Determine if $f(z) = log(e^z+1)$ is analytic and where












2












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I tried to substitute $z$ with $x+iy$ and then write it down as $f(z) = log(e^xcis(iy)+1) $ but it looks worse than the starting point. on original branch






complex-analysis analyticity







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Analytic where? The composition of analytic functions is analytic. But you have to be careful with $ln$: what branch are you using?
    $endgroup$
    – Robert Israel
    Jan 2 at 20:16












  • $begingroup$
    @RobertIsrael Actually it's a warmup problem for my next week exam, and i'm hell of a newbie to this section, so i think the point is to determine where is analytical area of this function.
    $endgroup$
    – no0ob
    Jan 2 at 20:20












  • $begingroup$
    @RobertIsrael after reading my note i think your answer is $-pi lt theta le pi$
    $endgroup$
    – no0ob
    Jan 2 at 20:29








  • 1




    $begingroup$
    Look at previous comment. $u=ln(r), v=theta$.
    $endgroup$
    – herb steinberg
    Jan 3 at 2:20






  • 1




    $begingroup$
    $theta$ has two possible values. To get the correct one, use $sintheta = frac{e^xsiny}{r}$ and $costheta=frac{e^xcosy+1}{r}$..
    $endgroup$
    – herb steinberg
    Jan 3 at 2:31
















2












$begingroup$


I tried to substitute $z$ with $x+iy$ and then write it down as $f(z) = log(e^xcis(iy)+1) $ but it looks worse than the starting point. on original branch






complex-analysis analyticity







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Analytic where? The composition of analytic functions is analytic. But you have to be careful with $ln$: what branch are you using?
    $endgroup$
    – Robert Israel
    Jan 2 at 20:16












  • $begingroup$
    @RobertIsrael Actually it's a warmup problem for my next week exam, and i'm hell of a newbie to this section, so i think the point is to determine where is analytical area of this function.
    $endgroup$
    – no0ob
    Jan 2 at 20:20












  • $begingroup$
    @RobertIsrael after reading my note i think your answer is $-pi lt theta le pi$
    $endgroup$
    – no0ob
    Jan 2 at 20:29








  • 1




    $begingroup$
    Look at previous comment. $u=ln(r), v=theta$.
    $endgroup$
    – herb steinberg
    Jan 3 at 2:20






  • 1




    $begingroup$
    $theta$ has two possible values. To get the correct one, use $sintheta = frac{e^xsiny}{r}$ and $costheta=frac{e^xcosy+1}{r}$..
    $endgroup$
    – herb steinberg
    Jan 3 at 2:31














2












2








2





$begingroup$


I tried to substitute $z$ with $x+iy$ and then write it down as $f(z) = log(e^xcis(iy)+1) $ but it looks worse than the starting point. on original branch






complex-analysis analyticity







share|cite|improve this question











$endgroup$




I tried to substitute $z$ with $x+iy$ and then write it down as $f(z) = log(e^xcis(iy)+1) $ but it looks worse than the starting point. on original branch






complex-analysis analyticity




complex-analysis analyticity






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 2 at 21:33







no0ob

















asked Jan 2 at 20:07









no0obno0ob

788




788








  • 1




    $begingroup$
    Analytic where? The composition of analytic functions is analytic. But you have to be careful with $ln$: what branch are you using?
    $endgroup$
    – Robert Israel
    Jan 2 at 20:16












  • $begingroup$
    @RobertIsrael Actually it's a warmup problem for my next week exam, and i'm hell of a newbie to this section, so i think the point is to determine where is analytical area of this function.
    $endgroup$
    – no0ob
    Jan 2 at 20:20












  • $begingroup$
    @RobertIsrael after reading my note i think your answer is $-pi lt theta le pi$
    $endgroup$
    – no0ob
    Jan 2 at 20:29








  • 1




    $begingroup$
    Look at previous comment. $u=ln(r), v=theta$.
    $endgroup$
    – herb steinberg
    Jan 3 at 2:20






  • 1




    $begingroup$
    $theta$ has two possible values. To get the correct one, use $sintheta = frac{e^xsiny}{r}$ and $costheta=frac{e^xcosy+1}{r}$..
    $endgroup$
    – herb steinberg
    Jan 3 at 2:31














  • 1




    $begingroup$
    Analytic where? The composition of analytic functions is analytic. But you have to be careful with $ln$: what branch are you using?
    $endgroup$
    – Robert Israel
    Jan 2 at 20:16












  • $begingroup$
    @RobertIsrael Actually it's a warmup problem for my next week exam, and i'm hell of a newbie to this section, so i think the point is to determine where is analytical area of this function.
    $endgroup$
    – no0ob
    Jan 2 at 20:20












  • $begingroup$
    @RobertIsrael after reading my note i think your answer is $-pi lt theta le pi$
    $endgroup$
    – no0ob
    Jan 2 at 20:29








  • 1




    $begingroup$
    Look at previous comment. $u=ln(r), v=theta$.
    $endgroup$
    – herb steinberg
    Jan 3 at 2:20






  • 1




    $begingroup$
    $theta$ has two possible values. To get the correct one, use $sintheta = frac{e^xsiny}{r}$ and $costheta=frac{e^xcosy+1}{r}$..
    $endgroup$
    – herb steinberg
    Jan 3 at 2:31








1




1




$begingroup$
Analytic where? The composition of analytic functions is analytic. But you have to be careful with $ln$: what branch are you using?
$endgroup$
– Robert Israel
Jan 2 at 20:16






$begingroup$
Analytic where? The composition of analytic functions is analytic. But you have to be careful with $ln$: what branch are you using?
$endgroup$
– Robert Israel
Jan 2 at 20:16














$begingroup$
@RobertIsrael Actually it's a warmup problem for my next week exam, and i'm hell of a newbie to this section, so i think the point is to determine where is analytical area of this function.
$endgroup$
– no0ob
Jan 2 at 20:20






$begingroup$
@RobertIsrael Actually it's a warmup problem for my next week exam, and i'm hell of a newbie to this section, so i think the point is to determine where is analytical area of this function.
$endgroup$
– no0ob
Jan 2 at 20:20














$begingroup$
@RobertIsrael after reading my note i think your answer is $-pi lt theta le pi$
$endgroup$
– no0ob
Jan 2 at 20:29






$begingroup$
@RobertIsrael after reading my note i think your answer is $-pi lt theta le pi$
$endgroup$
– no0ob
Jan 2 at 20:29






1




1




$begingroup$
Look at previous comment. $u=ln(r), v=theta$.
$endgroup$
– herb steinberg
Jan 3 at 2:20




$begingroup$
Look at previous comment. $u=ln(r), v=theta$.
$endgroup$
– herb steinberg
Jan 3 at 2:20




1




1




$begingroup$
$theta$ has two possible values. To get the correct one, use $sintheta = frac{e^xsiny}{r}$ and $costheta=frac{e^xcosy+1}{r}$..
$endgroup$
– herb steinberg
Jan 3 at 2:31




$begingroup$
$theta$ has two possible values. To get the correct one, use $sintheta = frac{e^xsiny}{r}$ and $costheta=frac{e^xcosy+1}{r}$..
$endgroup$
– herb steinberg
Jan 3 at 2:31










3 Answers
3






active

oldest

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$begingroup$

You're using the branch of logarithm that is analytic on the complement of the left half-line $B = (-infty, 0]$. So your function is analytic at $z$ as long as $e^z+1 notin B$, i.e. $e^z notin (-infty, -1]$.



Now if $z = x + i y$ with $x, y$ real, $e^z = e^x e^{i y}$. For this to be in $B$, you need $y = (2n+1) pi$ with for integer $n$ (so that $e^{iy} = -1$) and $x ge 0$ (so $|e^z| = e^x ge 1$). Thus your function is analytic on the complement of
the half-lines $y = (2n+1) pi$, $x ge 0$.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Summary of previous comments plus.



    Convert $e^z+1$ to a form $re^{itheta}$ Then $ln$ will be $ln(r)+itheta=u+iv$ for Cauchy-Riemann equations to check analytics. $e^z+1=e^xcosy+1+ie^xsiny$. Therefore $r^2=(e^xcosy+1)^2+(e^xsiny)^2$, $theta=arctan(frac{siny}{cosy+e^{−x}})$.



    For Cauchy-Riemann you could use $u=frac{ln(r^2)}{2}$. Also note that although $theta$ has two possible values, they differ by a constant ($pi$), so it doesn't effect the analysis.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I can't find relation between $r$ and $theta$ so i could write function as $u(r,theta) + iv(r,theta)$
      $endgroup$
      – no0ob
      Jan 3 at 12:41










    • $begingroup$
      $ln(e^z+1)=ln(r)+itheta$. Therefore $u=ln(r)=frac{ln(r^2)}{2}$ and $v=theta$
      $endgroup$
      – herb steinberg
      Jan 3 at 17:02










    • $begingroup$
      Aha Cauchy-Reimann is always correct for the $ln(r)+itheta$ now we have to check where does $ln(r)$ available for $r^2 = (e^xcos y+1)^2+(e^xsin y)^2$ and check domain of $arctan(frac{sin(y)}{cos(y)+e^{-x}}) $ and we are OK right?
      $endgroup$
      – no0ob
      Jan 3 at 19:58












    • $begingroup$
      I mean $ln(r)+itheta$ is analytic everywhere except $0$ and negative $x$ axis
      $endgroup$
      – no0ob
      Jan 3 at 20:05










    • $begingroup$
      This ends up as $z=x+ipi$ with $xle 0$.
      $endgroup$
      – herb steinberg
      Jan 3 at 21:12



















    0












    $begingroup$

    According to the determination of logarythm you saw, you should know on which domain the complex logarithm is holomorphic.
    The question would be : can the function exp(z) + 1 go outside of this domain ?






    share|cite|improve this answer









    $endgroup$









    • 1




      $begingroup$
      actually i better asked determine analytical area.
      $endgroup$
      – no0ob
      Jan 2 at 20:22










    • $begingroup$
      Then the question would be the following: for which z does exp(z) + 1 belong to the analytical area of the logarithm?
      $endgroup$
      – Cauchy is my master
      Jan 2 at 20:24










    • $begingroup$
      I think that's correct, and ln(z) is analytic in all of the plane but $xle0$ right? for the original branch
      $endgroup$
      – no0ob
      Jan 2 at 20:31






    • 1




      $begingroup$
      That depends on the determination of argument you chose, but I think you chose the usual one, so it's correct.
      $endgroup$
      – Cauchy is my master
      Jan 2 at 20:44










    • $begingroup$
      so now i should use function composition or sth? or is it possible to do it with the Cauchy-Reimann equation?
      $endgroup$
      – no0ob
      Jan 2 at 20:51












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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    You're using the branch of logarithm that is analytic on the complement of the left half-line $B = (-infty, 0]$. So your function is analytic at $z$ as long as $e^z+1 notin B$, i.e. $e^z notin (-infty, -1]$.



    Now if $z = x + i y$ with $x, y$ real, $e^z = e^x e^{i y}$. For this to be in $B$, you need $y = (2n+1) pi$ with for integer $n$ (so that $e^{iy} = -1$) and $x ge 0$ (so $|e^z| = e^x ge 1$). Thus your function is analytic on the complement of
    the half-lines $y = (2n+1) pi$, $x ge 0$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      You're using the branch of logarithm that is analytic on the complement of the left half-line $B = (-infty, 0]$. So your function is analytic at $z$ as long as $e^z+1 notin B$, i.e. $e^z notin (-infty, -1]$.



      Now if $z = x + i y$ with $x, y$ real, $e^z = e^x e^{i y}$. For this to be in $B$, you need $y = (2n+1) pi$ with for integer $n$ (so that $e^{iy} = -1$) and $x ge 0$ (so $|e^z| = e^x ge 1$). Thus your function is analytic on the complement of
      the half-lines $y = (2n+1) pi$, $x ge 0$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        You're using the branch of logarithm that is analytic on the complement of the left half-line $B = (-infty, 0]$. So your function is analytic at $z$ as long as $e^z+1 notin B$, i.e. $e^z notin (-infty, -1]$.



        Now if $z = x + i y$ with $x, y$ real, $e^z = e^x e^{i y}$. For this to be in $B$, you need $y = (2n+1) pi$ with for integer $n$ (so that $e^{iy} = -1$) and $x ge 0$ (so $|e^z| = e^x ge 1$). Thus your function is analytic on the complement of
        the half-lines $y = (2n+1) pi$, $x ge 0$.






        share|cite|improve this answer









        $endgroup$



        You're using the branch of logarithm that is analytic on the complement of the left half-line $B = (-infty, 0]$. So your function is analytic at $z$ as long as $e^z+1 notin B$, i.e. $e^z notin (-infty, -1]$.



        Now if $z = x + i y$ with $x, y$ real, $e^z = e^x e^{i y}$. For this to be in $B$, you need $y = (2n+1) pi$ with for integer $n$ (so that $e^{iy} = -1$) and $x ge 0$ (so $|e^z| = e^x ge 1$). Thus your function is analytic on the complement of
        the half-lines $y = (2n+1) pi$, $x ge 0$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 3 at 4:42









        Robert IsraelRobert Israel

        330k23218473




        330k23218473























            1












            $begingroup$

            Summary of previous comments plus.



            Convert $e^z+1$ to a form $re^{itheta}$ Then $ln$ will be $ln(r)+itheta=u+iv$ for Cauchy-Riemann equations to check analytics. $e^z+1=e^xcosy+1+ie^xsiny$. Therefore $r^2=(e^xcosy+1)^2+(e^xsiny)^2$, $theta=arctan(frac{siny}{cosy+e^{−x}})$.



            For Cauchy-Riemann you could use $u=frac{ln(r^2)}{2}$. Also note that although $theta$ has two possible values, they differ by a constant ($pi$), so it doesn't effect the analysis.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              I can't find relation between $r$ and $theta$ so i could write function as $u(r,theta) + iv(r,theta)$
              $endgroup$
              – no0ob
              Jan 3 at 12:41










            • $begingroup$
              $ln(e^z+1)=ln(r)+itheta$. Therefore $u=ln(r)=frac{ln(r^2)}{2}$ and $v=theta$
              $endgroup$
              – herb steinberg
              Jan 3 at 17:02










            • $begingroup$
              Aha Cauchy-Reimann is always correct for the $ln(r)+itheta$ now we have to check where does $ln(r)$ available for $r^2 = (e^xcos y+1)^2+(e^xsin y)^2$ and check domain of $arctan(frac{sin(y)}{cos(y)+e^{-x}}) $ and we are OK right?
              $endgroup$
              – no0ob
              Jan 3 at 19:58












            • $begingroup$
              I mean $ln(r)+itheta$ is analytic everywhere except $0$ and negative $x$ axis
              $endgroup$
              – no0ob
              Jan 3 at 20:05










            • $begingroup$
              This ends up as $z=x+ipi$ with $xle 0$.
              $endgroup$
              – herb steinberg
              Jan 3 at 21:12
















            1












            $begingroup$

            Summary of previous comments plus.



            Convert $e^z+1$ to a form $re^{itheta}$ Then $ln$ will be $ln(r)+itheta=u+iv$ for Cauchy-Riemann equations to check analytics. $e^z+1=e^xcosy+1+ie^xsiny$. Therefore $r^2=(e^xcosy+1)^2+(e^xsiny)^2$, $theta=arctan(frac{siny}{cosy+e^{−x}})$.



            For Cauchy-Riemann you could use $u=frac{ln(r^2)}{2}$. Also note that although $theta$ has two possible values, they differ by a constant ($pi$), so it doesn't effect the analysis.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              I can't find relation between $r$ and $theta$ so i could write function as $u(r,theta) + iv(r,theta)$
              $endgroup$
              – no0ob
              Jan 3 at 12:41










            • $begingroup$
              $ln(e^z+1)=ln(r)+itheta$. Therefore $u=ln(r)=frac{ln(r^2)}{2}$ and $v=theta$
              $endgroup$
              – herb steinberg
              Jan 3 at 17:02










            • $begingroup$
              Aha Cauchy-Reimann is always correct for the $ln(r)+itheta$ now we have to check where does $ln(r)$ available for $r^2 = (e^xcos y+1)^2+(e^xsin y)^2$ and check domain of $arctan(frac{sin(y)}{cos(y)+e^{-x}}) $ and we are OK right?
              $endgroup$
              – no0ob
              Jan 3 at 19:58












            • $begingroup$
              I mean $ln(r)+itheta$ is analytic everywhere except $0$ and negative $x$ axis
              $endgroup$
              – no0ob
              Jan 3 at 20:05










            • $begingroup$
              This ends up as $z=x+ipi$ with $xle 0$.
              $endgroup$
              – herb steinberg
              Jan 3 at 21:12














            1












            1








            1





            $begingroup$

            Summary of previous comments plus.



            Convert $e^z+1$ to a form $re^{itheta}$ Then $ln$ will be $ln(r)+itheta=u+iv$ for Cauchy-Riemann equations to check analytics. $e^z+1=e^xcosy+1+ie^xsiny$. Therefore $r^2=(e^xcosy+1)^2+(e^xsiny)^2$, $theta=arctan(frac{siny}{cosy+e^{−x}})$.



            For Cauchy-Riemann you could use $u=frac{ln(r^2)}{2}$. Also note that although $theta$ has two possible values, they differ by a constant ($pi$), so it doesn't effect the analysis.






            share|cite|improve this answer











            $endgroup$



            Summary of previous comments plus.



            Convert $e^z+1$ to a form $re^{itheta}$ Then $ln$ will be $ln(r)+itheta=u+iv$ for Cauchy-Riemann equations to check analytics. $e^z+1=e^xcosy+1+ie^xsiny$. Therefore $r^2=(e^xcosy+1)^2+(e^xsiny)^2$, $theta=arctan(frac{siny}{cosy+e^{−x}})$.



            For Cauchy-Riemann you could use $u=frac{ln(r^2)}{2}$. Also note that although $theta$ has two possible values, they differ by a constant ($pi$), so it doesn't effect the analysis.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 3 at 17:06

























            answered Jan 3 at 4:35









            herb steinbergherb steinberg

            3,0782311




            3,0782311












            • $begingroup$
              I can't find relation between $r$ and $theta$ so i could write function as $u(r,theta) + iv(r,theta)$
              $endgroup$
              – no0ob
              Jan 3 at 12:41










            • $begingroup$
              $ln(e^z+1)=ln(r)+itheta$. Therefore $u=ln(r)=frac{ln(r^2)}{2}$ and $v=theta$
              $endgroup$
              – herb steinberg
              Jan 3 at 17:02










            • $begingroup$
              Aha Cauchy-Reimann is always correct for the $ln(r)+itheta$ now we have to check where does $ln(r)$ available for $r^2 = (e^xcos y+1)^2+(e^xsin y)^2$ and check domain of $arctan(frac{sin(y)}{cos(y)+e^{-x}}) $ and we are OK right?
              $endgroup$
              – no0ob
              Jan 3 at 19:58












            • $begingroup$
              I mean $ln(r)+itheta$ is analytic everywhere except $0$ and negative $x$ axis
              $endgroup$
              – no0ob
              Jan 3 at 20:05










            • $begingroup$
              This ends up as $z=x+ipi$ with $xle 0$.
              $endgroup$
              – herb steinberg
              Jan 3 at 21:12


















            • $begingroup$
              I can't find relation between $r$ and $theta$ so i could write function as $u(r,theta) + iv(r,theta)$
              $endgroup$
              – no0ob
              Jan 3 at 12:41










            • $begingroup$
              $ln(e^z+1)=ln(r)+itheta$. Therefore $u=ln(r)=frac{ln(r^2)}{2}$ and $v=theta$
              $endgroup$
              – herb steinberg
              Jan 3 at 17:02










            • $begingroup$
              Aha Cauchy-Reimann is always correct for the $ln(r)+itheta$ now we have to check where does $ln(r)$ available for $r^2 = (e^xcos y+1)^2+(e^xsin y)^2$ and check domain of $arctan(frac{sin(y)}{cos(y)+e^{-x}}) $ and we are OK right?
              $endgroup$
              – no0ob
              Jan 3 at 19:58












            • $begingroup$
              I mean $ln(r)+itheta$ is analytic everywhere except $0$ and negative $x$ axis
              $endgroup$
              – no0ob
              Jan 3 at 20:05










            • $begingroup$
              This ends up as $z=x+ipi$ with $xle 0$.
              $endgroup$
              – herb steinberg
              Jan 3 at 21:12
















            $begingroup$
            I can't find relation between $r$ and $theta$ so i could write function as $u(r,theta) + iv(r,theta)$
            $endgroup$
            – no0ob
            Jan 3 at 12:41




            $begingroup$
            I can't find relation between $r$ and $theta$ so i could write function as $u(r,theta) + iv(r,theta)$
            $endgroup$
            – no0ob
            Jan 3 at 12:41












            $begingroup$
            $ln(e^z+1)=ln(r)+itheta$. Therefore $u=ln(r)=frac{ln(r^2)}{2}$ and $v=theta$
            $endgroup$
            – herb steinberg
            Jan 3 at 17:02




            $begingroup$
            $ln(e^z+1)=ln(r)+itheta$. Therefore $u=ln(r)=frac{ln(r^2)}{2}$ and $v=theta$
            $endgroup$
            – herb steinberg
            Jan 3 at 17:02












            $begingroup$
            Aha Cauchy-Reimann is always correct for the $ln(r)+itheta$ now we have to check where does $ln(r)$ available for $r^2 = (e^xcos y+1)^2+(e^xsin y)^2$ and check domain of $arctan(frac{sin(y)}{cos(y)+e^{-x}}) $ and we are OK right?
            $endgroup$
            – no0ob
            Jan 3 at 19:58






            $begingroup$
            Aha Cauchy-Reimann is always correct for the $ln(r)+itheta$ now we have to check where does $ln(r)$ available for $r^2 = (e^xcos y+1)^2+(e^xsin y)^2$ and check domain of $arctan(frac{sin(y)}{cos(y)+e^{-x}}) $ and we are OK right?
            $endgroup$
            – no0ob
            Jan 3 at 19:58














            $begingroup$
            I mean $ln(r)+itheta$ is analytic everywhere except $0$ and negative $x$ axis
            $endgroup$
            – no0ob
            Jan 3 at 20:05




            $begingroup$
            I mean $ln(r)+itheta$ is analytic everywhere except $0$ and negative $x$ axis
            $endgroup$
            – no0ob
            Jan 3 at 20:05












            $begingroup$
            This ends up as $z=x+ipi$ with $xle 0$.
            $endgroup$
            – herb steinberg
            Jan 3 at 21:12




            $begingroup$
            This ends up as $z=x+ipi$ with $xle 0$.
            $endgroup$
            – herb steinberg
            Jan 3 at 21:12











            0












            $begingroup$

            According to the determination of logarythm you saw, you should know on which domain the complex logarithm is holomorphic.
            The question would be : can the function exp(z) + 1 go outside of this domain ?






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              actually i better asked determine analytical area.
              $endgroup$
              – no0ob
              Jan 2 at 20:22










            • $begingroup$
              Then the question would be the following: for which z does exp(z) + 1 belong to the analytical area of the logarithm?
              $endgroup$
              – Cauchy is my master
              Jan 2 at 20:24










            • $begingroup$
              I think that's correct, and ln(z) is analytic in all of the plane but $xle0$ right? for the original branch
              $endgroup$
              – no0ob
              Jan 2 at 20:31






            • 1




              $begingroup$
              That depends on the determination of argument you chose, but I think you chose the usual one, so it's correct.
              $endgroup$
              – Cauchy is my master
              Jan 2 at 20:44










            • $begingroup$
              so now i should use function composition or sth? or is it possible to do it with the Cauchy-Reimann equation?
              $endgroup$
              – no0ob
              Jan 2 at 20:51
















            0












            $begingroup$

            According to the determination of logarythm you saw, you should know on which domain the complex logarithm is holomorphic.
            The question would be : can the function exp(z) + 1 go outside of this domain ?






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              actually i better asked determine analytical area.
              $endgroup$
              – no0ob
              Jan 2 at 20:22










            • $begingroup$
              Then the question would be the following: for which z does exp(z) + 1 belong to the analytical area of the logarithm?
              $endgroup$
              – Cauchy is my master
              Jan 2 at 20:24










            • $begingroup$
              I think that's correct, and ln(z) is analytic in all of the plane but $xle0$ right? for the original branch
              $endgroup$
              – no0ob
              Jan 2 at 20:31






            • 1




              $begingroup$
              That depends on the determination of argument you chose, but I think you chose the usual one, so it's correct.
              $endgroup$
              – Cauchy is my master
              Jan 2 at 20:44










            • $begingroup$
              so now i should use function composition or sth? or is it possible to do it with the Cauchy-Reimann equation?
              $endgroup$
              – no0ob
              Jan 2 at 20:51














            0












            0








            0





            $begingroup$

            According to the determination of logarythm you saw, you should know on which domain the complex logarithm is holomorphic.
            The question would be : can the function exp(z) + 1 go outside of this domain ?






            share|cite|improve this answer









            $endgroup$



            According to the determination of logarythm you saw, you should know on which domain the complex logarithm is holomorphic.
            The question would be : can the function exp(z) + 1 go outside of this domain ?







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 2 at 20:19









            Cauchy is my masterCauchy is my master

            413




            413








            • 1




              $begingroup$
              actually i better asked determine analytical area.
              $endgroup$
              – no0ob
              Jan 2 at 20:22










            • $begingroup$
              Then the question would be the following: for which z does exp(z) + 1 belong to the analytical area of the logarithm?
              $endgroup$
              – Cauchy is my master
              Jan 2 at 20:24










            • $begingroup$
              I think that's correct, and ln(z) is analytic in all of the plane but $xle0$ right? for the original branch
              $endgroup$
              – no0ob
              Jan 2 at 20:31






            • 1




              $begingroup$
              That depends on the determination of argument you chose, but I think you chose the usual one, so it's correct.
              $endgroup$
              – Cauchy is my master
              Jan 2 at 20:44










            • $begingroup$
              so now i should use function composition or sth? or is it possible to do it with the Cauchy-Reimann equation?
              $endgroup$
              – no0ob
              Jan 2 at 20:51














            • 1




              $begingroup$
              actually i better asked determine analytical area.
              $endgroup$
              – no0ob
              Jan 2 at 20:22










            • $begingroup$
              Then the question would be the following: for which z does exp(z) + 1 belong to the analytical area of the logarithm?
              $endgroup$
              – Cauchy is my master
              Jan 2 at 20:24










            • $begingroup$
              I think that's correct, and ln(z) is analytic in all of the plane but $xle0$ right? for the original branch
              $endgroup$
              – no0ob
              Jan 2 at 20:31






            • 1




              $begingroup$
              That depends on the determination of argument you chose, but I think you chose the usual one, so it's correct.
              $endgroup$
              – Cauchy is my master
              Jan 2 at 20:44










            • $begingroup$
              so now i should use function composition or sth? or is it possible to do it with the Cauchy-Reimann equation?
              $endgroup$
              – no0ob
              Jan 2 at 20:51








            1




            1




            $begingroup$
            actually i better asked determine analytical area.
            $endgroup$
            – no0ob
            Jan 2 at 20:22




            $begingroup$
            actually i better asked determine analytical area.
            $endgroup$
            – no0ob
            Jan 2 at 20:22












            $begingroup$
            Then the question would be the following: for which z does exp(z) + 1 belong to the analytical area of the logarithm?
            $endgroup$
            – Cauchy is my master
            Jan 2 at 20:24




            $begingroup$
            Then the question would be the following: for which z does exp(z) + 1 belong to the analytical area of the logarithm?
            $endgroup$
            – Cauchy is my master
            Jan 2 at 20:24












            $begingroup$
            I think that's correct, and ln(z) is analytic in all of the plane but $xle0$ right? for the original branch
            $endgroup$
            – no0ob
            Jan 2 at 20:31




            $begingroup$
            I think that's correct, and ln(z) is analytic in all of the plane but $xle0$ right? for the original branch
            $endgroup$
            – no0ob
            Jan 2 at 20:31




            1




            1




            $begingroup$
            That depends on the determination of argument you chose, but I think you chose the usual one, so it's correct.
            $endgroup$
            – Cauchy is my master
            Jan 2 at 20:44




            $begingroup$
            That depends on the determination of argument you chose, but I think you chose the usual one, so it's correct.
            $endgroup$
            – Cauchy is my master
            Jan 2 at 20:44












            $begingroup$
            so now i should use function composition or sth? or is it possible to do it with the Cauchy-Reimann equation?
            $endgroup$
            – no0ob
            Jan 2 at 20:51




            $begingroup$
            so now i should use function composition or sth? or is it possible to do it with the Cauchy-Reimann equation?
            $endgroup$
            – no0ob
            Jan 2 at 20:51


















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