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Find all solutions for $e^z = -2$. And we know there is no solution.

My trivial work:

Since if we take $\ln$ both sides, we simply have $z = \ln(-2)$ which isn't defined.

But, since my work looks very trivial, I want to add more.

Thank you.

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3  
there is no integer solution, but there are complex solutions. –  DanielY Sep 16 '13 at 6:10
    
I have to assume that it has no solution and prove why. –  therexists Sep 16 '13 at 6:12
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Considering the tag is complex analysis, it very well does have a solution in the complex domain –  Anthony Peter Sep 16 '13 at 6:14
    
then what you said is enough, assuming it's homework –  DanielY Sep 16 '13 at 6:14
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Please read all the comments on this page. The answer is you accepted may not lead you to learn the heart of the matter here. –  Vishal Sep 16 '13 at 6:45

3 Answers 3

up vote -2 down vote accepted

$e^z = -2$

$z = \ln(-2)$

Now since $e^{i\pi} = -1$, $\ln(-1)=i\pi$, and since $\log_z a + \log_z b = \log_z(ab)$,

$\ln(-2) = \ln(-1) + \ln(2)$

$\ln(-2) = i\pi + \ln(2)$

$z = i\pi +\ln(2)$

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1  
$e^{i\pi} = -1$ does not imply that $\ln(-1) = i\pi$. –  Vishal Sep 16 '13 at 6:34
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@Piman: You're incorrect. Do you think $e^{2\pi i}=1$ implies that $\ln(1)=2\pi i$? –  Zev Chonoles Sep 16 '13 at 6:40
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The logarithm in complex domain is more subtle than its real counterpart. First of all, we have to specify its domain and then on that domain too, the function is not injective. So we must be wary of complex logarithm. See the two solutions below for the "correct" approach. –  Vishal Sep 16 '13 at 6:44
    
The definition of a logarithm is that when $a^x = y$ then $\log_ya=x$, since $e^{iπ}=-1$ $\ln(-1) = iπ$ –  Yan Yau Sep 23 '13 at 5:38

By taking the natural log of both sides, one obtains $$ z = \log(2) + i\pi(2n+1)$$ $\forall n \in \mathbb{Z}$
Try to see if you can figure out why this is so.

If it truly is for a complex analysis course the following will be helpful:
$e^z = e^{x+iy} = e^xe^{iy} = e^x(\cos(y) + i\sin(y))$

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There is a solution. I won't give it away, but I'll give you a hint:

$e^{i\pi} = -1$

So $e^z = -1$ has a solution. Maybe you can jump from here to the solution for $e^z = -2$.

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