How do I solve e^z=4 and e^z=1, where z is complex number?

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- Dec 14th 2008, 01:10 PM #1

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- Dec 14th 2008, 01:22 PM #2
write $\displaystyle e^z$ as $\displaystyle e^{x + iy} = e^x \cdot e^{iy}$

now note that you can write $\displaystyle 4 = 4e^{i(2k \pi)}$ and $\displaystyle 1 = e^{i(2k \pi)}$ for some integer $\displaystyle k$.

so your first equation is: $\displaystyle e^x \cdot e^{iy} = 4e^{2k \pi i}$

and your second equation is: $\displaystyle e^x \cdot e^{iy} = e^{2k \pi i}$

can you finish?

- Dec 14th 2008, 01:59 PM #3

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So, here's what I'm having trouble with:

so for e^x+1=0, I knew that if I let e^x = 1 and e^(iy) or (cosy+ isiny) = 2(pi)n, I can have solutions. But how do I know that is the only solutions? That is, how do I know that there aren't some e^x(cosy) =1 that aren't listed by above method? For example, some e^x bigger than 1 and some cosy smaller than 1 and somehow make up the number 1? ... Oh wait... I think I just answered my own question.