# How do you write the expression (sqrt(2) - i)^6 in the standard form a + bi?

Mar 5, 2016

$- 23 + i 10 \sqrt{2} \cong - 23 + i \cdot 14.142136$

#### Explanation:

Calling
$C = \sqrt{2} - i$
We can express $C$ as $| C | \angle \phi$
With
$| C | = \sqrt{{\left(\sqrt{2}\right)}^{2} + {1}^{2}} = \sqrt{2 + 1} = \sqrt{3}$
And, since $- {90}^{\circ} < \phi < 0$
$\phi = - {\tan}^{- 1} \left(\frac{1}{\sqrt{2}}\right) = - {35.26439}^{\circ}$
=> $C = \sqrt{3} \text{ } \angle - {35.26439}^{\circ}$

Then

${C}^{6} = {\left(\sqrt{3}\right)}^{6} \text{ } \angle 6 \cdot \left(- {35.26439}^{\circ}\right)$
${C}^{6} = 27 \text{ } \angle - {211.58634}^{\circ}$
${C}^{6} = 27 \cos \left(- {211.58634}^{\circ}\right) + i \cdot 27 \sin \left(- {211.58634}^{\circ}\right)$
${C}^{6} = - 23 + i 14.142136$
If you like, reminding that $\sqrt{2} = 1.4142136$
${C}^{6} = - 23 + i 10 \sqrt{2}$