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

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

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Answer 1 · 2017-01-10T21:57:14

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

Explanation:

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

$e^{φ i} = cos φ + i sin φ$ $e^(varphii)=cos varphi+ isinvarphi$

In the given example we get:

$e^{\frac{π}{2} i} - e^{\frac{2 π}{3} i} = (cos (\frac{π}{2}) + i sin (\frac{π}{2})) - (cos (\frac{2 π}{3}) + i sin (\frac{2 π}{3})) =$ $e^(pi/2i)-e^((2pi)/3i)=(cos(pi/2)+isin(pi/2))-(cos((2pi)/3)+isin((2pi)/3))=$

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

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

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

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

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

1 Answer

Explanation:

$e^{φ i} = cos φ + i sin φ$ $e^(varphii)=cos varphi+ isinvarphi$

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How do you evaluate eπ2i−e2π3i e^( ( pi)/2 i) - e^( ( 2 pi)/3 i) using trigonometric functions?

1 Answer

Explanation:

eφi=cosφ+isinφe^(varphii)=cos varphi+ isinvarphi

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

$e^{φ i} = cos φ + i sin φ$ $e^(varphii)=cos varphi+ isinvarphi$