Originally Posted by

**Dfrtbx** Hello,

Trying to solve the second problem, I made a few manipulations, but I've seemed to arrive at a contradiction.

$\displaystyle (1+sin\frac{\pi}{5}+icos\frac{\pi}{5})^5+i(1+sin\f rac{\pi}{5}-icos\frac{\pi}{5})^5=0$

Using $\displaystyle i^5=i$

$\displaystyle (1+sin\frac{\pi}{5}+icos\frac{\pi}{5})^5+(i+isin\f rac{\pi}{5}+cos\frac{\pi}{5})^5=0$

$\displaystyle (1+sin\frac{\pi}{5}+icos\frac{\pi}{5})^5=-(i+isin\frac{\pi}{5}+cos\frac{\pi}{5})^5$

Using $\displaystyle \sqrt[5]{-1}=-1$

$\displaystyle 1+sin\frac{\pi}{5}+icos\frac{\pi}{5}=-(i+isin\frac{\pi}{5}+cos\frac{\pi}{5})$ The contradiction comes here, where you are taking the fifth root of both sides of an equation. A complex number has five fifth roots, and you can't assume that you are getting the same one on both sides of the equation. To take a much simpler example, (–1)^2 = 1^2, but you can't take the square root of both sides and conclude that –1 = 1.

$\displaystyle 1+sin\frac{\pi}{5}+icos\frac{\pi}{5}+i+isin\frac{\ pi}{5}+cos\frac{\pi}{5}=0$

$\displaystyle 1+sin\frac{\pi}{5}+cos\frac{\pi}{5} +i(1+sin\frac{\pi}{5}+cos\frac{\pi}{5})=0$

which only holds if the real and imaginary parts are equivalently 0. But because $\displaystyle 1+sin\frac{\pi}{5}+cos\frac{\pi}{5}$ is not, it's weird. My calculator says that I'm wrong.