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Convert $e^z$ to $a+bi$

I'm having trouble figuring out this very simple problem. Below is my attempt, but can you really have $1/e$ as the modulus of a complex number?

$$z=-1+\frac{i\pi}{4}$$ $$e^{-1+\frac{i\pi}{4}}=e^{-1}e^{\frac{i\pi}{4}}$$ $$=\frac{1}{e}\cos(\pi/4)+i \sin(\pi/4)$$ $$=\frac{1}{e}\left(\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}\right)$$

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  • $\begingroup$ Well, $\frac1e$ is a nonnegative real, so, yes, it can be a modulus. For example, it is its own modulus.... $\endgroup$
    – Cameron Buie
    Commented Feb 11, 2016 at 3:43
  • $\begingroup$ @CameronBuie isn't the real part of any complex number of the form $a+0i$ its own modulus? E.g. $2+0i$, the modulus is just 2. Mainly referring to the last part of your comment. $\endgroup$
    – whatwhatwhat
    Commented Feb 11, 2016 at 3:49
  • $\begingroup$ Nope. That's true only when $a\ge0$. $\endgroup$
    – Cameron Buie
    Commented Feb 11, 2016 at 5:20
  • $\begingroup$ @CameronBuie Ah ok, gotcha. $\endgroup$
    – whatwhatwhat
    Commented Feb 11, 2016 at 5:22

2 Answers 2

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You are right. The answer is correct. Modulus can be any positive value. So, $1\over e$ can surely be it.

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    $\begingroup$ This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review $\endgroup$
    – zz20s
    Commented Feb 11, 2016 at 4:38
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    $\begingroup$ @zz20s Did you read the question? It asks if 1/e can be modulus. I answered that it can be. $\endgroup$
    – Win Vineeth
    Commented Feb 11, 2016 at 4:40
  • $\begingroup$ Well, I mainly want to know how to convert this into Cartesian form (per the title of the question). The part about the modulus is where I got stuck. $\endgroup$
    – whatwhatwhat
    Commented Feb 11, 2016 at 5:23
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    $\begingroup$ @whatwhatwhat Well, You already got the answer. The last line of your question is the answer. That's y I said you are right. To be more precise, just take 1/e inside the brackets. you get your a and b values. $\endgroup$
    – Win Vineeth
    Commented Feb 11, 2016 at 5:25
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Set $e^z$ with $z=a+bi$ with $a,b\in\mathbb{R}$

$$e^{a+bi}=e^a\cdot e^{bi}=e^a\left(\cos(b)+\sin(b)i\right)=e^a\cos(b)+e^a\sin(b)i$$

So, in your problem $z=-1+\frac{\pi}{4}i$:

$$e^{-1+\frac{\pi}{4}i}=e^{-1}\cdot e^{\frac{\pi}{4}i}=$$ $$e^{-1}\left(\cos\left(\frac{\pi}{4}\right)+\sin\left(\frac{\pi}{4}\right)i\right)=\frac{\cos\left(\frac{\pi}{4}\right)}{e}+\frac{\sin\left(\frac{\pi}{4}\right)i}{e}=$$ $$\frac{\frac{1}{\sqrt{2}}}{e}+\frac{\frac{1}{\sqrt{2}}i}{e}=\frac{1}{e\sqrt{2}}+\frac{i}{e\sqrt{2}}$$

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  • $\begingroup$ Isn't all this already explained in the question? $\endgroup$
    – Did
    Commented Feb 11, 2016 at 16:19

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