Find the cube root of complex number ? Z=-√2 + i√2

Jan 17, 2018

multiply and divide LHS by -2
we get
$z = - 2 \left(\frac{1}{2} ^ \left(\frac{1}{2}\right) - i \frac{1}{2} ^ \left(\frac{1}{2}\right)\right) = - 2 \cdot {e}^{i - \frac{\pi}{4}} = | z | \cdot {e}^{i - \frac{\pi}{4}}$
therefore
${z}^{\frac{1}{3}} = {\left(| z | \cdot {e}^{i - \frac{\pi}{4}}\right)}^{\frac{1}{3}}$
$= - {2}^{\frac{1}{3}} \cdot {e}^{i - \frac{\pi}{12}}$
$= - {2}^{\frac{1}{3}} \left(\cos \left(- \frac{\pi}{12}\right) + i \sin \left(- \frac{\pi}{12}\right)\right)$
hence cube root of $z$ is
$- {2}^{\frac{1}{3}} \left(\cos \left(- \frac{\pi}{12}\right) + i \sin \left(- \frac{\pi}{12}\right)\right)$
hope u find it helpful :)

Jan 17, 2018

$\sqrt[3]{z} = {2}^{- \frac{1}{6}} + {2}^{- \frac{1}{6}} i$

Explanation:

Given:

$z = - \sqrt{2} + i \sqrt{2}$

$\textcolor{w h i t e}{z} = 2 \left(\cos \left(\frac{3 \pi}{4}\right) + i \sin \left(\frac{3 \pi}{4}\right)\right)$

The principal cube root of $z$ is:

${z}^{\frac{1}{3}} = {2}^{\frac{1}{3}} \left(\cos \left(\frac{\pi}{4}\right) + i \sin \left(\frac{\pi}{4}\right)\right)$

$\textcolor{w h i t e}{{z}^{\frac{1}{3}}} = {2}^{\frac{1}{3}} \left(\frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2}\right)$

$\textcolor{w h i t e}{{z}^{\frac{1}{3}}} = {2}^{- \frac{1}{6}} + {2}^{- \frac{1}{6}} i$

Jan 18, 2018

The solutions are $S = \left\{{2}^{\frac{1}{6}} \left(1 + i\right) , {2}^{\frac{1}{3}} \left(- \frac{\sqrt{2} + \sqrt{6}}{4} + i \frac{\sqrt{6} - \sqrt{2}}{4}\right) , {2}^{\frac{1}{3}} \left(- \frac{\sqrt{2} + \sqrt{6}}{4} - i \frac{\sqrt{6} + \sqrt{2}}{4}\right)\right\}$

Explanation:

Re write $z$ in the exponential form

$z = - \sqrt{2} + i \sqrt{2} = \sqrt{2} \left(- 1 + i\right)$

$= 2 \left(- \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} i\right)$

$= - 2 \left(\cos \left(\frac{3}{4} \pi\right) + i \sin \left(\frac{3}{4} \pi\right)\right)$

$= 2 {e}^{\frac{3}{4} \pi + 2 k \pi}$, $k \in \mathbb{Z}$

Therefore,

z^(1/3)=2^(1/3)e^(i(1/4pi+2/3kpi)

When $k = 0$

${z}_{0} = {2}^{\frac{1}{3}} {e}^{i \frac{1}{4} \pi} = {2}^{\frac{1}{3}} \left(\cos \left(\frac{\pi}{4}\right) + i \sin \left(\frac{\pi}{4}\right)\right)$

$= {2}^{\frac{1}{6}} \left(1 + i\right)$

When $k = 1$

${z}_{1} = {2}^{\frac{1}{3}} {e}^{i \frac{11}{12} \pi} = {2}^{\frac{1}{3}} \left(\cos \left(\frac{11}{12} \pi\right) + i \sin \left(\frac{11}{12} \pi\right)\right)$

$= {2}^{\frac{1}{3}} \left(- \frac{\sqrt{2} + \sqrt{6}}{4} + i \frac{\sqrt{6} - \sqrt{2}}{4}\right)$

When $k = 2$

${z}_{2} = {2}^{\frac{1}{3}} {e}^{i \frac{19}{12} \pi} = {2}^{\frac{1}{3}} \left(\cos \left(\frac{19}{12} \pi\right) + i \sin \left(\frac{19}{12} \pi\right)\right)$

$= {2}^{\frac{1}{3}} \left(- \frac{\sqrt{2} + \sqrt{6}}{4} - i \frac{\sqrt{6} + \sqrt{2}}{4}\right)$