Determine all solutions to $operatorname{Log}(z^2) = operatorname{Log}(z)$ (Complex Equation)












1












$begingroup$


I am new to complex analysis and i was trying to solve the given exercise




Determine all solutions to $operatorname{Log}(z^2) = operatorname{Log}(z)$




However, i am not sure whether my attempt is correct or complete.



My attempt:



It holds $$operatorname{Log}(z^2) = ln( vert z vert^2) + ioperatorname{Arg}(z^2) = 2ln(vert z vert) + 2ioperatorname{Arg}(z) $$



therefore



$$operatorname{Log}(z^2) = operatorname{Log}(z)$$ is true for all $z in mathbb{C}$ for which



$$2ln(vert z vert) + 2ioperatorname{Arg}(z) = ln(vert z vert) + ioperatorname{Arg}(z)$$
$$Leftrightarrow ln(vert z vert) + ioperatorname{Arg}(z) = 0$$



This is how far i got. Can i somehow solve this equation?



Thanks for any hints!






complex-analysis complex-numbers logarithms







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    I am new to complex analysis and i was trying to solve the given exercise




    Determine all solutions to $operatorname{Log}(z^2) = operatorname{Log}(z)$




    However, i am not sure whether my attempt is correct or complete.



    My attempt:



    It holds $$operatorname{Log}(z^2) = ln( vert z vert^2) + ioperatorname{Arg}(z^2) = 2ln(vert z vert) + 2ioperatorname{Arg}(z) $$



    therefore



    $$operatorname{Log}(z^2) = operatorname{Log}(z)$$ is true for all $z in mathbb{C}$ for which



    $$2ln(vert z vert) + 2ioperatorname{Arg}(z) = ln(vert z vert) + ioperatorname{Arg}(z)$$
    $$Leftrightarrow ln(vert z vert) + ioperatorname{Arg}(z) = 0$$



    This is how far i got. Can i somehow solve this equation?



    Thanks for any hints!






    complex-analysis complex-numbers logarithms







    share|cite|improve this question









    $endgroup$















      1












      1








      1


      1



      $begingroup$


      I am new to complex analysis and i was trying to solve the given exercise




      Determine all solutions to $operatorname{Log}(z^2) = operatorname{Log}(z)$




      However, i am not sure whether my attempt is correct or complete.



      My attempt:



      It holds $$operatorname{Log}(z^2) = ln( vert z vert^2) + ioperatorname{Arg}(z^2) = 2ln(vert z vert) + 2ioperatorname{Arg}(z) $$



      therefore



      $$operatorname{Log}(z^2) = operatorname{Log}(z)$$ is true for all $z in mathbb{C}$ for which



      $$2ln(vert z vert) + 2ioperatorname{Arg}(z) = ln(vert z vert) + ioperatorname{Arg}(z)$$
      $$Leftrightarrow ln(vert z vert) + ioperatorname{Arg}(z) = 0$$



      This is how far i got. Can i somehow solve this equation?



      Thanks for any hints!






      complex-analysis complex-numbers logarithms







      share|cite|improve this question









      $endgroup$




      I am new to complex analysis and i was trying to solve the given exercise




      Determine all solutions to $operatorname{Log}(z^2) = operatorname{Log}(z)$




      However, i am not sure whether my attempt is correct or complete.



      My attempt:



      It holds $$operatorname{Log}(z^2) = ln( vert z vert^2) + ioperatorname{Arg}(z^2) = 2ln(vert z vert) + 2ioperatorname{Arg}(z) $$



      therefore



      $$operatorname{Log}(z^2) = operatorname{Log}(z)$$ is true for all $z in mathbb{C}$ for which



      $$2ln(vert z vert) + 2ioperatorname{Arg}(z) = ln(vert z vert) + ioperatorname{Arg}(z)$$
      $$Leftrightarrow ln(vert z vert) + ioperatorname{Arg}(z) = 0$$



      This is how far i got. Can i somehow solve this equation?



      Thanks for any hints!






      complex-analysis complex-numbers logarithms




      complex-analysis complex-numbers logarithms






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 8 at 5:49









      ZestZest

      297113




      297113






















          2 Answers
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          $begingroup$

          Just put together the two conditions you have got right there and you're done.



          The $ln$ term is the real part of that last expression, and $itext{Arg}$ is the imaginary part, so they must both be $0$. $ln |z|$ is $0$ when $|z| = 1$, so you know such a $z$ must live on the unit circle. However $i text{Arg} z = 0 implies text{Arg} z = 0$, which means that $z$ is actually a positive real number.



          The only $z$ satisfying both conditions is $z = 1$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            thank you very much for your help!
            $endgroup$
            – Zest
            Jan 8 at 6:08



















          1












          $begingroup$

          If $z ne 0$ and $operatorname{Log}(z^2) = operatorname{Log}(z)$, then



          $z^2=e^{operatorname{Log}(z^2)}=e^{operatorname{Log}(z)}=z$, hence $z=1$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            thank you very much for your additional hint, highly appreciating it.
            $endgroup$
            – Zest
            Jan 8 at 6:07












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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          Just put together the two conditions you have got right there and you're done.



          The $ln$ term is the real part of that last expression, and $itext{Arg}$ is the imaginary part, so they must both be $0$. $ln |z|$ is $0$ when $|z| = 1$, so you know such a $z$ must live on the unit circle. However $i text{Arg} z = 0 implies text{Arg} z = 0$, which means that $z$ is actually a positive real number.



          The only $z$ satisfying both conditions is $z = 1$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            thank you very much for your help!
            $endgroup$
            – Zest
            Jan 8 at 6:08
















          2












          $begingroup$

          Just put together the two conditions you have got right there and you're done.



          The $ln$ term is the real part of that last expression, and $itext{Arg}$ is the imaginary part, so they must both be $0$. $ln |z|$ is $0$ when $|z| = 1$, so you know such a $z$ must live on the unit circle. However $i text{Arg} z = 0 implies text{Arg} z = 0$, which means that $z$ is actually a positive real number.



          The only $z$ satisfying both conditions is $z = 1$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            thank you very much for your help!
            $endgroup$
            – Zest
            Jan 8 at 6:08














          2












          2








          2





          $begingroup$

          Just put together the two conditions you have got right there and you're done.



          The $ln$ term is the real part of that last expression, and $itext{Arg}$ is the imaginary part, so they must both be $0$. $ln |z|$ is $0$ when $|z| = 1$, so you know such a $z$ must live on the unit circle. However $i text{Arg} z = 0 implies text{Arg} z = 0$, which means that $z$ is actually a positive real number.



          The only $z$ satisfying both conditions is $z = 1$.






          share|cite|improve this answer









          $endgroup$



          Just put together the two conditions you have got right there and you're done.



          The $ln$ term is the real part of that last expression, and $itext{Arg}$ is the imaginary part, so they must both be $0$. $ln |z|$ is $0$ when $|z| = 1$, so you know such a $z$ must live on the unit circle. However $i text{Arg} z = 0 implies text{Arg} z = 0$, which means that $z$ is actually a positive real number.



          The only $z$ satisfying both conditions is $z = 1$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 8 at 6:02









          Alfred YergerAlfred Yerger

          10.5k2249




          10.5k2249












          • $begingroup$
            thank you very much for your help!
            $endgroup$
            – Zest
            Jan 8 at 6:08


















          • $begingroup$
            thank you very much for your help!
            $endgroup$
            – Zest
            Jan 8 at 6:08
















          $begingroup$
          thank you very much for your help!
          $endgroup$
          – Zest
          Jan 8 at 6:08




          $begingroup$
          thank you very much for your help!
          $endgroup$
          – Zest
          Jan 8 at 6:08











          1












          $begingroup$

          If $z ne 0$ and $operatorname{Log}(z^2) = operatorname{Log}(z)$, then



          $z^2=e^{operatorname{Log}(z^2)}=e^{operatorname{Log}(z)}=z$, hence $z=1$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            thank you very much for your additional hint, highly appreciating it.
            $endgroup$
            – Zest
            Jan 8 at 6:07
















          1












          $begingroup$

          If $z ne 0$ and $operatorname{Log}(z^2) = operatorname{Log}(z)$, then



          $z^2=e^{operatorname{Log}(z^2)}=e^{operatorname{Log}(z)}=z$, hence $z=1$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            thank you very much for your additional hint, highly appreciating it.
            $endgroup$
            – Zest
            Jan 8 at 6:07














          1












          1








          1





          $begingroup$

          If $z ne 0$ and $operatorname{Log}(z^2) = operatorname{Log}(z)$, then



          $z^2=e^{operatorname{Log}(z^2)}=e^{operatorname{Log}(z)}=z$, hence $z=1$.






          share|cite|improve this answer









          $endgroup$



          If $z ne 0$ and $operatorname{Log}(z^2) = operatorname{Log}(z)$, then



          $z^2=e^{operatorname{Log}(z^2)}=e^{operatorname{Log}(z)}=z$, hence $z=1$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 8 at 6:06









          FredFred

          48.9k11849




          48.9k11849












          • $begingroup$
            thank you very much for your additional hint, highly appreciating it.
            $endgroup$
            – Zest
            Jan 8 at 6:07


















          • $begingroup$
            thank you very much for your additional hint, highly appreciating it.
            $endgroup$
            – Zest
            Jan 8 at 6:07
















          $begingroup$
          thank you very much for your additional hint, highly appreciating it.
          $endgroup$
          – Zest
          Jan 8 at 6:07




          $begingroup$
          thank you very much for your additional hint, highly appreciating it.
          $endgroup$
          – Zest
          Jan 8 at 6:07


















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