is the given function differentiable at origin












0












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it is given that f(z) = Im(z^2)/(z bar) when zis not equal to 0 and f(z) =0 when z=0



so when i find the limit of f(z) it depends on a parameter m
so limit doesnot exist uniquely and hence the given function is not differentiable at origin



but when i tried to find its limit by polar coordinates then its limit came out to b 0



i.e x=r cos theta and y = r sin theta
then



f(r, thetha) = (sin2 theta) r exp(i theta)



on putting limit r tends to 0 we get
zero



my query is that if the function is not differentiable at origin then whatever approach we apply its limit should not exist uniquely or finitely
but here wen i try to solve it by polar approach i m getting 0 ie unqiue and finite limit
& on the other hand by previous method i found that it depend on parameter m



is there any mistake i am making while solving it by latter method ? please solve










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    $begingroup$


    it is given that f(z) = Im(z^2)/(z bar) when zis not equal to 0 and f(z) =0 when z=0



    so when i find the limit of f(z) it depends on a parameter m
    so limit doesnot exist uniquely and hence the given function is not differentiable at origin



    but when i tried to find its limit by polar coordinates then its limit came out to b 0



    i.e x=r cos theta and y = r sin theta
    then



    f(r, thetha) = (sin2 theta) r exp(i theta)



    on putting limit r tends to 0 we get
    zero



    my query is that if the function is not differentiable at origin then whatever approach we apply its limit should not exist uniquely or finitely
    but here wen i try to solve it by polar approach i m getting 0 ie unqiue and finite limit
    & on the other hand by previous method i found that it depend on parameter m



    is there any mistake i am making while solving it by latter method ? please solve










    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      it is given that f(z) = Im(z^2)/(z bar) when zis not equal to 0 and f(z) =0 when z=0



      so when i find the limit of f(z) it depends on a parameter m
      so limit doesnot exist uniquely and hence the given function is not differentiable at origin



      but when i tried to find its limit by polar coordinates then its limit came out to b 0



      i.e x=r cos theta and y = r sin theta
      then



      f(r, thetha) = (sin2 theta) r exp(i theta)



      on putting limit r tends to 0 we get
      zero



      my query is that if the function is not differentiable at origin then whatever approach we apply its limit should not exist uniquely or finitely
      but here wen i try to solve it by polar approach i m getting 0 ie unqiue and finite limit
      & on the other hand by previous method i found that it depend on parameter m



      is there any mistake i am making while solving it by latter method ? please solve










      share|cite|improve this question











      $endgroup$




      it is given that f(z) = Im(z^2)/(z bar) when zis not equal to 0 and f(z) =0 when z=0



      so when i find the limit of f(z) it depends on a parameter m
      so limit doesnot exist uniquely and hence the given function is not differentiable at origin



      but when i tried to find its limit by polar coordinates then its limit came out to b 0



      i.e x=r cos theta and y = r sin theta
      then



      f(r, thetha) = (sin2 theta) r exp(i theta)



      on putting limit r tends to 0 we get
      zero



      my query is that if the function is not differentiable at origin then whatever approach we apply its limit should not exist uniquely or finitely
      but here wen i try to solve it by polar approach i m getting 0 ie unqiue and finite limit
      & on the other hand by previous method i found that it depend on parameter m



      is there any mistake i am making while solving it by latter method ? please solve







      complex-analysis






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      edited Jan 9 at 11:43







      Henry

















      asked Jan 9 at 11:26









      HenryHenry

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      327






















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          I'm a little confused about what you're trying to prove. Your calculation with polar coordinates shows correctly that $lim_{zrightarrow 0} fleft(zright) = 0$ . But this merely establishes that $f$ is continuous at $z = 0$, not that it's differentiable there. To determine whether it's differentiable you need to establish what happens to $frac{fleft(zright) - fleft(0right)}{z - 0} = frac{fleft(zright)}{z}$ as $zrightarrow 0$ .



          If you divide your expression for $fleft(zright)$ in terms of polar coordinates by $z = r e^{itheta}$ you're left with $sinleft(2thetaright)$ , independent of $r .$ This is equal to $1$ for $theta = frac{pi}{4}$ and $0$ for $theta = 0$, so there are values of $z$ arbitrarily close to $0$ for which $frac{fleft(zright)}{z} = 1$ and others for which $frac{fleft(zright)}{z} = 0$. It therefore can't have a limit as $zrightarrow 0$ , and consequently $f$ is not differentiable at $z= 0$ .






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            no no that i understood very well ..m query is about the limit ...wen i try to solve it with the help of polar coordinates ...limit exists and equal to zero ...but wen i try to solve it with an alternative approach like along different path ...it depends on parameter m n hence the limit doesnt exist in this case ....so why is this difference ? if the limit doesnt exist then no matter whatever approach u go with it will never exist
            $endgroup$
            – Henry
            Jan 9 at 14:43










          • $begingroup$
            ooops! my bad ...i just skipped a step in calculation ....how could i do such silly mistakes .....totally agree with u ....thankyou so much :)
            $endgroup$
            – Henry
            Jan 9 at 14:49



















          0












          $begingroup$

          The limit does not depend on a parameter !



          Since $|Im(z^2)| le |z|^2=|z|^2$, we get



          $|f(z)| le |z|$ for all $z $.






          share|cite|improve this answer









          $endgroup$













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            2 Answers
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            active

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            2 Answers
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            active

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            0












            $begingroup$

            I'm a little confused about what you're trying to prove. Your calculation with polar coordinates shows correctly that $lim_{zrightarrow 0} fleft(zright) = 0$ . But this merely establishes that $f$ is continuous at $z = 0$, not that it's differentiable there. To determine whether it's differentiable you need to establish what happens to $frac{fleft(zright) - fleft(0right)}{z - 0} = frac{fleft(zright)}{z}$ as $zrightarrow 0$ .



            If you divide your expression for $fleft(zright)$ in terms of polar coordinates by $z = r e^{itheta}$ you're left with $sinleft(2thetaright)$ , independent of $r .$ This is equal to $1$ for $theta = frac{pi}{4}$ and $0$ for $theta = 0$, so there are values of $z$ arbitrarily close to $0$ for which $frac{fleft(zright)}{z} = 1$ and others for which $frac{fleft(zright)}{z} = 0$. It therefore can't have a limit as $zrightarrow 0$ , and consequently $f$ is not differentiable at $z= 0$ .






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              no no that i understood very well ..m query is about the limit ...wen i try to solve it with the help of polar coordinates ...limit exists and equal to zero ...but wen i try to solve it with an alternative approach like along different path ...it depends on parameter m n hence the limit doesnt exist in this case ....so why is this difference ? if the limit doesnt exist then no matter whatever approach u go with it will never exist
              $endgroup$
              – Henry
              Jan 9 at 14:43










            • $begingroup$
              ooops! my bad ...i just skipped a step in calculation ....how could i do such silly mistakes .....totally agree with u ....thankyou so much :)
              $endgroup$
              – Henry
              Jan 9 at 14:49
















            0












            $begingroup$

            I'm a little confused about what you're trying to prove. Your calculation with polar coordinates shows correctly that $lim_{zrightarrow 0} fleft(zright) = 0$ . But this merely establishes that $f$ is continuous at $z = 0$, not that it's differentiable there. To determine whether it's differentiable you need to establish what happens to $frac{fleft(zright) - fleft(0right)}{z - 0} = frac{fleft(zright)}{z}$ as $zrightarrow 0$ .



            If you divide your expression for $fleft(zright)$ in terms of polar coordinates by $z = r e^{itheta}$ you're left with $sinleft(2thetaright)$ , independent of $r .$ This is equal to $1$ for $theta = frac{pi}{4}$ and $0$ for $theta = 0$, so there are values of $z$ arbitrarily close to $0$ for which $frac{fleft(zright)}{z} = 1$ and others for which $frac{fleft(zright)}{z} = 0$. It therefore can't have a limit as $zrightarrow 0$ , and consequently $f$ is not differentiable at $z= 0$ .






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              no no that i understood very well ..m query is about the limit ...wen i try to solve it with the help of polar coordinates ...limit exists and equal to zero ...but wen i try to solve it with an alternative approach like along different path ...it depends on parameter m n hence the limit doesnt exist in this case ....so why is this difference ? if the limit doesnt exist then no matter whatever approach u go with it will never exist
              $endgroup$
              – Henry
              Jan 9 at 14:43










            • $begingroup$
              ooops! my bad ...i just skipped a step in calculation ....how could i do such silly mistakes .....totally agree with u ....thankyou so much :)
              $endgroup$
              – Henry
              Jan 9 at 14:49














            0












            0








            0





            $begingroup$

            I'm a little confused about what you're trying to prove. Your calculation with polar coordinates shows correctly that $lim_{zrightarrow 0} fleft(zright) = 0$ . But this merely establishes that $f$ is continuous at $z = 0$, not that it's differentiable there. To determine whether it's differentiable you need to establish what happens to $frac{fleft(zright) - fleft(0right)}{z - 0} = frac{fleft(zright)}{z}$ as $zrightarrow 0$ .



            If you divide your expression for $fleft(zright)$ in terms of polar coordinates by $z = r e^{itheta}$ you're left with $sinleft(2thetaright)$ , independent of $r .$ This is equal to $1$ for $theta = frac{pi}{4}$ and $0$ for $theta = 0$, so there are values of $z$ arbitrarily close to $0$ for which $frac{fleft(zright)}{z} = 1$ and others for which $frac{fleft(zright)}{z} = 0$. It therefore can't have a limit as $zrightarrow 0$ , and consequently $f$ is not differentiable at $z= 0$ .






            share|cite|improve this answer











            $endgroup$



            I'm a little confused about what you're trying to prove. Your calculation with polar coordinates shows correctly that $lim_{zrightarrow 0} fleft(zright) = 0$ . But this merely establishes that $f$ is continuous at $z = 0$, not that it's differentiable there. To determine whether it's differentiable you need to establish what happens to $frac{fleft(zright) - fleft(0right)}{z - 0} = frac{fleft(zright)}{z}$ as $zrightarrow 0$ .



            If you divide your expression for $fleft(zright)$ in terms of polar coordinates by $z = r e^{itheta}$ you're left with $sinleft(2thetaright)$ , independent of $r .$ This is equal to $1$ for $theta = frac{pi}{4}$ and $0$ for $theta = 0$, so there are values of $z$ arbitrarily close to $0$ for which $frac{fleft(zright)}{z} = 1$ and others for which $frac{fleft(zright)}{z} = 0$. It therefore can't have a limit as $zrightarrow 0$ , and consequently $f$ is not differentiable at $z= 0$ .







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 9 at 14:06

























            answered Jan 9 at 13:22









            lonza leggieralonza leggiera

            89117




            89117












            • $begingroup$
              no no that i understood very well ..m query is about the limit ...wen i try to solve it with the help of polar coordinates ...limit exists and equal to zero ...but wen i try to solve it with an alternative approach like along different path ...it depends on parameter m n hence the limit doesnt exist in this case ....so why is this difference ? if the limit doesnt exist then no matter whatever approach u go with it will never exist
              $endgroup$
              – Henry
              Jan 9 at 14:43










            • $begingroup$
              ooops! my bad ...i just skipped a step in calculation ....how could i do such silly mistakes .....totally agree with u ....thankyou so much :)
              $endgroup$
              – Henry
              Jan 9 at 14:49


















            • $begingroup$
              no no that i understood very well ..m query is about the limit ...wen i try to solve it with the help of polar coordinates ...limit exists and equal to zero ...but wen i try to solve it with an alternative approach like along different path ...it depends on parameter m n hence the limit doesnt exist in this case ....so why is this difference ? if the limit doesnt exist then no matter whatever approach u go with it will never exist
              $endgroup$
              – Henry
              Jan 9 at 14:43










            • $begingroup$
              ooops! my bad ...i just skipped a step in calculation ....how could i do such silly mistakes .....totally agree with u ....thankyou so much :)
              $endgroup$
              – Henry
              Jan 9 at 14:49
















            $begingroup$
            no no that i understood very well ..m query is about the limit ...wen i try to solve it with the help of polar coordinates ...limit exists and equal to zero ...but wen i try to solve it with an alternative approach like along different path ...it depends on parameter m n hence the limit doesnt exist in this case ....so why is this difference ? if the limit doesnt exist then no matter whatever approach u go with it will never exist
            $endgroup$
            – Henry
            Jan 9 at 14:43




            $begingroup$
            no no that i understood very well ..m query is about the limit ...wen i try to solve it with the help of polar coordinates ...limit exists and equal to zero ...but wen i try to solve it with an alternative approach like along different path ...it depends on parameter m n hence the limit doesnt exist in this case ....so why is this difference ? if the limit doesnt exist then no matter whatever approach u go with it will never exist
            $endgroup$
            – Henry
            Jan 9 at 14:43












            $begingroup$
            ooops! my bad ...i just skipped a step in calculation ....how could i do such silly mistakes .....totally agree with u ....thankyou so much :)
            $endgroup$
            – Henry
            Jan 9 at 14:49




            $begingroup$
            ooops! my bad ...i just skipped a step in calculation ....how could i do such silly mistakes .....totally agree with u ....thankyou so much :)
            $endgroup$
            – Henry
            Jan 9 at 14:49











            0












            $begingroup$

            The limit does not depend on a parameter !



            Since $|Im(z^2)| le |z|^2=|z|^2$, we get



            $|f(z)| le |z|$ for all $z $.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              The limit does not depend on a parameter !



              Since $|Im(z^2)| le |z|^2=|z|^2$, we get



              $|f(z)| le |z|$ for all $z $.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                The limit does not depend on a parameter !



                Since $|Im(z^2)| le |z|^2=|z|^2$, we get



                $|f(z)| le |z|$ for all $z $.






                share|cite|improve this answer









                $endgroup$



                The limit does not depend on a parameter !



                Since $|Im(z^2)| le |z|^2=|z|^2$, we get



                $|f(z)| le |z|$ for all $z $.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 9 at 12:19









                FredFred

                46.9k1848




                46.9k1848






























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