# How do you evaluate  e^( ( pi)/2 i) - e^( ( 2 pi)/3 i) using trigonometric functions?

Jan 10, 2017

The value of this expression is $\frac{1}{2} + \left(1 - \frac{\sqrt{3}}{2}\right) i$

#### Explanation:

To evaluate this expression you have to write the complex numbers in algebraic form. To do this you use the identity:

## ${e}^{\varphi i} = \cos \varphi + i \sin \varphi$

In the given example we get:

${e}^{\frac{\pi}{2} i} - {e}^{\frac{2 \pi}{3} i} = \left(\cos \left(\frac{\pi}{2}\right) + i \sin \left(\frac{\pi}{2}\right)\right) - \left(\cos \left(\frac{2 \pi}{3}\right) + i \sin \left(\frac{2 \pi}{3}\right)\right) =$

$= i - \left(\cos \left(\pi - \frac{\pi}{3}\right) + i \sin \left(\pi - \frac{\pi}{3}\right)\right) =$

$= i - \left(- \cos \left(\frac{\pi}{3}\right) + i \sin \left(\frac{\pi}{3}\right)\right) =$

$= i + \cos \left(\frac{\pi}{3}\right) - i \sin \left(\frac{\pi}{3}\right) =$

$i + \frac{1}{2} - \frac{\sqrt{3}}{2} \cdot i = \frac{1}{2} + \left(1 - \frac{\sqrt{3}}{2}\right) i$