# Thread: [SOLVED] find analytic continuation

1. ## [SOLVED] find analytic continuation

Hello:

I have this function:

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

whose domain is the line:

$\displaystyle z=(1+i)t$

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

Thank you!

2. Originally Posted by Bop
Hello:

I have this function:

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

whose domain is the line:

$\displaystyle z=(1+i)t$

I need to find the analytic continuation for the whole complex plane. I don't know how to start it.
You need to express $\displaystyle f(z)=\frac{\cos t+i\sin t}{e^{t}}$ as a function of $\displaystyle z = \frac t{1+i}$. Start by using that fact that $\displaystyle \cos t+i\sin t = e^{it}$ and then use the index laws. Finally, substitute $\displaystyle z = \frac t{1+i}$ (and simplify the result).

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

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

as

$\displaystyle e^{it}=cost+isint$

so

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

we know

$\displaystyle t=(1+i)z$

so

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

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!

4. Originally Posted by Bop
Hello, thanks for answering, here what i've done:

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

as

$\displaystyle e^{it}=cost+isint$

so

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

we know

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

so

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

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 $\displaystyle f(z) = e^{iz}$, which certainly is analytic on the whole complex plane, and therefore gives the required analytic continuation for f(z).