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Write each of the given numbers in the form $a+bi$.

a.) $e^{\frac{-i\pi}{4}}$

b.) ${\frac{e^{1+i3\pi}}{e^{-1+i\pi /2}}}$

c.) $e^{e^i}$

For a, I got $(\frac{\sqrt 2}{2} -\frac{\sqrt 2}{2}i)$, which I think is right. I am stuck on how to do b and c.

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1 Answer 1

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You are right on $a$.

For $(b)$, $$\dfrac{e^{1+3 \pi i}}{e^{-1 + i \pi/2}} = \underbrace{e^{2+ 5 \pi i/2} = e^2 e^{i \pi/2}}_{\text{since }e^{2\pi i+ i \theta} = e^{i \theta}} = e^2 i$$

For $(c)$, note that $e^{i} = \cos(1) + i \sin(1)$. $$e^{e^i} = e^{\cos(1) + i\sin(1)} = e^{\cos(1)} e^{i\sin(1)} = e^{\cos(1)}\left(\cos(\sin(1)) + i \sin(\sin(1))\right)$$

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Thanks! I understand how to do c now but I am confused on part b. How did you $e^{2+ 5 \pi i/2} = e^2 e^{i \pi /2}$. I understand how you got $e^{2+ 5 \pi i/2}$ but not $e^2 e^{i \pi/2}$. –  Q.matin Jan 20 '13 at 3:57
    
@Q.matin As I mentioned in my post, we have $$e^{2+5 \pi i/2} = e^{2}e^{5 \pi i/2} = \underbrace{e^{2}e^{2 \pi i + \pi i/2} = e^2 e^{i \pi/2}}_{e^{2 \pi i + i \theta} = e^{i \theta}}$$ I used the fact that $$e^{2 \pi i + i \theta} = e^{i \theta}$$ since $e^{2 \pi i} = 1$. –  user17762 Jan 20 '13 at 3:59
    
Oh, so $e^{2 \pi i + \pi i/2}$ is an identity equal to $e^{i\theta}$? –  Q.matin Jan 20 '13 at 4:02
    
@Q.matin The main thing is $e^{2 k \pi i} = 1$ when $k$ is an integer and use the fact that $e^{a+b} = e^a \cdot e^b$. Hence, $e^{2 \pi i + i \theta} = e^{2 \pi i} \cdot e^{i \theta} =1 \cdot e^{i \theta} = e^{i \theta}$. –  user17762 Jan 20 '13 at 4:03
    
Thanks a lot, Marvis! –  Q.matin Jan 20 '13 at 4:06

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