[SOLVED] find analytic continuation

**Bop** · May 13th 2010, 05:21 PM

Hello:

I have this function:

$f(z)=\frac{cos(t)+isin(t)}{e^{t}}$

whose domain is the line:

$z=(1+i)t$

I need to find the analytic continuation for the whole complex plane. I don't know how to start it.

Thank you!

**Opalg** · May 14th 2010, 12:05 AM

Originally Posted by Bop

Hello:

I have this function:

$f(z)=\frac{\cos t+i\sin t}{e^{t}}$

whose domain is the line:

$z=(1+i)t$

I need to find the analytic continuation for the whole complex plane. I don't know how to start it.

You need to express $f(z)=\frac{\cos t+i\sin t}{e^{t}}$ as a function of $z = \frac t{1+i}$ . Start by using that fact that $\cos t+i\sin t = e^{it}$ and then use the index laws. Finally, substitute $z = \frac t{1+i}$ (and simplify the result).

**Bop** · May 14th 2010, 06:57 AM

Hello, thanks for answering, here what i've done:

$f(z)=\frac{cos(t)+isin(t)}{e^{t}}$

as

$e^{it}=cost+isint$

so

$f(z)=\frac{cos(t)+isin(t)}{e^{t}}=\frac{e^{it}}{e^ t}=e^{it-t}=e^{t(i-1)}$

we know

$t=(1+i)z$

so

$f(z)=e^{t(i-1)}=e^{-2z}$

This is what you meant? If it is, I don't understand why we have to do this to obtain the analytic continuation for the whole complex plane.

Thank you!

**Opalg** · May 14th 2010, 07:33 AM

Originally Posted by Bop

Hello, thanks for answering, here what i've done:

$f(z)=\frac{cos(t)+isin(t)}{e^{t}}$

as

$e^{it}=cost+isint$

so

$f(z)=\frac{cos(t)+isin(t)}{e^{t}}=\frac{e^{it}}{e^ t}=e^{it-t}=e^{t(i-1)}$

we know

$t=(1+i)z$ Wrong way round! You are told that $\color{red}z = (1+i)t$ .

so

$f(z)=e^{t(i-1)}=e^{-2z}$ In fact, $\color{red}t(i-1) = \frac{z(i-1)}{1+i} = iz$ .

This is what you meant? If it is, I don't understand why we have to do this to obtain the analytic continuation for the whole complex plane.

Yes, that is what I meant (apart from the mistake at the end). The reason for doing it that way is that if you want f to be an analytic function of z then you have to express it in terms of z. Fortunately, you end up with the expression $f(z) = e^{iz}$ , which certainly is analytic on the whole complex plane, and therefore gives the required analytic continuation for f(z).

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