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Express the following complex number in the standard form $x + iy$

$ie^{\frac{i\pi}{2} +3} $

I have made an attempt and got the answer $\cos(\frac{\pi}{2} +3) +i\sin(\frac{\pi}{2} +3)$. Is this an acceptable result? This is a non-calculator question.

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    $\begingroup$ Recall that $e^{ix}=\cos{x}+i\sin{x}$ so your expansion is incorrect. You have also forgotten about the $i$ multiplying the exponential. $\endgroup$ – user2850514 Dec 10 '14 at 15:59
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    $\begingroup$ $e^{i\frac\pi2+3} = e^3\cdot e^{i\frac\pi2}$, and the $e^3$ term is not a complex exponential. $\endgroup$ – peterwhy Dec 10 '14 at 16:02
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    $\begingroup$ The general approach is to write $e^{a+ib}$ as $e^a e^{ib}$. Then you can throw in the factor of $i$ afterwards. $\endgroup$ – MPW Dec 10 '14 at 16:14
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Here's what I did:

I used the identity $e^{ix}=\sin(x)+i\cos(x)$

I then split the probelm apart into this form: $$ie^{ \frac{i\pi}{2}}e^3$$ Then I applied the identity above: $$ie^3 \left( \cos \left( \frac{\pi}{2} \right) + i \sin \left( \frac{\pi}{2} \right) \right) $$ Evaluating the trig functions, I get: $$i^2e^3$$ And note that $i^2=-1$, so the final form is: $$-e^3$$

Note: if your teacher really wants it in the form $x+iy$, you can make it $-e^3+0i$

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So.. using Eulers identity for complex numbers, we know that for $z=a+ib$:

$e^z=e^a(\text{cos}(b)+i\text{sin}(b))$

So your number becomes:

$ie^{\frac{i\pi}{2}+3}=ie^3\left(\text{cos}(\frac{\pi}{2})+i\text{sin}(\frac{\pi}{2})\right)=e^3\left(i\text{cos}(\frac{\pi}{2})-\text{sin}(\frac{\pi}{2})\right)$

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