# Confusing on how to prove complex function is differentiable everywhere?

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.

• If it fails C-R (as this does) then it's not complex differentiable. Dec 1, 2019 at 0:45
• @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? Dec 1, 2019 at 0:47
• @user516076 complex differentiable, which implies analytical, is stronger than real differentiable in general.
– Vim
Dec 1, 2019 at 0:48
• So, it's not differentiable everywhere? Please let me know where is the point is not differentiable. Thanks Dec 1, 2019 at 0:50
• 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. Dec 1, 2019 at 0:54

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.

• Differentiability is a local property. The notion of being differentiable everywhere simply means that the function is differentiable at all points. Dec 1, 2019 at 1:46
• Hi, what do you think about my function? Is it not differentiable everywhere? Thanks Dec 1, 2019 at 1:59
• 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. 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 $$(x,y)$$.

• 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. Dec 1, 2019 at 4:47
• Then complex differentiable is different with differentiable everywhere? is that so? Dec 1, 2019 at 4:50
• @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. Dec 1, 2019 at 6:20
• Ok, then it's just differentiable at Real plane right? Not in the complex plane? Dec 1, 2019 at 6:22
• @user516076 You may say so, if you identify $f$ with $F$. Dec 1, 2019 at 6:24