Complex exponentials - homework

Poetria
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Could you give me a hint how to attack this problem?

Find a complex number z = a+i*b such that f(t)=Re e^(z*t) where f(t)=cos(2*pi*t)

I have begun as follows:

e^((a+i*b)*t)=e^(a*t)*(cos(b)+i*sin(b))

Re e^(z*t)= e^(a*t)*cos(b)

What to do now?
 
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Hi, ##t## is real? If is yes so you must find ##a,b## from ##e^{at}\cos{b}=cos(2\pi t)## ...
 
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Ssnow said:
Hi, ##t## is real? If is yes so you must find ##a,b## from ##e^{at}\cos{b}=cos(2\pi t)## ...

Yes, t is a real variable.I know I must but how? Thank you very much. :)
 
another hint: ## e^{at}cos{(b)}=e^{0t}cos{(2\pi t)}##, you can read now ##a=...## and ##b=...##
 
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Ssnow said:
another hint: ## e^{at}cos{(b)}=e^{0t}cos{(2\pi t)}##, you can read now ##a=...## and ##b=...##

Ok, I got it. :) Great. Thank you very much. :)
 
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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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