Question

determine a raiz quadrada do número complexo i

Answer 1

Determinar a raiz quadrada do número complexo $i$.

Colocando o número na forma polar, temos

$i=0+1i\\ \\ i=1\cdot \left(\cos \dfrac{\pi}{2}+i\cdot \mathrm{sen\,}\dfrac{\pi}{2} \right )$


Onde

$\left|i\right|=1\\ \\ \theta=\dfrac{\pi}{2}$


Utilizando a fórmula de De Moivre para o cálculo das raízes quadradas de $i$, temos

$\sqrt{i}=\sqrt{1}\cdot\left[\,\cos \left(\dfrac{\frac{\pi}{2}+k\cdot 2\pi}{2} \right ) +i\cdot \mathrm{sen}\left(\dfrac{\frac{\pi}{2}+k\cdot 2\pi}{2} \right )\,\right ]\\ \\ \sqrt{i}=1\cdot\left[\,\cos \left(\dfrac{\pi}{4}+k\pi \right ) +i\cdot \mathrm{sen}\left(\dfrac{\pi}{4}+k\pi \right )\,\right ]\\ \\ \sqrt{i}=\cos \left(\dfrac{\pi}{4}+k\pi \right ) +i\cdot \mathrm{sen}\left(\dfrac{\pi}{4}+k\pi \right )$

onde $k\in \mathbb{Z}$ e $k=0..1$.


$\bullet\;\;$ para $k=0$, temos

$\sqrt{i}=\cos \left(\dfrac{\pi}{4} \right ) +i\cdot \mathrm{sen}\left(\dfrac{\pi}{4} \right )\\ \\ \boxed{\sqrt{i}=\dfrac{\sqrt{2}}{2} +\dfrac{\sqrt{2}}{2}i}$


$\bullet\;\;$ para $k=1$, temos

$-\sqrt{i}=\cos \left(\dfrac{\pi}{4}+\pi \right ) +i\cdot \mathrm{sen}\left(\dfrac{\pi}{4}+\pi \right )\\ \\ \boxed{-\sqrt{i}=-\dfrac{\sqrt{2}}{2} -\dfrac{\sqrt{2}}{2}i}$