Evaluate $\left(1+\sin\frac{\pi}{5}+i\cos\frac{\pi}{5}\right)^5+i\left(1+\sin\frac{\pi}{5}-i\cos\frac{\pi}{5}\right)^5$ 
Evaluate
  $$\left(1+\sin\frac{\pi}{5}+i\cos\frac{\pi}{5}\right)^5+i\left(1+\sin\frac{\pi}{5}-i\cos\frac{\pi}{5}\right)^5$$

I did this by $$\left(1+\sin\frac{\pi}{5}+i\cos\frac{\pi}{5}\right)^5=\left(1+\cos\frac{3\pi}{10}+i\sin\frac{3\pi}{10}\right)^5$$ and  get $0$
Does anyone have another idea?
Thanks
 A: Say $z = \sin\frac{\pi}{5}+i\cos\frac{\pi}{5}$. 
$\big($We have $-iz = \cos\frac{\pi}{5}-i\sin\frac{\pi}{5} = \cos\frac{-\pi}{5}+i\sin\frac{-\pi}{5}$, so  $\boxed{z^5 = -i}$ by De'Moivre formula.$\big) $
Notice that $\bar{z}={1\over z}$.
Then your expression is \begin{eqnarray}w &=& (1+z)^5+i(1+\bar{z})^5\\
 &=& (1+z)^5+i(1+{1\over z})^5\\
 &=& (1+z)^5+i{(z+1)^5\over z^5}\\
 &=& (1+z)^5+i{(z+1)^5\over -i}\\
 &=& 0\\
\end{eqnarray}
A: $$\frac{1+\sin x+i\cos x}{1+\sin x-i\cos x}=\frac{(1+\sin x+i\cos x)^2}{(1+\sin x)^2+\cos^2 x}$$
$$=\frac{2(1+\sin x)(\sin x+i\cos x)}{2(1+\sin x)}=\cos\left(\frac\pi2-x\right)+i\sin\left(\frac\pi2-x\right).$$
Apply de Movire's Theorem, you can get
$$\left(\frac{1+\sin x+i\cos x}{1+\sin x-i\cos x}\right)^n=\cos n\left(\frac{\pi }{2}-x\right)+i\sin n\left(\frac{\pi}{2}-x\right).$$
In youe situation, take $x=\pi/5$.
$$\left(1+\sin\frac{\pi}{5}+i\cos\frac{\pi}{5}\right)^5+i\left(1+\sin\frac{\pi}{5}-i\cos\frac{\pi}{5}\right)^5$$
$$=\left(\frac{1+\sin\frac{\pi}{5}+i\cos\frac{\pi}{5}}{1+\sin\frac{\pi}{5}-i\cos\frac{\pi}{5}}\right)^5+i=(-i)^5+i=0.$$
A: Using the facts that $(e^{\pi i/5})^5=e^{\pi i}=-1$ and $i^5=i$, we have
$$\begin{align}
(1+\sin\pi/5+i\cos\pi/5)^5+i(1+\sin\pi/5-i\cos\pi/5)
&=(1+ie^{-\pi i/5})^5+i(1-ie^{\pi i/5})^5\\
&=-(e^{\pi i/5})^5(1+ie^{-\pi i/5})^5+i^5(1-ie^{\pi i/5})^5\\
&=-(e^{\pi i/5}+i)^5+(i+e^{\pi i/5})^5\\
&=0
\end{align}$$
