Limit of exp(z) (complex number)

rmcclurk
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Homework Statement



Find limit as z->infinity of exp(z) where z is complex

Homework Equations



See above

The Attempt at a Solution



The solution should be that the limit does not exist, but I don't know why. Any explanations?
 
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Put z=iy, where y is real. Now let y->infinity.
 
Is it because e^iy = r(cos(y)+i*sin(y)) and that equation simply oscillates and never goes to infinity no matter how large y gets?
 
rmcclurk said:
Is it because e^iy = r(cos(y)+i*sin(y)) and that equation simply oscillates and never goes to infinity no matter how large y gets?

Right, if you put r=1. Or consider z=x and z=(-x) for x real and let x->+infinity. One limit is infinity, and the other is zero. There is no definite single limit as z->infinity. Consider definite cases of z->infinity to get a feeling for what's going on.
 
Thanks a lot got it figured out
 
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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