Finding $z$ for a Complex Equation

fauboca
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How do I find all $z$ that satisfies:

z = \exp\left(2+3i\right)

I know the modulus has to be e^2 and the argument has to be 3 but where do I go from there?
 
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That's it. There's only one z equal to that z.
 
LCKurtz said:
That's it. There's only one z equal to that z.

That is really all there is to it?
 
fauboca said:
That is really all there is to it?

Maybe you are supposed to write it in a+bi form?
 
LCKurtz said:
Maybe you are supposed to write it in a+bi form?

Yes that is correct.
 
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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