Characterizing critical points of a complex valued function. Consider a function $f : \Bbb R^2 \longrightarrow \Bbb R^2$. A point $(x_0,y_0) \in \Bbb R^2$ is said to be a regular point of $f$ if $Df\ ((x_0,y_0)) : \Bbb R^2 \longrightarrow \Bbb R^2$ is an invertible linear operator. A point $(x_0,y_0) \in \Bbb R^2$ is said to be a critical point of $f$ if it is not a regular point of $f$. Since $\Bbb C$ is homeomorphic to $\Bbb R^2$ so I was very eager to know the concept of critical points in the complex plane and I found that the critical points of a complex valued function $f$ over the complex plane are precisely those points $z_0 = x_0 + iy_0 \in \Bbb C$ such that $f'(z_0) = 0$. How do I relate these two concepts? In fact how do I proof the theorem which states that 
"Let $f : \Bbb C \longrightarrow \Bbb C$ be a differentiable function. Then a point $z_0 \in \Bbb C$ is a critical point of $f$ (in the multivariable sense) if and only if $f'(z_0) = 0$."
Please help me in proving this theorem. Then it will be really very helpful for me.
Thank you very much.
 A: To relate the two concepts, it is helpful to write $f$ in terms of its real and imaginary parts:
$$ f(x + iy) := u(x, y) + i v(x,y).$$
We can now construct a function $g : \mathbb R^2 \to \mathbb R^2$ as follows:
$$ g : \begin{bmatrix} x \\ y\end{bmatrix} \mapsto \begin{bmatrix} u(x,y) \\ v(x,y) \end{bmatrix}.$$
The claim is that $f'(x_0 + i y_0) = 0$ if and only if $Dg(x_0, y_0) $ is not invertible. (Of course, $f'$ denotes the complex derivative of $f : \mathbb C \to \mathbb C$, whereas $Dg$ denotes the real multivariate derivative of $g : \mathbb R^2 \to \mathbb R^2$.)
Now the derivative of $f$ is given by
$$ f' = \frac{\partial u}{\partial x} + i\frac{\partial v}{\partial x}.$$
Meanwhile, the derivative of $g$ is given by
$$ Dg = \begin{bmatrix} \frac{\partial u}{\partial x} & \frac{\partial u }{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v }{\partial y}  \end{bmatrix},$$
and its determinant is
$$ \det Dg = \frac{\partial u}{\partial x}\frac{\partial v}{\partial y} - \frac{\partial v }{\partial x}\frac{\partial u }{\partial y} = \left( \frac{\partial u }{\partial x}\right)^2 + \left( \frac{\partial v}{\partial x}\right)^2$$
[To derive the second expression, I used the Cauchy-Riemann equations $\frac{\partial v}{\partial y } = \frac{\partial u }{\partial x}$ and  $\frac{\partial u}{\partial y } = - \frac{\partial v }{\partial x}$]
Thus
$$ |f'|^2 =  \det Dg.$$
which means that $f'$ is zero if and only if $Dg$ is singular.
