I don't know the relationship about complex differentiable and differentiable everywhere, actually.
From WolframMathWorld : Complex Differentiable,
The function is complex differentiable when $f(z)$ has derivative, has continuous partial derivative, and satisfies the Riemann Equation. It said that analytic is equivalent with Complex differentiable.
Suppose i have :
$$f(z)=\cos x \cosh y +i \sin x \sinh y$$
The function is indeed continuous since it has no singularity. But it fails on Cauchy-Riemann Equation.
Is that means it's not complex differentiable? Then, how about differentiable everywhere?
Actually the first thing i thought of before i don't know the definition when i heard "differentiable everywhere" was just involving derivative on the complex plane and has no singularity.
Best Answer
I believe the key to understanding the link between complex differentiability and Cauchy-Riemann equations, is to understand why the Cauchy-Riemann equations must hold.
Your last statement is exactly what it means to be complex differentiable. A function $f$ is complex differentiable at $z$, exactly when $lim_{h\rightarrow 0} \frac{f(z+h)-f(z)}{h}$ exists. Or equivalently if $$f(z+h)= f(z) + hf'(z) + h \phi(h) $$ for some function $\phi$ satisfying that $lim_{h\rightarrow 0}\phi(h)=0$. Notice that $h$ can approach $0$ from infinitely many different paths, so complex differentiability is a much stronger criteria than regular differentiation.
Now a function $f:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ is said to be differentiable at $x\in \mathbb{R}^2$ if $$ f(x+h) = f(x) + Jh + \phi(h)||h|| $$ again with $\phi$ continous at $0$ and J is a linear map (more precisely the Jacobian matrix).
If we have a function $f=(u,v)^T$, then for $\tilde{f}=u+iv$ to be complex differentiable, the equations above suggests that $J$ needs to correspond to multiplication by a complex number. But how does multiplication of complex numbers look as linear maps on $\mathbb{R}^2$? Since $(a+bi)(c+di)=(ac-bd)+(ad+cb)i$, we get for the vectorized computations that $$\begin{pmatrix} a & -b \\ b & a \end{pmatrix}\begin{pmatrix} c \\ d \end{pmatrix}= \begin{pmatrix} ac-bd \\ ad+cb \end{pmatrix}$$ Now plugging in the definition of the Jacobian gives $$J=\begin{pmatrix} \frac{du}{dx} & \frac{du}{dy} \\ \frac{dv}{dx} & \frac{dv}{dy} \end{pmatrix} = \begin{pmatrix} a & -b \\ b & a \end{pmatrix} $$ which gives us exactly the Cauchy-Riemann equations.