How do you evaluate $e^{\frac{π}{2} i} - e^{\frac{23 π}{3} i}$ $e^( ( pi)/2 i) - e^( ( 23 pi)/3 i)$ using trigonometric functions?

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How do you evaluate $e^{\frac{π}{2} i} - e^{\frac{23 π}{3} i}$ $e^( ( pi)/2 i) - e^( ( 23 pi)/3 i)$ using trigonometric functions?

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Answer 1 · 2016-12-17T19:47:11

With the help of Euler's formula we can take this right down to
0.5 + $i$ $i$ (1 + $\frac{\sqrt{3}}{2}$ $(sqrt3)/2$ )

Explanation:

Start with Euler's formula: $e^{i x}$ $e^(ix)$ = $cos x + i sin x$ $cos x + i Sin x$

Your expression becomes:

cos $\frac{π}{2}$ $pi/2$ + $i$ $i$ sin $\frac{π}{2}$ $pi/2$ - cos $\frac{23 π}{3}$ $(23pi)/3$ - $i$ $i$ sin $\frac{23 π}{3}$ $(23pi)/3$

To simplify, note cos $\frac{π}{2}$ $pi/2$ = 0, sin $\frac{π}{2}$ $pi/2$ = 1

and since each complete revolution of a unit circle is 2 $π$ $pi$ , which we will write as $\frac{6 π}{3}$ $(6pi)/3$ , cos $\frac{23 π}{3}$ $(23pi)/3$ = cos $\frac{5 π}{3}$ $(5pi)/3$
and likewise for sin $\frac{23 π}{3}$ $(23pi)/3$ = sin $\frac{5 π}{3}$ $(5pi)/3$

So, we now have the original expression simplified down to

$i$ $i$ - cos $\frac{5 π}{3}$ $(5pi)/3$ - $i$ $i$ sin $\frac{5 π}{3}$ $(5pi)/3$

where cos $\frac{5 π}{3}$ $(5pi)/3$ = 0.5 and sin $\frac{5 π}{3}$ $(5pi)/3$ = - $(\frac{\sqrt{3}}{2})$ $(sqrt3/2)$

So, finally, the result is

0.5 + $i$ $i$ (1 + $\frac{\sqrt{3}}{2}$ $(sqrt3)/2$ )

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How do you evaluate $e^{\frac{π}{2} i} - e^{\frac{23 π}{3} i}$ $e^( ( pi)/2 i) - e^( ( 23 pi)/3 i)$ using trigonometric functions?

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