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I understand that the Cauchy-Riemann equations $$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$$ and $$\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$$ are necessary for a complex function to be complex-differentiable, but I would like to see a proof that they are also sufficient for this to be true. I am assuming that these partial derivatives exist and are continuous everywhere that they are satisfied.

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  • $\begingroup$ They aren't. You also need conditions on the partial derivatives involved (continuity, for example). Note that, for example, $f(x+iy) = x^2$ satisfies the equations at $z=0$, but is not analytic. The general sufficient conditions are either rather complicated or not actually known, depending on who you ask. Either way, your question needs a more precise formulation. $\endgroup$
    – Chappers
    Jun 23, 2015 at 1:33
  • $\begingroup$ Is there an example of a function that is not complex-differentiable at a point where the C-R equations are satisfied and the partial derivatives exist and are continuous? $\endgroup$ Jun 23, 2015 at 2:02
  • $\begingroup$ No, see my answer. $\endgroup$ Jun 23, 2015 at 2:07

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Well, let be provide some inspiration based on linear algebra.

Remember in multivarible culculus, we always deem derivative as a linear mapping, or more concretely, a matrix. Now let's apply it in the case of complex linear space.

Let $f$ be a complex function on $ \mathbb{C} $, then $f$ is homolophic at point $z_0\in \mathbb{C}$ if and only if $Df(z_0)$ is a complex linear map from $\mathbb{C}$ to $\mathbb{C} $. Then identify $i$ to a matrix $\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$, i.e., we deem $i$ as a complex linear map which rotate the plane 90 degrees anticlockwise. Next, $Df(z_0)$ is a complex linear map if and only if $Df(z_0)$ commutes with $i$(think it why?), i.e., $Df(z_0)=\begin{pmatrix} u_x & u_y \\ v_x & v_y \end{pmatrix}$

$$\begin{pmatrix} u_x & u_y \\ v_x & v_y \end{pmatrix}\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}=\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} u_x & u_y \\ v_x & v_y \end{pmatrix}$$

if and only if $u_x=v_y, u_y=-v_x.$

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  • $\begingroup$ An more interesting thing is that by the same way you may find $f$ is anti-homolophic at point $z_0\in \mathbb{C}$ if only if $u_x=-v_y, u_y=-v_x.$ at point $z_0$ $\endgroup$
    – AG learner
    Jun 23, 2015 at 3:12
  • $\begingroup$ Oh wow, that's really insightful. Thank you for that. $\endgroup$ Jun 23, 2015 at 3:37
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    $\begingroup$ Sorry, let me fix my previous comment: $f$ is anti-holomorphic at point $z_0\in \mathbb{C}$ if only if $u_x=-v_y, u_y=v_x$ at $z_0$. Hint: In this time, the condition is $Df(z_0)i=-iDf(z_0)$. @JacksonFitzsimmons $\endgroup$
    – AG learner
    Jun 23, 2015 at 4:17
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A function $f:\mathbb{R}^2 \to \mathbb{R}^2$ represents a complex-linear map (with respect to the standard complex-linear structure on $\mathbb{R}^2$ given by $i\cdot(u,v) = (-v,u)$) iff its matrix with respect to the standard $\mathbb{R}$-basis has the form $\begin{pmatrix} a & -b \\ b & a \end{pmatrix}$ (easy exercise). If $f(x,y) = (u(x,y),v(x,y))$ is $C^1$ then its derivative (again in the standard basis) is the matrix $\begin{pmatrix} u_x & u_y \\ v_x & v_y \end{pmatrix}.$

Now note that $f$ is complex-differentiable at a point if and only if it is real-differentiable there and its derivative is a complex-linear map. (Proof: $f$ is complex-differentiable at $z$ with $f'(z) = a \in \mathbb{C}$ if and only if $f(z+h) = f(z) + ah + o(h),$ and any $\mathbb{C}$-linear map $\mathbb{C} \to \mathbb{C}$ has the form $h \mapsto ah$ for some $a \in \mathbb{C}.$) Combining this with the remarks above we see that a function $f:\mathbb{R}^2 \to \mathbb{R}^2$ is complex-differentiable if and only if it is real-differentiable and its real and imaginary parts satisfy the C-R equations. In particular if $f$ is of class $C^1$ then $f$ is complex-differentiable if and only if its real and imaginary parts satisfy the C-R equations.

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The integral around every single triangle is zero by Green's theorem. If the integral around every single triangle is zero then the function is complex differentiable by Morera's theorem. Done.

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  • $\begingroup$ But now I have to see a proof of Morera's theorem, thank you though. I am new to complex analysis. $\endgroup$ Jun 23, 2015 at 2:01
  • $\begingroup$ "If the integral around any triangle is zero, then$\ldots$" is ambiguous: it could mean "If there is any triangle around which the integral is zero, then$\ldots$". But "If the integral around every triangle is zero" is unambiguous. ${}\qquad{}$ $\endgroup$ Jun 23, 2015 at 3:11
  • $\begingroup$ can i recommend Priestley's Introduction to Complex Analysis? $\endgroup$
    – Mark Joshi
    Jun 23, 2015 at 4:38

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