What is the trigonometric form of $9 e^{\frac{3 π}{2} i}$ $9 e^( (3pi)/2 i )$ ?

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What is the trigonometric form of $9 e^{\frac{3 π}{2} i}$ $9 e^( (3pi)/2 i )$ ?

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Answer 1 · 2017-11-27T13:40:51

$9 \cdot (cos (\frac{3 π}{2}) + i sin (\frac{3 π}{2}))$ $9*(cos((3pi)/2)+isin((3pi)/2))$

Explanation:

Recall Eulers formula:
$e^{i z} = cos z + i sin z$ $e^(iz)=cosz+isinz$

In this case $z = \frac{3 π}{2}$ $z=(3pi)/2$

Therefore $9 e^{\frac{3 π}{2} i} = 9 \cdot (cos (\frac{3 π}{2}) + i sin (\frac{3 π}{2}))$ $9e^((3pi)/2i)=9*(cos((3pi)/2)+isin((3pi)/2))$

If you wish to simplify this further, $cos (\frac{3 π}{2}) = 0$ $cos((3pi)/2)=0$ , and $sin (\frac{3 π}{2}) = - 1$ $sin((3pi)/2)=-1$ , therefore:

$9 e^{\frac{3 π}{2} i} = 9 \cdot (cos (\frac{3 π}{2}) + i sin (\frac{3 π}{2})) = 9 \cdot (0 + i \cdot (- 1)) = - 9 i$ $9e^((3pi)/2i)=9*(cos((3pi)/2)+isin((3pi)/2))=9*(0+i*(-1))=-9i$

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What is the trigonometric form of $9 e^{\frac{3 π}{2} i}$ $9 e^( (3pi)/2 i )$ ?

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What is the trigonometric form of $9 e^{\frac{3 π}{2} i}$ $9 e^( (3pi)/2 i )$ ?