How can you use trigonometric functions to simplify $6 e^{\frac{23 π}{12} i}$ $6 e^( ( 23 pi)/12 i )$ into a non-exponential complex number?

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How can you use trigonometric functions to simplify $6 e^{\frac{23 π}{12} i}$ $6 e^( ( 23 pi)/12 i )$ into a non-exponential complex number?

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Answer 1 · 2016-11-15T13:15:14

$6 e^{i \frac{23 π}{12}} = \frac{3 (\sqrt{6} + \sqrt{2})}{2} - i \frac{3 (\sqrt{6} - \sqrt{2})}{2}$ $6e^(i(23pi)/12)=(3(sqrt6+sqrt2))/2-i(3(sqrt6-sqrt2))/2$

Explanation:

A complex number $a + i b$ $a+ib$ in polar form can be written as $r cos θ + i r sin θ$ $rcostheta+irsintheta$

and using series expansion, it can also be written as $r e^{i θ}$ $re^(itheta)$

Hence $6 e^{i \frac{23 π}{12}} = 6 cos (\frac{23 π}{12}) + i 6 sin (\frac{23 π}{12})$ $6e^(i(23pi)/12)=6cos((23pi)/12)+i6sin((23pi)/12)$

= $6 cos (2 π - \frac{π}{12}) + i 6 sin (2 π - \frac{π}{12})$ $6cos(2pi-pi/12)+i6sin(2pi-pi/12)$

= $6 cos (\frac{π}{12}) - i 6 sin (\frac{π}{12})$ $6cos(pi/12)-i6sin(pi/12)$

= $6 \times \frac{\sqrt{6} + \sqrt{2}}{4} - 6 i \frac{\sqrt{6} - \sqrt{2}}{4}$ $6xx(sqrt6+sqrt2)/4-6i(sqrt6-sqrt2)/4$

= $\frac{3 (\sqrt{6} + \sqrt{2})}{2} - i \frac{3 (\sqrt{6} - \sqrt{2})}{2}$ $(3(sqrt6+sqrt2))/2-i(3(sqrt6-sqrt2))/2$

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How can you use trigonometric functions to simplify $6 e^{\frac{23 π}{12} i}$ $6 e^( ( 23 pi)/12 i )$ into a non-exponential complex number?

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Explanation:

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How can you use trigonometric functions to simplify 6 e^( ( 23 pi)/12 i ) 6e23π12i 6 e^( ( 23 pi)/12 i ) into a non-exponential complex number?

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How can you use trigonometric functions to simplify $6 e^{\frac{23 π}{12} i}$ $6 e^( ( 23 pi)/12 i )$ into a non-exponential complex number?