continuity of $f(z)=(z-i)log(z^2+1)$

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I am trying to determine where $f(z)=(z-i)log(z^2+1)$ is continuous.




My attempt:



Well, $z-i$ is a complex polynomial, so it is continuous in all of $mathbbC$. Hence, the continuity of $f(z)$ is dependent on $log(z^2+1)$. I know that $log(w)$ is continuous on $mathbbC$ $(-infty,0]$, so i figured that $log(z^2+1)$ is continuous when $z^2+1>0$. So,
beginalign
z^2+1&>0 \
x^2-y^2+2xyi&>-1 \
endalign
Equating real and imaginary, $$x^2-y^2>-1 textand xy=0$$
There are two cases to consider.



Case 1: $$x^2-y^2>-1 , x=0$$
this yields $-1<y<1$.



Case 2: $$x^2-y^2>-1 , y=0$$
this yields the empty set.



But the answer is for $ygeq 1, yleq -1$. Where have I gone wrong?




complex-analysis proof-verification continuity




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    up vote
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    down vote

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    I am trying to determine where $f(z)=(z-i)log(z^2+1)$ is continuous.




    My attempt:



    Well, $z-i$ is a complex polynomial, so it is continuous in all of $mathbbC$. Hence, the continuity of $f(z)$ is dependent on $log(z^2+1)$. I know that $log(w)$ is continuous on $mathbbC$ $(-infty,0]$, so i figured that $log(z^2+1)$ is continuous when $z^2+1>0$. So,
    beginalign
    z^2+1&>0 \
    x^2-y^2+2xyi&>-1 \
    endalign
    Equating real and imaginary, $$x^2-y^2>-1 textand xy=0$$
    There are two cases to consider.



    Case 1: $$x^2-y^2>-1 , x=0$$
    this yields $-1<y<1$.



    Case 2: $$x^2-y^2>-1 , y=0$$
    this yields the empty set.



    But the answer is for $ygeq 1, yleq -1$. Where have I gone wrong?




    complex-analysis proof-verification continuity




    share|cite|improve this question
























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite












      I am trying to determine where $f(z)=(z-i)log(z^2+1)$ is continuous.




      My attempt:



      Well, $z-i$ is a complex polynomial, so it is continuous in all of $mathbbC$. Hence, the continuity of $f(z)$ is dependent on $log(z^2+1)$. I know that $log(w)$ is continuous on $mathbbC$ $(-infty,0]$, so i figured that $log(z^2+1)$ is continuous when $z^2+1>0$. So,
      beginalign
      z^2+1&>0 \
      x^2-y^2+2xyi&>-1 \
      endalign
      Equating real and imaginary, $$x^2-y^2>-1 textand xy=0$$
      There are two cases to consider.



      Case 1: $$x^2-y^2>-1 , x=0$$
      this yields $-1<y<1$.



      Case 2: $$x^2-y^2>-1 , y=0$$
      this yields the empty set.



      But the answer is for $ygeq 1, yleq -1$. Where have I gone wrong?




      complex-analysis proof-verification continuity




      share|cite|improve this question















      I am trying to determine where $f(z)=(z-i)log(z^2+1)$ is continuous.




      My attempt:



      Well, $z-i$ is a complex polynomial, so it is continuous in all of $mathbbC$. Hence, the continuity of $f(z)$ is dependent on $log(z^2+1)$. I know that $log(w)$ is continuous on $mathbbC$ $(-infty,0]$, so i figured that $log(z^2+1)$ is continuous when $z^2+1>0$. So,
      beginalign
      z^2+1&>0 \
      x^2-y^2+2xyi&>-1 \
      endalign
      Equating real and imaginary, $$x^2-y^2>-1 textand xy=0$$
      There are two cases to consider.



      Case 1: $$x^2-y^2>-1 , x=0$$
      this yields $-1<y<1$.



      Case 2: $$x^2-y^2>-1 , y=0$$
      this yields the empty set.



      But the answer is for $ygeq 1, yleq -1$. Where have I gone wrong?





      complex-analysis proof-verification continuity





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      share|cite|improve this question




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      edited Aug 21 at 10:01

























      asked Aug 21 at 9:54









      Bell

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          1 Answer
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          The condition for continuity is not $z^2+1>0$. The correct condition (with the chosen branch of the logarithm) is $z^2+1notin(-infty,0]$ (you will have to give a special look to $z=i$.)






          share|cite|improve this answer




















          • Does this condition imply that $Re(z^2+1)leq 0$ and $Im(z^2+1)=0$?
            – Bell
            Aug 21 at 10:09










          • Yes, that is the discontinuity set. However,have a closer look at $z=i$
            – Julián Aguirre
            Aug 21 at 10:11










          • the function is discontinuous at $z=i$ (as this would yield $log(0)$)?
            – Bell
            Aug 21 at 10:20










          • But $lim_zto if(z)=0$
            – Julián Aguirre
            Aug 21 at 10:29











          • Ah crap, so it is continuous at $z=pm i$.
            – Bell
            Aug 21 at 10:40










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          1 Answer
          1






          active

          oldest

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          active

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          up vote
          1
          down vote



          accepted










          The condition for continuity is not $z^2+1>0$. The correct condition (with the chosen branch of the logarithm) is $z^2+1notin(-infty,0]$ (you will have to give a special look to $z=i$.)






          share|cite|improve this answer




















          • Does this condition imply that $Re(z^2+1)leq 0$ and $Im(z^2+1)=0$?
            – Bell
            Aug 21 at 10:09










          • Yes, that is the discontinuity set. However,have a closer look at $z=i$
            – Julián Aguirre
            Aug 21 at 10:11










          • the function is discontinuous at $z=i$ (as this would yield $log(0)$)?
            – Bell
            Aug 21 at 10:20










          • But $lim_zto if(z)=0$
            – Julián Aguirre
            Aug 21 at 10:29











          • Ah crap, so it is continuous at $z=pm i$.
            – Bell
            Aug 21 at 10:40














          up vote
          1
          down vote



          accepted










          The condition for continuity is not $z^2+1>0$. The correct condition (with the chosen branch of the logarithm) is $z^2+1notin(-infty,0]$ (you will have to give a special look to $z=i$.)






          share|cite|improve this answer




















          • Does this condition imply that $Re(z^2+1)leq 0$ and $Im(z^2+1)=0$?
            – Bell
            Aug 21 at 10:09










          • Yes, that is the discontinuity set. However,have a closer look at $z=i$
            – Julián Aguirre
            Aug 21 at 10:11










          • the function is discontinuous at $z=i$ (as this would yield $log(0)$)?
            – Bell
            Aug 21 at 10:20










          • But $lim_zto if(z)=0$
            – Julián Aguirre
            Aug 21 at 10:29











          • Ah crap, so it is continuous at $z=pm i$.
            – Bell
            Aug 21 at 10:40












          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          The condition for continuity is not $z^2+1>0$. The correct condition (with the chosen branch of the logarithm) is $z^2+1notin(-infty,0]$ (you will have to give a special look to $z=i$.)






          share|cite|improve this answer












          The condition for continuity is not $z^2+1>0$. The correct condition (with the chosen branch of the logarithm) is $z^2+1notin(-infty,0]$ (you will have to give a special look to $z=i$.)







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Aug 21 at 10:05









          Julián Aguirre

          65.2k23894




          65.2k23894











          • Does this condition imply that $Re(z^2+1)leq 0$ and $Im(z^2+1)=0$?
            – Bell
            Aug 21 at 10:09










          • Yes, that is the discontinuity set. However,have a closer look at $z=i$
            – Julián Aguirre
            Aug 21 at 10:11










          • the function is discontinuous at $z=i$ (as this would yield $log(0)$)?
            – Bell
            Aug 21 at 10:20










          • But $lim_zto if(z)=0$
            – Julián Aguirre
            Aug 21 at 10:29











          • Ah crap, so it is continuous at $z=pm i$.
            – Bell
            Aug 21 at 10:40
















          • Does this condition imply that $Re(z^2+1)leq 0$ and $Im(z^2+1)=0$?
            – Bell
            Aug 21 at 10:09










          • Yes, that is the discontinuity set. However,have a closer look at $z=i$
            – Julián Aguirre
            Aug 21 at 10:11










          • the function is discontinuous at $z=i$ (as this would yield $log(0)$)?
            – Bell
            Aug 21 at 10:20










          • But $lim_zto if(z)=0$
            – Julián Aguirre
            Aug 21 at 10:29











          • Ah crap, so it is continuous at $z=pm i$.
            – Bell
            Aug 21 at 10:40















          Does this condition imply that $Re(z^2+1)leq 0$ and $Im(z^2+1)=0$?
          – Bell
          Aug 21 at 10:09




          Does this condition imply that $Re(z^2+1)leq 0$ and $Im(z^2+1)=0$?
          – Bell
          Aug 21 at 10:09












          Yes, that is the discontinuity set. However,have a closer look at $z=i$
          – Julián Aguirre
          Aug 21 at 10:11




          Yes, that is the discontinuity set. However,have a closer look at $z=i$
          – Julián Aguirre
          Aug 21 at 10:11












          the function is discontinuous at $z=i$ (as this would yield $log(0)$)?
          – Bell
          Aug 21 at 10:20




          the function is discontinuous at $z=i$ (as this would yield $log(0)$)?
          – Bell
          Aug 21 at 10:20












          But $lim_zto if(z)=0$
          – Julián Aguirre
          Aug 21 at 10:29





          But $lim_zto if(z)=0$
          – Julián Aguirre
          Aug 21 at 10:29













          Ah crap, so it is continuous at $z=pm i$.
          – Bell
          Aug 21 at 10:40




          Ah crap, so it is continuous at $z=pm i$.
          – Bell
          Aug 21 at 10:40












           

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