# How do you use DeMoivre's Theorem to simplify (3-2i)^5?

How do you use DeMoivre's Theorem to simplify (3-2i)^5?

Answer 1

${\left(3 - 2 i\right)}^{5} = 69 \sqrt{13} \left(\cos 5 \alpha + i \sin 5 \alpha\right)$,

where $\alpha = {\tan}^{- 1} \left(- \frac{2}{3}\right)$

#### Explanation:

DeMoivre's Theorem states that if a complex number $z$ in polar form is given by $z = r \left(\cos \theta + i \sin \theta\right)$

then ${z}^{n} = {r}^{n} \left(\cos n \theta + i \sin n \theta\right)$

Here $| 3 - 2 i | = \sqrt{{3}^{2} + {2}^{2}} = \sqrt{13}$, hence we can write
$3 - 2 i$ as $\sqrt{13} \left(\cos \alpha + i \sin \alpha\right)$, where $\alpha = {\tan}^{- 1} \left(- \frac{2}{3}\right)$

Hence ${\left(3 - 2 i\right)}^{5}$

= ${\left(\sqrt{13}\right)}^{5} \left(\cos 5 \alpha + i \sin 5 \alpha\right)$

= $169 \sqrt{13} \left(\cos 5 \alpha + i \sin 5 \alpha\right)$

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