Prove that if $f:\Bbb C\rightarrow C$ is a complex function whose components are of class $C^1$ then $f$ is differentiable.

Question

If $f:\Bbb C\rightarrow\Bbb C$ is a complex function then there exist two real scalar function $u,v:\Bbb R^2\rightarrow \Bbb R$ such that $$ f(x+yi)=u(x,y)+v(x,y)i $$ for any $(x+yi)\in\Bbb C$. Now for the uniqueness of the limit if $f$ is derivable then the Cauchy-Riemann idetities $$ \frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}\,\,\,\text{and}\,\,\,\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x} $$ hold. Conversely if the functions $u$ and $v$ are of class $C^1$ then it is possibile to show that $f$ is derivable but untortunately I was not able to show this: in particular I tried to prove the statement using the following argumentations

but unfortunately I was not able to prove that $$ f(z+h)-f(z)=\Big(\frac{\partial u}{\partial x}-\frac{\partial u}{\partial y}i\Big)(h_1+h_2i)+|h|\psi(h) $$ moreover I tried to prove the above identity using the mean value theorem for which there exist $\xi_h,\eta_h\in\Bbb R^2$ such that $$ u(x+h_1,x+h_2)-u(x,y)=Du(\xi_h)\cdot h\,\,\,\text{and}\,\,\,v(x+h_1,y+h_2)-v(x,y)=Dv(\eta_h)\cdot h $$ and so such that $$ f\big((x+yi)-(h_1+h_2i)\big)-f\big(x+yi\big)=\\ \Big(\frac{\partial u}{\partial x}(\xi_h)\cdot h_1+\frac{\partial u}{\partial y}(\xi_h)\cdot h_2\Big)+\Big(\frac{\partial v}{\partial x}(\eta_h)\cdot h_1+\frac{\partial v}{\partial y}(\eta_h)\cdot h_2\Big)i $$ but I was not able to proceed further. So could someone help me, please?

Answer 1

You have using Cauchy-Riemann equations

$$ \begin{aligned} \left(\frac{\partial u}{\partial x}h_1 + \frac{\partial u}{\partial y}h_2\right) + i \left(\frac{\partial v}{\partial x}h_1 + \frac{\partial v}{\partial y}h_2\right) &= \left(\frac{\partial u}{\partial x} + i\frac{\partial v}{\partial x}\right)h_1 + \left(\frac{\partial u}{\partial y} + i\frac{\partial v}{\partial y}\right)h_2\\ &= \left(\frac{\partial u}{\partial x} - i\frac{\partial u}{\partial y}\right)h_1 + \left(\frac{\partial u}{\partial y} + i\frac{\partial u}{\partial x}\right)h_2\\ &=\left(\frac{\partial u}{\partial x} - i\frac{\partial u}{\partial y}\right)(h_1+ih_2) \end{aligned}$$

for the first term. And then just set $\psi(h) = \psi_1(h) + i \psi_2(h)$.

Answer 2

You can't prove it without assuming Cauchy-Riemman identity. The canonical example is $$f(z) = \bar{z}.$$ In this case $u(x,y)=x$ and $v(x,y)=-y$, so both $u,v$ are even of $C^\infty$ class. But $f$ is not differentiable in the complex sense.