Dosta često me ljudi znaju pitati kako se rješava sljedeći zadatak i njemu slični:
Zapiši u trigonometrijskom obliku kompleksni broj
Rješenje.
Treba primjetiti da se u zadatku, kao argument trigonometrijskih funkcija, pojavljuje 
dok drugi, nakon malo razmišljanja, lako izvedemo
Sada je
![\begin{array}{rcl} z &=& \displaystyle \frac{i-1}{i(1-\cos\frac{2\pi}{5})+\sin\frac{2\pi}{5}}\\[15pt] &=& \displaystyle \frac{\sqrt{2}(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4})}{i(2\sin^2\frac{\pi}{5})+2\sin\frac{\pi}{5}\cos\frac{\pi}{5}}\\[15pt] &=& \displaystyle \frac{\sqrt{2}(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4})}{2\sin\frac{\pi}{5}(\cos\frac{\pi}{5} + i\sin\frac{\pi}{5})}\\[15pt] &=& \displaystyle\frac{\sqrt{2}}{2\sin\frac{\pi}{5}}\left(\cos\frac{11\pi}{20}+i\sin\frac{11\pi}{20}\right) \end{array}](http://s0.wp.com/latex.php?latex=%5Cbegin%7Barray%7D%7Brcl%7D++z+%26%3D%26+%5Cdisplaystyle+%5Cfrac%7Bi-1%7D%7Bi%281-%5Ccos%5Cfrac%7B2%5Cpi%7D%7B5%7D%29%2B%5Csin%5Cfrac%7B2%5Cpi%7D%7B5%7D%7D%5C%5C%5B15pt%5D++%26%3D%26+%5Cdisplaystyle+%5Cfrac%7B%5Csqrt%7B2%7D%28%5Ccos%5Cfrac%7B3%5Cpi%7D%7B4%7D%2Bi%5Csin%5Cfrac%7B3%5Cpi%7D%7B4%7D%29%7D%7Bi%282%5Csin%5E2%5Cfrac%7B%5Cpi%7D%7B5%7D%29%2B2%5Csin%5Cfrac%7B%5Cpi%7D%7B5%7D%5Ccos%5Cfrac%7B%5Cpi%7D%7B5%7D%7D%5C%5C%5B15pt%5D++%26%3D%26+%5Cdisplaystyle+%5Cfrac%7B%5Csqrt%7B2%7D%28%5Ccos%5Cfrac%7B3%5Cpi%7D%7B4%7D%2Bi%5Csin%5Cfrac%7B3%5Cpi%7D%7B4%7D%29%7D%7B2%5Csin%5Cfrac%7B%5Cpi%7D%7B5%7D%28%5Ccos%5Cfrac%7B%5Cpi%7D%7B5%7D+%2B+i%5Csin%5Cfrac%7B%5Cpi%7D%7B5%7D%29%7D%5C%5C%5B15pt%5D++%26%3D%26+%5Cdisplaystyle%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%5Csin%5Cfrac%7B%5Cpi%7D%7B5%7D%7D%5Cleft%28%5Ccos%5Cfrac%7B11%5Cpi%7D%7B20%7D%2Bi%5Csin%5Cfrac%7B11%5Cpi%7D%7B20%7D%5Cright%29++%5Cend%7Barray%7D&bg=ffffff&fg=333333&s=0 "\begin{array}{rcl} z &=& \displaystyle \frac{i-1}{i(1-\cos\frac{2\pi}{5})+\sin\frac{2\pi}{5}}\\[15pt] &=& \displaystyle \frac{\sqrt{2}(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4})}{i(2\sin^2\frac{\pi}{5})+2\sin\frac{\pi}{5}\cos\frac{\pi}{5}}\\[15pt] &=& \displaystyle \frac{\sqrt{2}(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4})}{2\sin\frac{\pi}{5}(\cos\frac{\pi}{5} + i\sin\frac{\pi}{5})}\\[15pt] &=& \displaystyle\frac{\sqrt{2}}{2\sin\frac{\pi}{5}}\left(\cos\frac{11\pi}{20}+i\sin\frac{11\pi}{20}\right) \end{array}")
Dakako, zadnji niz je napisan kao da znate pretvoriti kompleksni broj iz algebarskog u trigonometrijski oblik, te podijeliti kompleksne brojeve u trigonometrijskom zapisu.
