What Are the Solutions for e^z in Complex Numbers?

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Homework Statement


e^z = e
e^z = e^-z


Homework Equations


they are the first 2 questions on my homework and my book is so bad i don't even know how to get the right answers, I've been online for over an hour I'm i'm sure they are easy.

The Attempt at a Solution


ffirst one just guessing ?
z = 1 ?
second in the back of the book says z = i*k*pi k = 1, 2, 3..
 
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What can you say about ex+iy for x and y real numbers?
 
e^x(cos(y) + isin(y)) = e??
i got that ones, i feel like I'm being really dumb and just not seeing something really simple
 
What? That doesn't make any sense. You should review how to calculate ex+iy by splitting it up into exeiy and calculating eiy
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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