Dúvida de triângulos e números complexos.

[latex]i^{\dfrac{1}{x}} =e^{\ln i^{\frac{1}{x}}} = e^{\frac{\ln i}{x}} [/latex]

Temos [latex] i = \cos \left(\dfrac{\pi}{2} \right) + i \sin \left(\dfrac{\pi}{2}\right) =\cos \left(\dfrac{\pi}{2}+2k\pi \right) + i \sin \left(\dfrac{\pi}{2}+2k\pi \right) = e^{\left(\dfrac{\pi}{2} + 2k\pi\right)\cdot i} [/latex]

Logo [latex] \ln i = \ln e^{\left(\dfrac{\pi}{2} + 2k\pi\right)\cdot i} =  \left(\dfrac{\pi}{2} + 2k\pi\right)\cdot i [/latex].

Logo [latex] i^{\dfrac{1}{x}} =e^{\frac{\ln i}{x}} = e^{\frac{\left(\frac{\pi}{2} + 2k\pi\right)\cdot i}{x}} =\exp\left( \dfrac{\left(\frac{\pi}{2} + 2k\pi\right)\cdot i}{x}\right) =\exp\left( \dfrac{\left(\frac{\pi}{2} + 2k\pi\right)}{x}\cdot i\right) = \cos\left(\dfrac{\left(\frac{\pi}{2} + 2k\pi\right)}{x}\right) + i \sin\left(\dfrac{\left(\frac{\pi}{2} + 2k\pi\right)}{x}\right)   [/latex]