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?


  • $\begingroup$ Did you use the complex e-power? $\endgroup$ – imranfat Oct 29 '19 at 13:00
  • $\begingroup$ "Does anyone have another idea?" Could you show us your method? $\endgroup$ – The Jade Reaper Oct 29 '19 at 13:01
  • $\begingroup$ math.stackexchange.com/questions/936196/… this post maybe help you! $\endgroup$ – Riemann Oct 29 '19 at 13:05
  • $\begingroup$ @The Jade Reaper, I did this by $\left(1+\sin\frac{\pi}{5}+i\cos\frac{\pi}{5}\right)^5=\left(1+\cos\frac{3\pi}{10}+i\sin\frac{3\pi}{10}\right)^5$ $\endgroup$ – user62498 Oct 29 '19 at 13:06

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}

  • $\begingroup$ Aqua,I know this way but I don't use binomial theorem $\endgroup$ – user62498 Oct 29 '19 at 13:09
  • 1
    $\begingroup$ @ Aqua, this is great! thanks $\endgroup$ – user62498 Oct 29 '19 at 13:25

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

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


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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.