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

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- December 14th 2008, 02:10 PM #1

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- December 14th 2008, 02:22 PM #2

- December 14th 2008, 02:59 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.