# How can you use trigonometric functions to simplify  2 e^( ( pi)/8 i )  into a non-exponential complex number?

Apr 11, 2016

$1.8478 + 0.7654 i$

#### Explanation:

We can use the identity $r {e}^{i \theta} = r \cos \theta + i r \sin \theta$

Hence, $2 {e}^{\frac{\pi}{8} i} = 2 \cos \left(\frac{\pi}{8}\right) + 2 \sin \left(\frac{\pi}{8}\right) i$

= $2 \times \frac{\sqrt{2 + \sqrt{2}}}{2} + 2 \times \frac{\sqrt{2 - \sqrt{2}}}{2} i$

or in decimal notation approximately

= $2 \times 0.9239 + 2 \times 0.3827 i$

= $1.8478 + 0.7654 i$