It will be shown that a complex function $\,f = u + i.v\,$ is complex
differentiable to $\,z = x + i.y\,$ if and only if it is real differentiable at $\,z\,$ and if the partial derivatives of $\,u\,$ and $\,v\,$ to $\,x\,$ and $\,y\,$ obey the so
called Cauchy-Riemann Equations:
$\;\partial u /\partial x = \partial v /\partial y\;$ and
$\;\partial u /\partial y = - \partial v /\partial x$ .
Two special cases
The complex derivative $f'(z)$ of a complex function $f(z)$ with $z$ complex is
defined as follows (with $\Delta z$ complex as well).
$$
f'(z) = \lim_{\Delta z\rightarrow 0} \frac{f(z + \Delta z) - f(z)}{\Delta z}
$$
A complex function can always be written in real valued components $(x,y,u,v)$
as:
$$
f(z) = u(x,y) + i.v(x,y) \qquad \mbox{where} \quad z = x + i.y
$$
Complex differentiation is independent of the direction, the way $z$ approaches
zero. Two special cases are distinguished: differentiation in the $x$-direction
and differentiation in the $y$-direction. In the $x$-direction it is found that
$\Delta z = \Delta x$ and the complex derivative is:
$$
f'(z) = \lim_{\Delta x\rightarrow 0}
\frac{u(x + \Delta x,y) - u(x,y) + i\left[v(x + \Delta x,y) - v(x,y)\right]}
{\Delta x} = \frac{\partial u}{\partial x} + i.\frac{\partial v}{\partial x}
$$
In the $y$-direction it is found that $\Delta z = i.\Delta y$ and the complex
derivative is:
$$
f'(z) = \lim_{\Delta y\rightarrow 0}
\frac{u(x,y + \Delta y) - u(x,y) + i\left[v(x,y + \Delta y) - v(x,y)\right]}
{i.\Delta y} = -i.\frac{\partial u}{\partial y} + \frac{\partial v}{\partial y}
$$
Two complex numbers are equal if and only if the real and the imaginary parts
are equal:
$$
f'(z) = \frac{\partial u}{\partial x} + i.\frac{\partial v}{\partial x} =
-i.\frac{\partial u}{\partial y} + \frac{\partial v}{\partial y} \quad \Longrightarrow \quad
\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \quad \mbox{and} \quad
\frac{\partial v}{\partial x} = - \frac{\partial u}{\partial y}
$$
These are the well-known Cauchy-Riemann equations. Conclusion: if a function $\,f = u + i.v\;$ is complex differentiable, then the real and imaginary parts $\,u\,$ and $\,v\,$ of $\,f\,$ satisfy the Cauchy-Riemann equations at $\,z = x + i.y$ .
Real Differentiable
The complex derivative $f'(z)$ of a complex function $f(z)$ with $z$ complex is
defined as follows (with $\Delta z$ complex as well).
$$
f'(z) = \lim_{\Delta z\rightarrow 0} \frac{f(z + \Delta z) - f(z)}{\Delta z}
$$
Consequently:
$$
f(z + \Delta z) = f(z) + f'(z) \Delta z \qquad \mbox{for} \quad z \rightarrow 0
$$
A complex function can always be written in real valued components $(x,y,u,v)$
as:
$$
\begin{cases} f(z) = u(x,y) + i.v(x,y) \\ f'(z) = u'(x,y) + i.v'(x,y) \end{cases}
\qquad \mbox{where} \quad z = x + i.y
$$
Consequently, for $\Delta x \rightarrow 0$ and $\Delta y \rightarrow 0$ :
$$
u(x + \Delta x,y + \Delta y) + i . v(x + \Delta x,y + \Delta y) =
u(x,y) + i . v(x,y) $$ $$ + \left[ u'(x,y) + i . v'(x,y) \right]
\left[ \Delta x + i \Delta y \right] \quad \Longrightarrow \quad
$$ $$
u(x + \Delta x,y + \Delta y) = u(x,y) + u'(x,y)\Delta x - v'(x,y)\Delta y
$$ $$
v(x + \Delta x,y + \Delta y) = v(x,y) + v'(x,y)\Delta x + u'(x,y)\Delta y
$$
Meaning that, if a function $f = u + i.v$ is complex differentiable at $ z =
x + i.y$ , then its real and imaginary parts $(u,v)$ are real differentiable
at ($x,y)$ . On the other hand it is known from real analysis that:
$$
u(x + \Delta x,y + \Delta y) = u(x,y) + \frac{\partial u}{\partial x}\Delta x
+ \frac{\partial u}{\partial y}\Delta y
\qquad \mbox{for} \quad (\Delta x,\Delta y) \rightarrow (0,0)
$$ $$
v(x + \Delta x,y + \Delta y) = v(x,y) + \frac{\partial v}{\partial x}\Delta x
+ \frac{\partial v}{\partial y}\Delta y
\qquad \mbox{for} \quad (\Delta x,\Delta y) \rightarrow (0,0)
$$
This can only be consistent if the Cauchy-Riemann equations are indeed valid:
$$
\frac{\partial u}{\partial x} = u'(x,y) = \frac{\partial v}{\partial y} \qquad ; \qquad
\frac{\partial v}{\partial x} = v'(x,y) = - \frac{\partial u}{\partial y}
$$
It is known from real analysis that a function $\,f(x,y)\,$ is total differentiable at $(a,b)$ , if and only if the partial derivatives $\,\partial f /\partial x\,$ and $\,\partial f /\partial y\,$ exist in a neighborhood of $(a,b)$ and both are continuous there. Thus making the picture complete.
Independent of angle
The reverse question is: if the Cauchy-Riemann equations hold for the real and
imaginary parts of a complex function $\,f$ , is $\,f\,$ complex differentiable then ? Let's write the complex derivative with $\,\Delta z =
r.e^{i\theta}$ . The reason is that the complex derivative must be independent
of any (real) angle $\theta$ while the (real) distance $r$ from $z + \Delta z$
to $z$ approaches zero. So this is what we do:
$$
f'(z) = \lim_{r\rightarrow 0}
\frac{f(z + r.e^{i\theta}) - f(z)}{r.e^{i\theta}}
$$
Remember that $f(z) = u(x,y) + i.v(x,y)$ where $z = x + i.y$ . Also remember
the jewel formula by Euler $\;e^{i\theta} = \cos(\theta) + i.\sin(\theta)$ .
Giving:
$$
f'(z) = \lim_{r\rightarrow 0}
\frac{u(x + r\cos(\theta),y + r\sin(\theta)) - u(x,y)}
{r\cos(\theta) + i.r\sin(\theta)}
$$ $$
+ \; i \; \lim_{r\rightarrow 0}
\frac{v(x + r\cos(\theta),y + r\sin(\theta)) - v(x,y)}
{r\cos(\theta) + i.r\sin(\theta)}
$$
Given the real differentiable functions $[u,v](x,y)$ , for $[\Delta x,\Delta y] \rightarrow 0$ :
$$
[u,v](x+\Delta x,y+\Delta y) - [u,v](x,y) = \frac{\partial [u,v]}{\partial x}(x,y)\, \Delta x + \frac{\partial [u,v]}{\partial y}(x,y)\, \Delta y
$$
With $\left[\Delta x,\Delta y\right] = \left[r\cos(\theta), r\sin(\theta)\right]$ :
$$
f'(z) = \lim_{r\rightarrow 0}
\frac{\partial u /\partial x.r\cos(\theta) + \partial u /\partial y.r\sin(\theta)}
{r\cos(\theta) + i.r\sin(\theta)}
$$ $$
+ \, i \, \lim_{r\rightarrow 0}
\frac{\partial v /\partial x.r\cos(\theta) + \partial v /\partial y.r\sin(\theta)}
{r\cos(\theta) + i.r\sin(\theta)}
$$
Taking the limit for $r\rightarrow 0$ is easy:
$$
f'(z) =
\frac{\partial u /\partial x.\cos(\theta) + \partial u /\partial y.\sin(\theta)}
{\cos(\theta) + i\sin(\theta)}
\; + \; i \; \frac{\partial v /\partial x.\cos(\theta) + \partial v /\partial y.\sin(\theta)}
{\cos(\theta) + i\sin(\theta)}
$$
Substitute herein the Cauchy-Riemann equations (by a copy and paste from the
preceding subsection):
$$
\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \quad \mbox{and} \quad
\frac{\partial v}{\partial x} = - \frac{\partial u}{\partial y}
$$
Just doit:
$$
f'(z) =
\frac{\partial u /\partial x.\cos(\theta) - \partial v /\partial x.\sin(\theta)}
{\cos(\theta) + i\sin(\theta)}
\; + \; i \; \frac{\partial v /\partial x.\cos(\theta) + \partial u /\partial x.\sin(\theta)}
{\cos(\theta) + i\sin(\theta)}
$$
$$
f'(z) = \frac{\partial u}{\partial x} \; \frac{\cos(\theta) + i\sin(\theta)}
{\cos(\theta) + i\sin(\theta)} \; + i \;
\frac{\partial v}{\partial x} \; \frac{\cos(\theta) + i\sin(\theta)}
{\cos(\theta) + i\sin(\theta)}
= \frac{\partial u}{\partial x} + i \, \frac{\partial v}{\partial x}
$$
Which indeed is independent of any angle $\theta$ . Consequently, if the function $\,f = u + i.v\,$ is real differentiable at $\,z = x + i.y\,$ and if the partial derivatives of $\,u\,$ and $\,v\,$ to $\,x\,$ and $\,y\,$ obey the Cauchy-Riemann Equations, then the complex derivative of $\,f\,$ is independent of the direction in which we differentiate.
