# Why isn't $f(z)=\bar{z}$ complex differentiable

It is quite easy to see why $f(z)=\bar{z}$ isn't complex differentiable.

$\frac{\partial{u}}{\partial{x}}=1\neq-1=\frac{\partial{v}}{\partial{y}}$

But I struggle to see why this is the case.

Visualization of the conjugate function in my head is simply flipping the complex plane upside down. I don't see how this creates any discontinuities.

Am I having the wrong visualization or have I misunderstood complex differentiability?

• The map $f$ is not complex differentiable because it reverses orientation, just like the mirror in your room. What could be more visual? Feb 28, 2018 at 12:20
• "It is quite easy to see why $f(z)=\bar{z}$ isn't complex differentiable [...] But I struggle to see why " - Ok. Mar 19, 2018 at 3:26

Of course it doesn't satisfy one of the possible definitions (Cauchy-Riemann equations, the specific limit does not exist), but in my opinion that doesn't give you much intuition - at least of the geometric kind. Hopefully the following will compensate for that.

TL;DR:

If $f$ is differentiable at a point $z_0$, then there's the linear function $$(Df)(z_0)\colon \mathbb C \to \mathbb C;\quad z \mapsto f'(z_0)z$$ that approximates $f$ well around $z_0$. Multiplication by a complex number is a rotation or a scaling of the complex plane, thus it keeps orientation.

These imply that $f$ has to keep orientation locally, around $z_0$. Conjugation is a reflection so it flips orientation, therefore it cannot be differentiable at any point in the complex sense.

If $f\colon \mathbb C \to \mathbb C$ is complex-differentiable, then it is differentiable as an $\mathbb R^2 \to \mathbb R^2$ function, and it locally almost keeps angles and circles. To see this, look at one equivalent version of the complex-differentiability: $$f(z_0 + re^{it}) = f(z_0) + re^{it}f'(z_0) + \omicron(r),$$ where $z_0 \in \mathbb C, 0 \leq r, 0 \leq t \leq 2\pi$. In fact, as a demonstration have a picture of the image of two circles around $1 + i$ by $\operatorname{id_{\mathbb C}}^2$.

The black,$\{z_0 + re^{it} \mid t \in [0,2\pi]\}$ ones are our original circles ,
the reds, $\,\,\color{red}{f(\{z_0 + re^{it} \mid t \in [0,2\pi]\})}$ are the blacks' images by $f$, and
the blues, $\color{blue}{\{f(z_0) + re^{it}f'(z_0) \mid t \in [0,2\pi] \}}$ are the first order approximation of the red ones.

If we pick $z_0 = 1 + i, r=0.1$ and $0.2$, then we get, in order: As you can see in the $r=0.1$ case the red and blue circles are almost equal. As $r \to 0$, the difference becomes smaller and smaller. That's what I meant by locally almost keeping circles.

Here's another picture, displaying the image of a point from each of the two black circles. The black and blue circles represent the same as before, the green circles are due to the $\omicron(r)$ part.

Look for the points $\color{red}{a}, \color{blue}{a'}, \color{red}{f(a)}$ and $\color{purple}{b},\color{blue}{b'}, \color{purple}{f(b)}$. In fact, in the $r=0.1$ case the green circle is so small, you can barely see any of it (this is why the red and blue circles from a moment ago were almost equal).

As you can see, the image of a circle is essentially the sum of multiple circles, in this case, two; however, if $f$ has lots of non-zero derivatives, then there'll be lots of circles, and each "new" circle will have smaller radius, but higher frequency than the previous one (assuming $r < 1$).

Anyway, this is not so surprising: if you consider the Taylor series of a holomorphic $f$ at a point, and plug in a point from a circle around that point, you'll get $$f(z_0 + re^{it}) = \sum\limits_{n=0}^\infty \frac{f^{(n)}(z_0)}{n!}r^ne^{int}.$$ Here the $n$th term will correspond to the circle $\left\{\frac{f^{(n)}(z_0)}{n!}r^ne^{int} \mid t \in [0,2\pi]\right\}$. OK, back to the original question.

Hopefully I've convinced you that the $\omicron(r)$ part is not significant. (Of course only so if the first derivative is non-zero, but in the other cases we can say something similar too.)

Let's say a word about locally almost keeping angles. Remember $$f(z_0 + re^{it}) = f(z_0) + re^{it}f'(z_0) + \omicron(r).$$ Picking two numbers, $z_0 + re^{it}, z_0 + re^{is}$ the angle between them that is kept locally is $(t-s)$. Their images are $$f(z_0) + re^{it}f'(z_0) + \omicron(r), \text{and } f(z_0) + re^{is}f'(z_0) + \omicron(r).$$ As you can see, the local (i.e. looking from the point $f(z_0)$) angle between them is again, approximately $(t-s)$.

Now let's really get back to your original question. For $f(z) = \bar z$, assuming differentiability at $z_0$, the following would be true for all $t\in [0,2\pi]$: \begin{align} f(z_0 + re^{it}) &= f(z_0) + re^{it}f'(z_0) + \omicron(r) \\ \iff re^{-it} &= re^{it}f'(z_0) + \omicron(r). \end{align} This is absurd. Of course we could also note that conjugation locally flips angles. Going back to our points $z_0 + re^{it}, z_0 + re^{is}$, the angle is still $(t-s)$. Their images now are $$\bar{z_0} + re^{-it}, \text{ and } \bar{z_0} + re^{-is},$$ meaning that the appropriate angle is $-(t-s)$. Thus again, conjugation can not be complex-differentiable.

• Very nice answer, +1. Mar 27, 2017 at 14:08

Viewed as a real map the complex conjugation map $$z \mapsto \bar z$$ is differentiable (in fact, linear): Under the usual identification $$\Bbb R^2 \leftrightarrow \Bbb C$$, $$(x, y) \leftrightarrow x + iy$$, it is given by $$(x, y) \to (x, -y)$$.

On the other hand, it is not complex-differentiable, which is a much stronger property. By definition, $$f$$ is complex-differentiable at $$z_0$$ if the usual limit $$\lim_{h \to 0} \frac{f(z_0 + h) - f(z_0)}{h}$$ of the difference quotient of $$f$$ exists, in which case we define $$f'(z_0)$$ to be its value. Critically, $$h$$ is here a complex variable: In the case of the conjugation map, the limit simplifies to $$\lim_{h \to 0} \frac{\bar h}{h},$$ and for real $$h$$ we have $$\frac{\bar h}{h} = \frac{h}{h} = 1$$, but for imaginary $$h$$ we have $$\frac{\bar h}{h} = \frac{-h}{h} = -1$$. Thus, the limit does not exist for any $$z_0$$, and the map is not complex-differentiable anywhere.

• quick question why h bar become -h?
– EM4
Oct 1, 2020 at 23:18
• @EM4 When $h$ is purely imaginary, complex conjugation simply flips its sign, since $\operatorname{Re}(h) = 0$. Oct 8, 2020 at 6:07

$f$ does not create any discontinuities ! $f$ is continuous, but not complex differentiable.

Let $z_0 \in \mathbb C$ and $t \in \mathbb R$. Then:

$\frac{f(z_0+t)-f(z_0)}{t}=1 \to 1$ for $t \to 0$,

but

$\frac{f(z_0+it)-f(z_0)}{it}=-1 \to -1$ for $t \to 0$.

To be complex differentiable (holomorphic) require more than differentiability for a real function (that, anyway, is more than continuity) because we need that the limit of the incremental quotient have to be the same ''from any direction'' in the Argand Plane. You can see a good illustration of the case $f(z)=\bar z$ here.