# How do you use DeMoivre's Theorem to find (1+i)^20 in standard form?

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May 19, 2016

Divide by the norm of $1 + i$ and apply De Moivre's theorem to find that

${\left(1 + i\right)}^{20} = - 1024$

#### Explanation:

De Moivre's formula, which can be derived from Euler's formula that ${e}^{i \theta} = \cos \left(\theta\right) + i \sin \left(\theta\right)$, states that

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

However, as $1 + i$ does not lie on the unit circle, we cannot use the formula directly. To fix this, we can divide $1 + i$ by its norm $\sqrt{2}$ and factor out the necessary constant:

${\left(1 + i\right)}^{20} = {\left(\sqrt{2} \left(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} i\right)\right)}^{20}$

$= {\left(\sqrt{2}\right)}^{20} {\left(\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} i\right)}^{20}$

$= 1024 {\left(\cos \left(\frac{\pi}{4}\right) + i \sin \left(\frac{\pi}{4}\right)\right)}^{20}$

$= 1024 \left(\cos \left(20 \cdot \frac{\pi}{4}\right) + i \sin \left(20 \cdot \frac{\pi}{4}\right)\right)$

$= 1024 \left(\cos \left(5 \pi\right) + i \sin \left(5 \pi\right)\right)$

$= 1024 \left(- 1 + i \cdot 0\right)$

$= - 1024$

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