How do you evaluate $e^{\frac{π}{4} i} - e^{\frac{35 π}{4} i}$ $e^( ( pi)/4 i) - e^( ( 35 pi)/4 i)$ using trigonometric functions?

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How do you evaluate $e^{\frac{π}{4} i} - e^{\frac{35 π}{4} i}$ $e^( ( pi)/4 i) - e^( ( 35 pi)/4 i)$ using trigonometric functions?

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Answer 1 · 2016-03-28T09:18:34

$e^{\frac{π}{4} i} - e^{\frac{35 π}{4} i} = \sqrt{2}$ $e^((pi)/4i)-e^((35pi)/4i)=sqrt2$

Explanation:

As $e^{i θ} = cos θ + i sin θ$ $e^(itheta)=costheta+isintheta$ , we have

$e^{\frac{π}{4} i} = cos (\frac{π}{4}) + i sin (\frac{π}{4}) = \frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}}$ $e^((pi)/4i)=cos((pi)/4)+isin((pi)/4)=1/sqrt2+i1/sqrt2$

$e^{\frac{35 π}{4} i} = cos (\frac{35 π}{4}) + i sin (\frac{35 π}{4})$ $e^((35pi)/4i)=cos((35pi)/4)+isin((35pi)/4)$

= $cos (8 π + \frac{3 π}{4}) + i sin (8 π + \frac{3 π}{4})$ $cos(8pi+(3pi)/4)+isin(8pi+(3pi)/4)$

= $cos (\frac{3 π}{4}) + i sin (\frac{3 π}{4})$ $cos((3pi)/4)+isin((3pi)/4)$

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

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

= $- \frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}}$ $-1/sqrt2+i1/sqrt2$

Hence $e^{\frac{π}{4} i} - e^{\frac{35 π}{4} i} = (\frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}}) - (- \frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}})$ $e^((pi)/4i)-e^((35pi)/4i)=(1/sqrt2+i1/sqrt2)-(-1/sqrt2+i1/sqrt2)$

= $(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) + i (\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}})$ $(1/sqrt2+1/sqrt2)+i(1/sqrt2-1/sqrt2)$

= $\frac{2}{\sqrt{2}} + 0 = \sqrt{2}$ $2/sqrt2+0=sqrt2$

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How do you evaluate $e^{\frac{π}{4} i} - e^{\frac{35 π}{4} i}$ $e^( ( pi)/4 i) - e^( ( 35 pi)/4 i)$ using trigonometric functions?

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Explanation:

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