# How do you find z, z^2, z^3, z^4 given z=sqrt2/2(1+i)?

Dec 17, 2016

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

#### Explanation:

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

= $\left(\frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2}\right)$ and in trigonometric form

= $\left(\cos \left(\frac{\pi}{4}\right) + i \sin \left(\frac{\pi}{4}\right)\right)$ or in exponential

= ${e}^{i \frac{\pi}{4}}$

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

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

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

and

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