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

Question

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

Answer 1

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}

Answer 2

$$\frac{1+\sin x+i\cos x}{1+\sin x-i\cos x}=\frac{(1+\sin x+i\cos x)^2}{(1+\sin x)^2+\cos^2 x}$$

$$=\frac{2(1+\sin x)(\sin x+i\cos x)}{2(1+\sin x)}=\cos\left(\frac\pi2-x\right)+i\sin\left(\frac\pi2-x\right).$$ 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 $x=\pi/5$. $$\left(1+\sin\frac{\pi}{5}+i\cos\frac{\pi}{5}\right)^5+i\left(1+\sin\frac{\pi}{5}-i\cos\frac{\pi}{5}\right)^5$$ $$=\left(\frac{1+\sin\frac{\pi}{5}+i\cos\frac{\pi}{5}}{1+\sin\frac{\pi}{5}-i\cos\frac{\pi}{5}}\right)^5+i=(-i)^5+i=0.$$

Answer 3

Using the facts that $(e^{\pi i/5})^5=e^{\pi i}=-1$ and $i^5=i$, we have

$$\begin{align} (1+\sin\pi/5+i\cos\pi/5)^5+i(1+\sin\pi/5-i\cos\pi/5) &=(1+ie^{-\pi i/5})^5+i(1-ie^{\pi i/5})^5\\ &=-(e^{\pi i/5})^5(1+ie^{-\pi i/5})^5+i^5(1-ie^{\pi i/5})^5\\ &=-(e^{\pi i/5}+i)^5+(i+e^{\pi i/5})^5\\ &=0 \end{align}$$