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.

  • $\begingroup$ If it fails C-R (as this does) then it's not complex differentiable. $\endgroup$ Dec 1, 2019 at 0:45
  • $\begingroup$ @LordSharktheUnknown Actually my tittle is should be how to prove function is differentiable everywhere which the function is complex. Is differentiable everywhere the same as complex differentiable? $\endgroup$
    – user516076
    Dec 1, 2019 at 0:47
  • $\begingroup$ @user516076 complex differentiable, which implies analytical, is stronger than real differentiable in general. $\endgroup$
    – Vim
    Dec 1, 2019 at 0:48
  • $\begingroup$ So, it's not differentiable everywhere? Please let me know where is the point is not differentiable. Thanks $\endgroup$
    – user516076
    Dec 1, 2019 at 0:50
  • $\begingroup$ Anyplace where Cauchy-Riemann fails is a place where it's not differentiable. Cauchy-Riemann is the definition of differentiability for functions of a complex variable. $\endgroup$ Dec 1, 2019 at 0:54

2 Answers 2


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.

  • $\begingroup$ Differentiability is a local property. The notion of being differentiable everywhere simply means that the function is differentiable at all points. $\endgroup$ Dec 1, 2019 at 1:46
  • $\begingroup$ Hi, what do you think about my function? Is it not differentiable everywhere? Thanks $\endgroup$
    – user516076
    Dec 1, 2019 at 1:59
  • 1
    $\begingroup$ The question has somewhat been answered in the other answer, but just to clarify: If you say that a complex function is "differentiable everywhere", then from the context it has be interpreted as "complex differentiable everywhere", hence cauchy-riemann equations must hold everywhere. Saying that $(Re(f),Im(f))$ is differentiable everywhere on $\mathbb{R}^2$ is a different statement. $\endgroup$ Dec 1, 2019 at 11:42

I would like to write something about the geometric intuition.

A complex function $f(x+iy)$ could be viewed as a function $F: \mathbb{R}^2 \to \mathbb{R}^2$ defined by $F(x,y) = (u(x,y),v(x,y))$ where $u,v$ are the real and imaginary parts of $f$.

The Cauchy-Riemann equations say that $f$ is complex differential at a point $z_0 = x_0 + iy_0$ iff $F$ is differential at $(x_0,y_0)$ AND the Jacobian matrix $J$ of $F$ at $(x_0,y_0)$ represents a 'conformal' linear transformation (i.e., if $\vec{a},\vec{b} \in \mathbb{R}^2$ then $\angle(\vec{a},\vec{b}) = \angle(J\vec{a},J\vec{b})$).

A reason for this, is that if $f'(z_0)$ is defined then it is a complex number $a + ib$ and $f(z_0 + \Delta z) - f(z_0)$ can be approximated by $(a+ib)\Delta z$. Multiplication by $(a+ib)$ could be viewed as a linear transform on $\mathbb{R}^2$. Then it is clear from Euler's formula that such a linear transform is a composition of scalar multiplication and rotation, thus 'conformal'. Conversely, by easy linear algebra, any conformal linear transformation on the plane is of the form $$ \begin{pmatrix} a & -b \\ b & a \end{pmatrix}, $$ so can be represented by a complex number.

As for $f(x+iy) = \cos x \cosh y + i \sin x \sinh y$, the corresponding $F$ on $\mathbb{R}^2$ is differential on the whole plane and at a point $(x,y)$ has a Jacobian matrix like $$ \begin{pmatrix} -\sin x \cosh y & \cos x \sinh y \\ \cos x \sinh y & \sin x \cosh y \end{pmatrix} $$ which is not conformal at every $(x,y)$.

  • $\begingroup$ As for $f(x+iy) = \cos x \cosh y + i \sin x \sinh y$, the corresponding $F$ on $\mathbb{R}^2$ is differential on the whole plane and at a point $(x,y)$ . So, my function is differentiable everywhere but it's not complex differentiable? Sorry for my retard. $\endgroup$
    – user516076
    Dec 1, 2019 at 4:47
  • $\begingroup$ Then complex differentiable is different with differentiable everywhere? is that so? $\endgroup$
    – user516076
    Dec 1, 2019 at 4:50
  • $\begingroup$ @user516076 $f$ being complex differentiable at $z = x + iy$ is different from the corresponding $F$ on the plane being differentiable at $(x,y)$. An extra condition ($J$ being conformal) is needed. $\endgroup$
    – Wei Wang
    Dec 1, 2019 at 6:20
  • $\begingroup$ Ok, then it's just differentiable at Real plane right? Not in the complex plane? $\endgroup$
    – user516076
    Dec 1, 2019 at 6:22
  • $\begingroup$ @user516076 You may say so, if you identify $f$ with $F$. $\endgroup$
    – Wei Wang
    Dec 1, 2019 at 6:24

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