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Math Help - [SOLVED] find analytic continuation

  1. #1
    Bop
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    [SOLVED] find analytic continuation

    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!
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  2. #2
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    Quote Originally Posted by Bop View Post
    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).
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  3. #3
    Bop
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    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!
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  4. #4
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    Quote Originally Posted by Bop View Post
    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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