# How do you evaluate  e^( ( pi)/2 i) - e^( ( 5 pi)/3 i) using trigonometric functions?

Oct 27, 2016

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

#### Explanation:

The expression is evaluated by applying Euler's formula:
$\textcolor{red}{{e}^{i x} = \cos x + i \sin x}$

We will use the following trigonometric identities:
color(blue)(cos(2pi+alpha)=cosalpha
$\textcolor{b l u e}{\cos \left(- \alpha\right) = \cos \alpha}$
$\textcolor{b l u e}{\sin \left(- \alpha\right) = - \sin \alpha}$
Let us compute ${e}^{\frac{\pi}{2} i} \mathmr{and} {e}^{\frac{5 \pi}{3} i}$ by applying the above formula

${e}^{\frac{\pi}{2} i} = \cos \left(\frac{\pi}{2}\right) + i \sin \left(\frac{\pi}{2}\right)$
$\Rightarrow {e}^{\frac{\pi}{2} i} = 0 + i \left(1\right)$
$\Rightarrow {e}^{\frac{\pi}{2} i} = i$

${e}^{\frac{5 \pi}{3} i} = \cos \left(\frac{5 \pi}{3}\right) + i \sin \left(\frac{5 \pi}{3}\right)$
$\Rightarrow {e}^{\frac{5 \pi}{3} i} = \cos \left(\frac{6 \pi}{3} - \frac{\pi}{3}\right) + i \sin \left(\frac{6 \pi}{3} - \frac{\pi}{3}\right)$
$\Rightarrow {e}^{\frac{5 \pi}{3} i} = \cos \left(2 \pi - \frac{\pi}{3}\right) + i \sin \left(2 \pi - \frac{\pi}{3}\right)$
$\Rightarrow {e}^{\frac{5 \pi}{3} i} = \textcolor{b l u e}{\cos \left(- \frac{\pi}{3}\right) + i \sin \left(- \frac{\pi}{3}\right)}$
rArre^((5pi)/3i)=color(blue)(cos(pi/3)-isin(pi/3)
$\Rightarrow {e}^{\frac{5 \pi}{3} i} = \frac{1}{2} - i \frac{\sqrt{3}}{2}$

${e}^{\frac{\pi}{2} i} - {e}^{\frac{5 \pi}{3} i}$

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

Therefore,

${e}^{\frac{\pi}{2} i} - {e}^{\frac{5 \pi}{3} i} = - \frac{1}{2} + i \left(1 + \frac{\sqrt{3}}{2}\right)$