# Thread: Complex numbers question (Addition)

1. ## Complex numbers question (Addition)

$(\frac{1+i}{1-i})^{16}+(\frac{1-i}{1+i})^8$

Progress
$(\frac{1+i}{1-i})^{16}+(\frac{1-i}{1+i})^8=\frac{(1+i)^{16}(1+i)^8+(1-i)^{16}(1-i)^8}{(1-i)^{16}(1+i)^8}$

$\frac{(1+i)^{24}+(1-i)^{24}}{(1-i)^{16}(1+i)^8}$

$\frac{(1+i)^{24}+(1-i)^{24}}{2^8(1-i)^{8}}$

Any hints on continuing, other than ugly expanding.
Thanks in advance

2. Originally Posted by I-Think
$(\frac{1+i}{1-i})^{16}+(\frac{1-i}{1+i})^8$

Progress
$(\frac{1+i}{1-i})^{16}+(\frac{1-i}{1+i})^8=\frac{(1+i)^{16}(1+i)^8+(1-i)^{16}(1-i)^8}{(1-i)^{16}(1+i)^8}$

$\frac{(1+i)^{24}+(1-i)^{24}}{(1-i)^{16}(1+i)^8}$

$\frac{(1+i)^{24}+(1-i)^{24}}{2^8(1-i)^{8}}$

Any hints on continuing, other than ugly expanding.
Thanks in advance

Hints:

(1+i)^2=?
(1-i)^2=?

3. Exponentiation of Complex Numbers is always easiest in polar form...

Note that $1 + i = e^{\frac{\pi i}{4}}$ and $1 - i = e^{-\frac{\pi i}{4}}$.

So $\left(\frac{1 + i}{1 - i}\right)^{16} + \left(\frac{1 - i}{1 + i}\right)^8 = \frac{(1 + i)^{16}}{(1 - i)^{16}} + \frac{(1 - i)^8}{(1 + i)^8}$

$= \frac{\left(e^{\frac{\pi i}{4}}\right)^{16}}{\left(e^{-\frac{\pi i}{4}}\right)^{16}} + \frac{\left(e^{-\frac{\pi i}{4}}\right)^{8}}{\left(e^{\frac{\pi i}{4}}\right)^{8}}$

$= \frac{e^{4\pi i}}{e^{-4\pi i}} + \frac{e^{-2\pi i}}{e^{2\pi i}}$

$= e^{8\pi i} + e^{-4\pi i}$

$= \left(e^{\pi i}\right)^8 + \left(e^{\pi i}\right)^{-4}$

$= (-1)^8 + (-1)^{-4}$

$= 1 + 1$

$= 2$.