# How do I use the modulus and argument to square (1+i)?

Apr 27, 2018

You should use the de Moivre's Theorem. See explanation.

#### Explanation:

For any complex number $z$ given in trigonometric form (having modulus $| z |$ and argument $\varphi$) we can calculate any natural power using the formula:

${z}^{n} = | z {|}^{n} \cdot \left(\cos n \varphi + i \cdot \sin n \varphi\right)$

Here we get:

$| z | = \sqrt{{1}^{2} + {1}^{2}} = \sqrt{2}$

$\cos \varphi = \frac{R e \left(z\right)}{|} z | = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \implies \varphi = {45}^{o}$

Now we can calculate the square:

${z}^{2} = | z {|}^{2} \cdot \left(\cos \left(2 \cdot 45\right) + i \sin \left(2 \cdot 45\right)\right) =$

$= 2 \cdot \left(\cos 90 + i \sin 90\right) = 2 \cdot \left(0 + 1 i\right) = 2 i$