How do you evaluate $e^{\frac{π}{2} i} - e^{\frac{4 π}{3} i}$ $e^( ( pi)/2 i) - e^( ( 4 pi)/3 i)$ using trigonometric functions?

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How do you evaluate $e^{\frac{π}{2} i} - e^{\frac{4 π}{3} i}$ $e^( ( pi)/2 i) - e^( ( 4 pi)/3 i)$ using trigonometric functions?

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Answer 1 · 2016-10-20T15:51:57

the answer is $= \frac{1}{2} + i (1 + \frac{\sqrt{3}}{2})$ $=1/2+i(1+sqrt3/2)$

Explanation:

Let $z = e^{i \frac{π}{2}} - e^{4 i \frac{π}{3}}$ $z= e^(ipi/2)-e^(4ipi/3)$
Use the relation $e^{i θ} = cos θ + i sin θ$ $e^(itheta)=costheta + isintheta$
So $e^{i \frac{π}{2}} = cos (\frac{π}{2}) + i sin (\frac{π}{2}) = 0 + i = i$ $e^(ipi/2)=cos(pi/2)+isin(pi/2)=0+i=i$
and $e^{4 i \frac{π}{3}} = cos (\frac{4 π}{3}) + i sin (\frac{4 π}{3}) = - \frac{1}{2} - i \frac{\sqrt{3}}{2}$ $e^(4ipi/3)=cos((4pi)/3)+isin((4pi)/3)=-1/2-isqrt3/2$
so $z = i + \frac{1}{2} + i \frac{\sqrt{3}}{2} = \frac{1}{2} + i (1 + \frac{\sqrt{3}}{2})$ $z=i+1/2+isqrt3/2=1/2+i(1+sqrt3/2)$

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How do you evaluate $e^{\frac{π}{2} i} - e^{\frac{4 π}{3} i}$ $e^( ( pi)/2 i) - e^( ( 4 pi)/3 i)$ using trigonometric functions?

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