# How do you write the following in trigonometric form and perform the operation given (2+2i)(1-i)?

Oct 20, 2016

$\left(2 + 2 i\right) \left(1 - i\right) = 4$

#### Explanation:

Let ${z}_{1} = 2 + 2 i$ And ${z}_{2} = 1 - i$
Then the moduli are ∣z_1∣ = sqrt (2^2+2^2)=2sqrt2
and ${z}_{2} = \sqrt{1 + 1} = \sqrt{2}$
Rewrite the numbers
${z}_{1} = 2 + 2 i = 2 \sqrt{2} \left(\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}\right) = 2 \sqrt{2} \left(\cos {\theta}_{1} + i \sin {\theta}_{1}\right)$
Here ${\theta}_{1} = \frac{\pi}{4}$
so ${z}_{1} = 2 \sqrt{2} \left(\cos \left(\frac{\pi}{4}\right) + i \sin \left(\frac{\pi}{4}\right)\right) = 2 \sqrt{2} {e}^{i \frac{\pi}{4}}$
Similarly
${z}_{2} = 1 - i = \sqrt{2} \left(\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}\right) = \sqrt{2} \left(\cos {\theta}_{2} - i \sin {\theta}_{2}\right)$
${\theta}_{2} = - \frac{\pi}{4}$

so ${z}_{2} = \sqrt{2} \left(\cos \left(- \frac{\pi}{4}\right) + i \sin \left(- \frac{\pi}{4}\right)\right) = \sqrt{2} {e}^{- i \frac{\pi}{4}}$
so the result is ${z}_{1} \cdot {z}_{2} = 2 \sqrt{2} {e}^{i \frac{\pi}{4}} \cdot \sqrt{2} {e}^{- i \frac{\pi}{4}} = 4$