# Question #064a0

Apr 7, 2017

${\left(1 + i\right)}^{6} = 0 - 8 i$

#### Explanation:

Convert, $1 + i$, from $a + b i$ to polar form:

$r = \sqrt{{a}^{2} + {b}^{2}}$

$r = \sqrt{{1}^{2} + {1}^{2}}$

$r = \sqrt{2}$

$\theta = {\tan}^{-} 1 \left(\frac{b}{a}\right)$

$\theta = {\tan}^{-} 1 \left(\frac{1}{1}\right)$

$\theta = \frac{\pi}{4} \text{ radians}$

$1 + i = \sqrt{2} \left(\cos \left(\frac{\pi}{4}\right) + i \sin \left(\frac{\pi}{4}\right)\right)$

Raise both to the sixth power:

${\left(1 + i\right)}^{6} = {\left(\sqrt{2}\right)}^{6} \left(\cos \left(6 \frac{\pi}{4}\right) + i \sin \left(6 \frac{\pi}{4}\right)\right)$

Please observe that raising the polar form to the 6 power consists of raising the radius to the sixth power and multiplying the angle by 6.

Simplify:

${\left(1 + i\right)}^{6} = 8 \left(\cos \left(3 \frac{\pi}{2}\right) + i \sin \left(3 \frac{\pi}{2}\right)\right)$

We can convert the right side back to $a + b i$ form by substituting the know values for the trigonometric functions:

${\left(1 + i\right)}^{6} = 8 \left(0 + i \left(- 1\right)\right)$

${\left(1 + i\right)}^{6} = 0 - 8 i$