Look's Complex

$\begin{aligned} \frac {(1+i)^{2011}}{(1-i)^{2009}} & = \frac {(1+i)^{2\times 1005+1}}{(1-i)^{2\times 1004+1}} & \small \color{#3D99F6} \text{Note that }(1\pm i)^2 = \pm 2i \\ & = \frac {(2i)^{1005}(1+i)}{(-2i)^{1004}(1-i)} \\ & = \frac {2^{1005}i^{4\times 251+1}(1+i)}{(-2)^{1004}i^{4\times 251}(1-i)} & \small \color{#3D99F6} \text{and }i^2 = -1 \implies i^4 = 1 \\ & = \frac {2i(1+i)}{1-i} \\ & = \frac {2(i-1)}{1-i} \\ & = \boxed{-2} \end{aligned}$

Or using Euler's formula : $e^{ix} = \cos x + i \sin x$ .

$\begin{aligned} \frac {(1+i)^{2011}}{(1-i)^{2009}} & = \frac {\left(\sqrt 2\left(\frac 1{\sqrt 2}+\frac i{\sqrt 2}\right)\right)^{2011}}{\left(\sqrt 2\left(\frac 1{\sqrt 2}-\frac i{\sqrt 2}\right)\right)^{2009}} \\ & = \frac {\left(\sqrt 2\left(\cos \frac \pi 4+i\sin \frac \pi 4\right)\right)^{2011}}{\left(\sqrt 2\left(\cos \frac \pi 4-i\sin \frac \pi 4\right)\right)^{2009}} \\ & = \frac {\left(\sqrt 2e^{\frac \pi 4i}\right)^{2011}}{\left(\sqrt 2e^{-\frac \pi 4i}\right)^{2009}} \\ & = \left(\sqrt 2\right)^{2011-2009} e^{\frac {2011+2009}4 \pi i} \\ & = \left(\sqrt 2\right)^2 e^{1005 \pi i} & \small \color{#3D99F6} \text{Note that }e^{\pi i} = -1 \\ & = 2 (-1)^{1005} \\ & = \boxed{-2} \end{aligned}$

$\newcommand{\iu}{{i\mkern1mu}} \frac{(1+i)^{2011} \times i^{2009}}{(1-i)^{2009} \times i^{2009}} = \frac{(1+i)^{2011} \times i^{2009}}{((1-i) \times i)^{2009}} = \frac{(1+i)^{2011} \times i^{2009}}{(i+1)^{2009}}$ = $(1+i)^2 \times ((i^2)^2)^{502} \times i = (1 + 2i + i^2) \times i=-2$ (since $i^2 = -1$ )

There's no comment section, so I'm forced to post a comment here. It is not a valid explanation but a question. I believe I've solved the problem illegitimately and I'd like to know whether I'm right to be suspicious, and also why it worked.

I started working on the problem right after reading the wiki section about powers of i, so I tried to apply what I had just read.

To simplify larger powers of i, take the last two digits of the number and divide it by 4. Find the remainder and if the remainder is k, then the value is $i^{k}$

Then I commited something horrible: I ignored the fact that the problem has (1+i) and (1-i) instead of i's only and simplified it like so: $\frac{(1+i)^{2011}}{(1-i)^{2009}}=\frac{(1+i)^{3}}{(1-i)^{1}}=\frac{(1+i)^{2}(1+i)}{1-i}=\frac{2i(1+i)}{1-i}=\frac{2i+2i^{2}}{1-i}=\frac{-2+2i}{1-i}=\frac{-2(1-i)}{1-i}=-2$

I think I shouldn't have simplified the exponents, because I wasn't working with pure i's, but I'm surprised that it worked anyway. Why?

I first found out the logarithm of the given expression to the base $e$ and then raised $e$ to the power of the result.

$\log \frac{(i+1)^{2011}}{(1-i)^{2009}} \\ = 2011 \log (1+i) - 2009 \log (1-i) \\$ Now, using $\text{The principal value of } \log z = \log |z| + i arg(z),$ where arg(z) is the principal argument of z (which is also known as its amplitude), we get,

$\log (1+i) = \log \sqrt 2 + i(\frac{\pi}{4})$ and $\\ log (1-i) = \log \sqrt 2 - i(\frac{\pi}{4}).$

Substituting these values in the first equation, we get

$\log \frac{(1+i)^{2011}}{(1-i)^{2009}} \\ = 2011(\log \sqrt 2 + i \frac {\pi}{4}) - 2009(\log \sqrt 2 - i \frac{\pi}{4}) \\ = 2 \log \sqrt 2 + 4020 i \frac{\pi}{4} \\ = \log 2 + 1005 i \pi$

To get the value of original expression, we raise $e$ to the power of $\log 2 + 1005 i \pi$ .

$\therefore e^{\log 2 + 1005 i \pi} \\ = e^{\log 2} \times e^{1005 i \pi} \\ = 2[\cos(1005\pi) + i\sin(1005\pi] \ldots \textcolor{#3D99F6}{\text{Using Euler's Identity}} \\ = 2[-1 + i (0)] \\ = \boxed{-2}$ Feel free to comment on this solution and I apologise for any mistakes in LaTex as I am new to it.