Complex analysis derivatives












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$begingroup$


If a function $f(x,y)=u(x,y)+icdot v(x,y)$ is $mathbb{C}$-differentiable,how can I prove that
for $u$ and $v$ the partial derivatives exist?
I have figured out if f is continuously differentiable then the derivative value must be $f_x$,but $f_x$ exists only when $u_x,v_x$ exist.So how can I ensure their existence?






complex-analysis derivatives







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    -1












    $begingroup$


    If a function $f(x,y)=u(x,y)+icdot v(x,y)$ is $mathbb{C}$-differentiable,how can I prove that
    for $u$ and $v$ the partial derivatives exist?
    I have figured out if f is continuously differentiable then the derivative value must be $f_x$,but $f_x$ exists only when $u_x,v_x$ exist.So how can I ensure their existence?






    complex-analysis derivatives







    share|cite|improve this question











    $endgroup$















      -1












      -1








      -1





      $begingroup$


      If a function $f(x,y)=u(x,y)+icdot v(x,y)$ is $mathbb{C}$-differentiable,how can I prove that
      for $u$ and $v$ the partial derivatives exist?
      I have figured out if f is continuously differentiable then the derivative value must be $f_x$,but $f_x$ exists only when $u_x,v_x$ exist.So how can I ensure their existence?






      complex-analysis derivatives







      share|cite|improve this question











      $endgroup$




      If a function $f(x,y)=u(x,y)+icdot v(x,y)$ is $mathbb{C}$-differentiable,how can I prove that
      for $u$ and $v$ the partial derivatives exist?
      I have figured out if f is continuously differentiable then the derivative value must be $f_x$,but $f_x$ exists only when $u_x,v_x$ exist.So how can I ensure their existence?






      complex-analysis derivatives




      complex-analysis derivatives






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      edited Jan 19 at 11:42









      idriskameni

      755321




      755321










      asked Jan 19 at 10:57









      ANJAN SAMANTAANJAN SAMANTA

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          $begingroup$

          Let $z_0inmathbb C$. Asserting that $f$ is differentiable at $z_0$ means that the lime$$lim_{hto0}frac{f(z_0+h)-f(z_0)}h$$ exists. In particular, the limit$$lim_{hto0, hinmathbb{R}}frac{u(z_0+h)+iv(z_0+h)-u(z_0)-if(z_0)}h$$exists, which is the same thing as asserting that both limits$$lim_{hto0, hinmathbb{R}}frac{u(z_0+h)-u(z_0)}htext{ and }lim_{hto0, hinmathbb{R}}frac{v(z_0+h)-v(z_0)}h$$exist. But this means that $u_x$ and $v_x$ exist at $z_0$.



          A similar argument shows that $u_y$ and $v_y$ exist at $z_0$.






          share|cite|improve this answer









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            $begingroup$

            Unpack the definition; the function is $mathbb{C}$-differentiable if $lim_{zto z_0} frac{f(z)-f(z_0)}{z-z_0} = f'(z_0)$ exists.
            $$lim_{zto z_0} frac{f(z)-f(z_0)}{z-z_0} - f'(z_0) = 0$$
            $$lim_{zto z_0} frac1{z-z_0}left(f(z)-f(z_0) - f'(z_0)(z-z_0)right) = 0$$
            $$lim_{zto z_0} frac1{|z-z_0|}left|f(z)-f(z_0) - f'(z_0)(z-z_0)right| = 0$$
            $$lim_{(x,y)to (x_0,y_0)}frac1{|(x,y)-(x_0,y_0)|}left|u(x,y)+iv(x,y)-u(x_0,y_0)-iv(x_0,y_0) - L_{f'(z_0)}(x-x_0,y-y_0)right| = 0$$
            There $L_{zeta}$ is the linear map from $mathbb{R}^2$ to $mathbb{C}$ that acts the same as multiplication by $zeta$. Split this into real and imaginary parts; we get
            $$lim_{(x,y)to (x_0,y_0)}frac1{|(x,y)-(x_0,y_0)|}left|u(x,y)-u(x_0,y_0)- L(x-x_0,y-y_0)right| = 0$$
            for some linear transformation $L$, and similarly for the imaginary part. That's the definition of differentiability for a function from $mathbb{R}^2$ to $mathbb{R}$; we have proved, straight from the definition, that if a function is $mathbb{C}$-differentiable, its real and imaginary parts are $mathbb{R}$-differentiable. A $mathbb{R}$-differentiable function has partial derivatives, and we're done.



            Next, of course, is verifying the relation between those partial derivatives (the Cauchy-Riemann equations). Related, the theorem to remember from this: a function is $mathbb{C}$-differentiable at a point if and only if it's $mathbb{R}$-differentiable at that point and its partial derivatives there satisfy the Cauchy-Riemann equations.



            Yes, we could have gotten just the partial derivatives more easily. The full power of differentiability at that point is worth a little extra effort.






            share|cite|improve this answer









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              2 Answers
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              $begingroup$

              Let $z_0inmathbb C$. Asserting that $f$ is differentiable at $z_0$ means that the lime$$lim_{hto0}frac{f(z_0+h)-f(z_0)}h$$ exists. In particular, the limit$$lim_{hto0, hinmathbb{R}}frac{u(z_0+h)+iv(z_0+h)-u(z_0)-if(z_0)}h$$exists, which is the same thing as asserting that both limits$$lim_{hto0, hinmathbb{R}}frac{u(z_0+h)-u(z_0)}htext{ and }lim_{hto0, hinmathbb{R}}frac{v(z_0+h)-v(z_0)}h$$exist. But this means that $u_x$ and $v_x$ exist at $z_0$.



              A similar argument shows that $u_y$ and $v_y$ exist at $z_0$.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                Let $z_0inmathbb C$. Asserting that $f$ is differentiable at $z_0$ means that the lime$$lim_{hto0}frac{f(z_0+h)-f(z_0)}h$$ exists. In particular, the limit$$lim_{hto0, hinmathbb{R}}frac{u(z_0+h)+iv(z_0+h)-u(z_0)-if(z_0)}h$$exists, which is the same thing as asserting that both limits$$lim_{hto0, hinmathbb{R}}frac{u(z_0+h)-u(z_0)}htext{ and }lim_{hto0, hinmathbb{R}}frac{v(z_0+h)-v(z_0)}h$$exist. But this means that $u_x$ and $v_x$ exist at $z_0$.



                A similar argument shows that $u_y$ and $v_y$ exist at $z_0$.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Let $z_0inmathbb C$. Asserting that $f$ is differentiable at $z_0$ means that the lime$$lim_{hto0}frac{f(z_0+h)-f(z_0)}h$$ exists. In particular, the limit$$lim_{hto0, hinmathbb{R}}frac{u(z_0+h)+iv(z_0+h)-u(z_0)-if(z_0)}h$$exists, which is the same thing as asserting that both limits$$lim_{hto0, hinmathbb{R}}frac{u(z_0+h)-u(z_0)}htext{ and }lim_{hto0, hinmathbb{R}}frac{v(z_0+h)-v(z_0)}h$$exist. But this means that $u_x$ and $v_x$ exist at $z_0$.



                  A similar argument shows that $u_y$ and $v_y$ exist at $z_0$.






                  share|cite|improve this answer









                  $endgroup$



                  Let $z_0inmathbb C$. Asserting that $f$ is differentiable at $z_0$ means that the lime$$lim_{hto0}frac{f(z_0+h)-f(z_0)}h$$ exists. In particular, the limit$$lim_{hto0, hinmathbb{R}}frac{u(z_0+h)+iv(z_0+h)-u(z_0)-if(z_0)}h$$exists, which is the same thing as asserting that both limits$$lim_{hto0, hinmathbb{R}}frac{u(z_0+h)-u(z_0)}htext{ and }lim_{hto0, hinmathbb{R}}frac{v(z_0+h)-v(z_0)}h$$exist. But this means that $u_x$ and $v_x$ exist at $z_0$.



                  A similar argument shows that $u_y$ and $v_y$ exist at $z_0$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 19 at 11:08









                  José Carlos SantosJosé Carlos Santos

                  164k22131235




                  164k22131235























                      0












                      $begingroup$

                      Unpack the definition; the function is $mathbb{C}$-differentiable if $lim_{zto z_0} frac{f(z)-f(z_0)}{z-z_0} = f'(z_0)$ exists.
                      $$lim_{zto z_0} frac{f(z)-f(z_0)}{z-z_0} - f'(z_0) = 0$$
                      $$lim_{zto z_0} frac1{z-z_0}left(f(z)-f(z_0) - f'(z_0)(z-z_0)right) = 0$$
                      $$lim_{zto z_0} frac1{|z-z_0|}left|f(z)-f(z_0) - f'(z_0)(z-z_0)right| = 0$$
                      $$lim_{(x,y)to (x_0,y_0)}frac1{|(x,y)-(x_0,y_0)|}left|u(x,y)+iv(x,y)-u(x_0,y_0)-iv(x_0,y_0) - L_{f'(z_0)}(x-x_0,y-y_0)right| = 0$$
                      There $L_{zeta}$ is the linear map from $mathbb{R}^2$ to $mathbb{C}$ that acts the same as multiplication by $zeta$. Split this into real and imaginary parts; we get
                      $$lim_{(x,y)to (x_0,y_0)}frac1{|(x,y)-(x_0,y_0)|}left|u(x,y)-u(x_0,y_0)- L(x-x_0,y-y_0)right| = 0$$
                      for some linear transformation $L$, and similarly for the imaginary part. That's the definition of differentiability for a function from $mathbb{R}^2$ to $mathbb{R}$; we have proved, straight from the definition, that if a function is $mathbb{C}$-differentiable, its real and imaginary parts are $mathbb{R}$-differentiable. A $mathbb{R}$-differentiable function has partial derivatives, and we're done.



                      Next, of course, is verifying the relation between those partial derivatives (the Cauchy-Riemann equations). Related, the theorem to remember from this: a function is $mathbb{C}$-differentiable at a point if and only if it's $mathbb{R}$-differentiable at that point and its partial derivatives there satisfy the Cauchy-Riemann equations.



                      Yes, we could have gotten just the partial derivatives more easily. The full power of differentiability at that point is worth a little extra effort.






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        Unpack the definition; the function is $mathbb{C}$-differentiable if $lim_{zto z_0} frac{f(z)-f(z_0)}{z-z_0} = f'(z_0)$ exists.
                        $$lim_{zto z_0} frac{f(z)-f(z_0)}{z-z_0} - f'(z_0) = 0$$
                        $$lim_{zto z_0} frac1{z-z_0}left(f(z)-f(z_0) - f'(z_0)(z-z_0)right) = 0$$
                        $$lim_{zto z_0} frac1{|z-z_0|}left|f(z)-f(z_0) - f'(z_0)(z-z_0)right| = 0$$
                        $$lim_{(x,y)to (x_0,y_0)}frac1{|(x,y)-(x_0,y_0)|}left|u(x,y)+iv(x,y)-u(x_0,y_0)-iv(x_0,y_0) - L_{f'(z_0)}(x-x_0,y-y_0)right| = 0$$
                        There $L_{zeta}$ is the linear map from $mathbb{R}^2$ to $mathbb{C}$ that acts the same as multiplication by $zeta$. Split this into real and imaginary parts; we get
                        $$lim_{(x,y)to (x_0,y_0)}frac1{|(x,y)-(x_0,y_0)|}left|u(x,y)-u(x_0,y_0)- L(x-x_0,y-y_0)right| = 0$$
                        for some linear transformation $L$, and similarly for the imaginary part. That's the definition of differentiability for a function from $mathbb{R}^2$ to $mathbb{R}$; we have proved, straight from the definition, that if a function is $mathbb{C}$-differentiable, its real and imaginary parts are $mathbb{R}$-differentiable. A $mathbb{R}$-differentiable function has partial derivatives, and we're done.



                        Next, of course, is verifying the relation between those partial derivatives (the Cauchy-Riemann equations). Related, the theorem to remember from this: a function is $mathbb{C}$-differentiable at a point if and only if it's $mathbb{R}$-differentiable at that point and its partial derivatives there satisfy the Cauchy-Riemann equations.



                        Yes, we could have gotten just the partial derivatives more easily. The full power of differentiability at that point is worth a little extra effort.






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          Unpack the definition; the function is $mathbb{C}$-differentiable if $lim_{zto z_0} frac{f(z)-f(z_0)}{z-z_0} = f'(z_0)$ exists.
                          $$lim_{zto z_0} frac{f(z)-f(z_0)}{z-z_0} - f'(z_0) = 0$$
                          $$lim_{zto z_0} frac1{z-z_0}left(f(z)-f(z_0) - f'(z_0)(z-z_0)right) = 0$$
                          $$lim_{zto z_0} frac1{|z-z_0|}left|f(z)-f(z_0) - f'(z_0)(z-z_0)right| = 0$$
                          $$lim_{(x,y)to (x_0,y_0)}frac1{|(x,y)-(x_0,y_0)|}left|u(x,y)+iv(x,y)-u(x_0,y_0)-iv(x_0,y_0) - L_{f'(z_0)}(x-x_0,y-y_0)right| = 0$$
                          There $L_{zeta}$ is the linear map from $mathbb{R}^2$ to $mathbb{C}$ that acts the same as multiplication by $zeta$. Split this into real and imaginary parts; we get
                          $$lim_{(x,y)to (x_0,y_0)}frac1{|(x,y)-(x_0,y_0)|}left|u(x,y)-u(x_0,y_0)- L(x-x_0,y-y_0)right| = 0$$
                          for some linear transformation $L$, and similarly for the imaginary part. That's the definition of differentiability for a function from $mathbb{R}^2$ to $mathbb{R}$; we have proved, straight from the definition, that if a function is $mathbb{C}$-differentiable, its real and imaginary parts are $mathbb{R}$-differentiable. A $mathbb{R}$-differentiable function has partial derivatives, and we're done.



                          Next, of course, is verifying the relation between those partial derivatives (the Cauchy-Riemann equations). Related, the theorem to remember from this: a function is $mathbb{C}$-differentiable at a point if and only if it's $mathbb{R}$-differentiable at that point and its partial derivatives there satisfy the Cauchy-Riemann equations.



                          Yes, we could have gotten just the partial derivatives more easily. The full power of differentiability at that point is worth a little extra effort.






                          share|cite|improve this answer









                          $endgroup$



                          Unpack the definition; the function is $mathbb{C}$-differentiable if $lim_{zto z_0} frac{f(z)-f(z_0)}{z-z_0} = f'(z_0)$ exists.
                          $$lim_{zto z_0} frac{f(z)-f(z_0)}{z-z_0} - f'(z_0) = 0$$
                          $$lim_{zto z_0} frac1{z-z_0}left(f(z)-f(z_0) - f'(z_0)(z-z_0)right) = 0$$
                          $$lim_{zto z_0} frac1{|z-z_0|}left|f(z)-f(z_0) - f'(z_0)(z-z_0)right| = 0$$
                          $$lim_{(x,y)to (x_0,y_0)}frac1{|(x,y)-(x_0,y_0)|}left|u(x,y)+iv(x,y)-u(x_0,y_0)-iv(x_0,y_0) - L_{f'(z_0)}(x-x_0,y-y_0)right| = 0$$
                          There $L_{zeta}$ is the linear map from $mathbb{R}^2$ to $mathbb{C}$ that acts the same as multiplication by $zeta$. Split this into real and imaginary parts; we get
                          $$lim_{(x,y)to (x_0,y_0)}frac1{|(x,y)-(x_0,y_0)|}left|u(x,y)-u(x_0,y_0)- L(x-x_0,y-y_0)right| = 0$$
                          for some linear transformation $L$, and similarly for the imaginary part. That's the definition of differentiability for a function from $mathbb{R}^2$ to $mathbb{R}$; we have proved, straight from the definition, that if a function is $mathbb{C}$-differentiable, its real and imaginary parts are $mathbb{R}$-differentiable. A $mathbb{R}$-differentiable function has partial derivatives, and we're done.



                          Next, of course, is verifying the relation between those partial derivatives (the Cauchy-Riemann equations). Related, the theorem to remember from this: a function is $mathbb{C}$-differentiable at a point if and only if it's $mathbb{R}$-differentiable at that point and its partial derivatives there satisfy the Cauchy-Riemann equations.



                          Yes, we could have gotten just the partial derivatives more easily. The full power of differentiability at that point is worth a little extra effort.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 19 at 11:24









                          jmerryjmerry

                          11.3k1426




                          11.3k1426






























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