How do you evaluate $e^{\frac{π}{4} i} - e^{\frac{7 π}{3} i}$ $e^( ( pi)/4 i) - e^( ( 7 pi)/3 i)$ using trigonometric functions?

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How do you evaluate $e^{\frac{π}{4} i} - e^{\frac{7 π}{3} i}$ $e^( ( pi)/4 i) - e^( ( 7 pi)/3 i)$ using trigonometric functions?

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Answer 1 · 2018-07-13T19:55:46

Euler's Formula states that

$e^{i θ} = cos θ + i sin θ$ $e^(icolor(red)theta) = coscolor(red)theta+isincolor(red)theta$

Which is the trigonometric form of a complex number. Hence,

$e^{i \frac{π}{4}} = cos (\frac{π}{4}) + i sin (\frac{π}{4}) = \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2}$ $e^(icolor(red)(pi/4))=cos(color(red)(pi/4)) +isin(color(red)(pi/4))=sqrt2/2+isqrt2/2$

$e^{i \frac{7 π}{3}} = cos (\frac{7 π}{3}) + i sin (\frac{7 π}{3})$ $e^(icolor(red)((7pi)/3))=cos(color(red)((7pi)/3)) + isin(color(red)((7pi)/3))$
$= cos (2 π + \frac{π}{3}) + i sin (2 π + \frac{π}{3})$ $=cos(2pi+pi/3)+isin(2pi+pi/3)$
$= cos (\frac{π}{3}) + i sin (\frac{π}{3})$ $=cos(pi/3) + i sin(pi/3)$
$= \frac{1}{2} + i \frac{\sqrt{3}}{2}$ $=1/2+isqrt3/2$

$:. e^(ipi/4) - e^(i(7pi)/3)=sqrt2/2+isqrt2/2-1/2-isqrt3/2$

$=(sqrt2/2-1/2)+i(sqrt2/2-sqrt3/2)$

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How do you evaluate $e^{\frac{π}{4} i} - e^{\frac{7 π}{3} i}$ $e^( ( pi)/4 i) - e^( ( 7 pi)/3 i)$ using trigonometric functions?

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