frac{49\pi}{4}\right) \\

&= 8\sqrt{2} \left(\cos \frac{\pi}{4} + i\sin \frac{\pi}{4}\right)

\end{aligned}$

(49pi/4 and pi/4 are coterminal angles.)

When you divide two complex numbers in polar form, you divide the two moduli and subtract the arguments:

$\displaystyle \begin{aligned}

\frac {(1+i)^5}{(1-i)^7} &= \frac{4\sqrt{2} \left(\cos \frac{5\pi}{4} + i\sin \frac{5\pi}{4}\right)}{8\sqrt{2} \left(\cos \frac{\pi}{4} + i\sin \frac{\pi}{4}\right)} \\

&= \frac{4\sqrt{2}}{8\sqrt{2}}\left[\cos \left(\frac{5\pi}{4} - \frac{\pi}{4}\right) + i\sin \left(\frac{5\pi}{4} - \frac{\pi}{4}\right)\right] \\

&= \frac{1}{2}(\cos \pi + i \sin \pi) \\

&= -\frac{1}{2}

\end{aligned}$

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