<s>[INST] $(1+i)$ to the power $n$ 
Possible Duplicate:
Complex number: calculate $(1 + i)^n$. 

I came across a difficult problem which I would like to ask you about:
Compute
$ (1+i)^n $ for $ n \in \mathbb{Z}$
My ideas so far were to write out what this expression gives for $n=1,2,\ldots,8$, but I see no pattern such that I can come up with a closed formula.
Then I remember that one can write any complex number $a+bi$ like:
$$(a+bi)=\sqrt{a^2+b^2} \cdot \left( \frac{a}{\sqrt{a^2+b^2}}+\frac{b}{\sqrt{a^2+b^2}}\cdot i\right)$$
and $\frac{a}{\sqrt{a^2+b^2}} = \cos(\phi)$ and $\frac{b}{\sqrt{a^2+b^2}} = \sin(\phi)$  where $\phi$ is $\arctan{\frac{b}{a}} $
So it becomes, 
$(a+bi)=\sqrt{a^2+b^2} \cdot ( \cos(\phi)+\sin(\phi)\cdot i)$ Taking this entire thing to the power $n$ using De Moivre
$$(a+bi)^n=(\sqrt{a^2+b^2})^n \cdot ( \cos(n\phi)+\sin(n\phi)\cdot i)$$
Substituting my $a=1$ and $b=1$
$(1+i)^n=(\sqrt{2})^n \cdot ( \cos(n\cdot\frac{\pi}{4})+\sin(n\cdot\frac{\pi}{4})\cdot i)$
$\phi$ is 45 degrees hence $\frac{\pi}{4}$
But now I don't know how to continue further and I would really appreciate any help! Again, Im looking for a closed formula depending on n.
Best regards
 [/INST] You got to this expression:
$$(1+i)^n=\left(\sqrt{2}\right)^n  \left(\cos\left(n\frac{\pi}{4}\right)+i\sin\left(n\frac{\pi}{4}\right)\right)$$
Now, you should note that this expression (without $\left(\sqrt{2}\right)^n$) has periodicity of $8$, i.e. for $n=k$ and $n=k+8$ the value will be the same. So you can just look at $n=0,1,...,7$ and write the answer in the form:
$$(1+i)^n=\left(\sqrt{2}\right)^n\left\{ \begin{array}{cc} 1 & n\equiv0(\mod8)\\ \vdots & \end{array}\right.$$
</s>