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

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

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Answer 1 · 2017-10-17T13:40:53

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

Explanation:

$e^{\frac{π}{4} \cdot i}$ $e^(pi/4*i)$

= $cos (\frac{π}{4}) + i \cdot sin (\frac{π}{4})$ $cos(pi/4)+i*sin(pi/4)$

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

$e^{\frac{π}{3} \cdot i}$ $e^(pi/3*i)$

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

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

Hence,

$e^{\frac{π}{4} \cdot i} - e^{\frac{π}{3} \cdot i} = \frac{\sqrt{2} - 1}{2} + i \cdot \frac{\sqrt{2} - \sqrt{3}}{2}$ $e^(pi/4*i)-e^(pi/3*i)=(sqrt(2)-1)/2+i*(sqrt(2)-sqrt(3))/2$

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

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

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