How can you use trigonometric functions to simplify $16 e^{\frac{7 π}{6} i}$ $16 e^( ( 7 pi)/6 i )$ into a non-exponential complex number?

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How can you use trigonometric functions to simplify $16 e^{\frac{7 π}{6} i}$ $16 e^( ( 7 pi)/6 i )$ into a non-exponential complex number?

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Answer 1 · 2016-07-29T13:44:34

I used the exponential converion relationship:

Explanation:

You may remember that your complex number can be written as:
$z = ρ e^{θ i} = ρ [cos (θ) + i sin (θ)]$ $z=rhoe^(thetai)=rho[cos(theta)+isin(theta)]$
in your case:
$z = 16 e^{\frac{7}{6} π i} = 16 [cos (\frac{7}{6} π) + i sin (\frac{7}{6} π)]$ $z=16e^(7/6pii)=16[cos(7/6pi)+isin(7/6pi)]$
we can now evaluate the above trig functions and multiply by $16$ $16$ to get to the form: $z = a + i b$ $z=a+ib$ :

$z = - 13.8 - 8 i$ $z=-13.8-8i$

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How can you use trigonometric functions to simplify $16 e^{\frac{7 π}{6} i}$ $16 e^( ( 7 pi)/6 i )$ into a non-exponential complex number?

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Explanation:

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Science

Math

Humanities

... and beyond

How can you use trigonometric functions to simplify 16e7π6i 16 e^( ( 7 pi)/6 i ) into a non-exponential complex number?

1 Answer

Explanation:

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Impact of this question

How can you use trigonometric functions to simplify $16 e^{\frac{7 π}{6} i}$ $16 e^( ( 7 pi)/6 i )$ into a non-exponential complex number?