Proof $(1+i)^{n}$ = $2^{n/2}(\cos(nπ/4)+i\sin(nπ/4))$ While  binomial problem I struck here
$$(1+i)^{n} = 2^{\frac{n}{2}}\left(\cos(\frac{n\pi}{4})+i\sin(\frac{n\pi}{4})\right).$$
Please proof this equation .
Any help will be appreciated.
And Thanks  for help
 A: $$1+i=\sqrt{2}\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right)$$ and the rest for you.
A: Hint:
$1+i=\sqrt{2}\exp\left(\dfrac{\pi i}{4}\right)$ and $e^{i\theta}=\cos\theta+i\sin\theta$.
A: This is not a binomial question,

Using Taylor series,
  $$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots$$
  $$\cos x=1-\frac{x^2}{2!} +\frac{x^4}{4!} -\cdots $$
  $$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!} +\cdots$$

$$e^{ix}=1+ix+\frac{(ix)^2}{2!}+\frac{(ix)^3}{3!} +\cdots$$
$$e^{ix}=1+ix-\frac{(x)^2}{2!}+i\frac{(x)^3}{3!} +\cdots$$
$$e^{ix}=(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots)+i(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots)$$

$$e^{ix}=\cos x+i\sin x$$

And,
We can write
$$a+ib=r\cos t+ir\sin t$$

$$a+ib=r e^{it}$$

Hint

$$r=|a+ib|$$
  $$t=\tan^{-1} (\frac{b}{a})$$

If you want answer

So(wait a minute...)\begin{align}(1+i)=\sqrt{1+1}e^{\frac{iπ}{4}}\end{align} \begin{align} (1+i)^n = 2^{\frac{n}{2}} e^{\frac{inπ}{4}}\end{align} \begin{align} (1+i)^n= 2^{\frac{n}{2}} (\cos {\frac{nπ}{4}} +i\sin {\frac{nπ}{4}})\end{align}

