I think I am a bit confused about the definition of (complex) differentiability. Yes, I know that's stupid, but I am hoping that someone could clear it up for me. I know that the definition of (complex) differentiability is when $\lim\limits_{h\to 0}{f(z+h)-f(z)\over h}$ exists.

So, is $|z|^2$ considered differentiable? What I think is it is only differentiable at $z=0$ since at any other point if we take $f(z+h)-f(z)\over h$ as $h\to 0$ along a contour line of $|z|^2$ then the limit is $0$ whereas if we take a path say perpendicular to the contour lines, the "gradient" wouldn't be $0$, right? But then if this is true then all complex functions that are "not flat" would not be differentiable, so I must be wrong. Could someone kindly explain to me what is going on? Sorry for my stupidity!

  • 1
    $\begingroup$ Well, what do you mean by a countour line of $|z|^2$? If you mean $|h|^2=C$, than this doesn't approaches zero.... $\endgroup$ – N. S. Jan 30 '12 at 15:05
  • 3
    $\begingroup$ Being a little nit picky, you do not need to add "no matter what path h takes to tend to 0." This is implicit in the definition of a limit. $\endgroup$ – Henrik Jan 30 '12 at 15:18
  • 1
    $\begingroup$ @user22705: Thanks for pointing that out. $\endgroup$ – James C Jan 30 '12 at 15:24
  • 1
    $\begingroup$ The difference in the case of $Im(z)$ is that you make $h \to 0$ on two different paths which GO to $0$. And the problem with your approach is that when you think of level curves, you have the picture of real functions in mind, but I think that picture only is accurate if your function takes REAL values.... What is a level curve of the function $f(z)=z^2$? Keep in mind that the graph of this is in $C^2$ which is actually four dimensional (as space over the reals)... And the equation $z^2=C$ is not really a curve (over the reals).... $\endgroup$ – N. S. Jan 30 '12 at 15:56
  • 1
    $\begingroup$ BTW: It is true that if $f: C \rightarrow R$ is differentiable, then it must be constant, and in that case your intuition, is probably right... $\endgroup$ – N. S. Jan 30 '12 at 15:58

The function $z\mapsto|z|^2$ is not the typical "complex function" that aspires to be analytic, because it is real-valued to begin with. The latter fact makes it possible to talk about contour-lines, while a truly complex function $f:\ {\mathbb C}\to{\mathbb C}$ has no contour lines: The solutions to an equation of the form $f(z)=w_0\in{\mathbb C}$ typically form a set of isolated points in the $z$-plane.

Any "complex function" $$f:\quad {\mathbb C}\to{\mathbb C}, \qquad z\mapsto w:=f(z)$$ can be viewed as a vector-valued function $${\bf f}:\quad{\mathbb R}^2\to{\mathbb R}^2\ , \qquad{\bf z}\mapsto{\bf w}={\bf f}({\bf z})$$ resp. as a pair of functions $$(x,y)\ \mapsto \bigl(u(x,y),v(x,y)\bigr)$$ via the identifications ${\bf z}:=(x,y)=x+iy=:z$, and similarly for ${\bf w}$. The Jacobian $$J_{\bf f}({\bf z}_0) =\left[\matrix{u_x(x_0,y_0) & u_y(x_0,y_0) \cr v_x(x_0,y_0) & v_y(x_0,y_0) \cr}\right]$$ of such an ${\bf f}$ at a given point ${\bf z}_0=(x_0,y_0)$ can be any $(2\times2)$-matrix and describes a certain linear map from the tangent space at ${\bf z}_0$ to the tangent space at ${\bf w}_0={\bf f}({\bf z}_0)$.

When such a function $f$ resp. ${\bf f}$ is analytic then the Jacobian of ${\bf f}$ at a point ${\bf z}_0$ can no longer be an arbitrary matrix. The fact that one has an approximation of the sort $$f(z_0+h)-f(z_0)= C\ h + o(|h|)\qquad (h\to 0\in{\mathbb C})$$ for some complex factor $C=:f'(z_0)\in{\mathbb C}$ implies that $J_{\bf f}({\bf z}_0)$ is a matrix of the form $$\left[\matrix{A&-B\cr B & A\cr}\right]\ .$$ Geometrically this means that ${\bf f}'({\bf z}_0)$ is a (proper) similarity with stretching factor $\sqrt{A^2+B^2}$ and turning angle $\phi:=\arg(A,B)$. The $A$ and $B$ appearing in this matrix are related to $f'(z_0)$ via $f'(z_0)=A+iB$.

For an analytic function $f$ these facts must be true not only at a single point $z_0$ in the domain of $f$ but for all points $z_0$ in the domain of $f$. This is expressed in the so-called Cauchy-Riemann differential equations $u_x=v_y$, $u_y=-v_x$.

  • 1
    $\begingroup$ That's a great first line :) $\endgroup$ – Bruno Stonek Jan 30 '12 at 20:39
  • $\begingroup$ Thank you very much, this is a very good explanation! $\endgroup$ – James C Jan 30 '12 at 20:42
  • 6
    $\begingroup$ I don't see how this answered the question... $\endgroup$ – AmadeusDrZaius Jul 10 '14 at 15:14
  • $\begingroup$ Christian, what do you think is a good sense of differentiability of a function from the complex numbers to the real numbers? Recasting as a function from $\mathbb R^2$ to $\mathbb R$ seems a bit crude. $\endgroup$ – shuhalo Mar 18 '17 at 21:52

Try what happens if you take $f(z) = |z|^2 = z\overline{z}$ in the definition of differentiability. You get $\frac{(z+h)\overline{(z+h)} - z\overline{z}}{h}$ which simplifies to $z\frac{\overline{h}}{h} + \overline{z} + \overline{h}$. Now, the last term $\overline{h}$ has the same absolute value as $h$ does, so it will tend to zero, when h goes to zero. So the only possible problem here would be $\frac{\overline{h}}{h}$ in the first term. This is the same as $(\frac{|h|}{h})^2$. But we can write every $h$ uniquely in the form $r e^{i\phi}$, so the fraction $\frac{|h|}{h}$ simplifies to $e^{-2 i \phi}$, where $\phi$ is determined by $h$. So, no matter how close to zero $h$ gets, $(\frac{|h|}{h})^2$ will describe a whole unit circle in the plane and thus in can't possibly have a limit as $h\to 0$, so the function is not complex-differentiable at any point, except indeed at the point $z=0$, where the first term is $0 \frac{\overline{h}}{h}$ and thus equal to zero. (I hope this helps.)

Complex differentiability is quite a strong condition, so many not-so-ugly functions are in fact not complex differentiable.

  • $\begingroup$ Thanks, Dejan, so is it true that all functions that are not flat are not (complex) differentiable? But there are functions like $\cos(z)$ which is analytic so must be differentiable but is not "flat" so we could again choose to go along a contour along another path and not get a limit, no? $\endgroup$ – James C Jan 30 '12 at 15:50
  • $\begingroup$ @James: well all polynomials in $z$ are complex-differentiable for example. The proof is the same as in the real-valued case. The function $cos(z)$ is complex-differentiable, so it does not matter which path you take, you will always get the same limit. (I'm not completely sure what you mean by these "contours" though.) $\endgroup$ – Dejan Govc Jan 30 '12 at 15:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.