# Prove complex function is not holomorphic using definition (without Cauchy-Riemann equations)

I'm just starting to learn complex analysis and currently reading the book "Complex analysis" by Stein and Shakarchi.

Since the fact of the function being holomorphic appears to be pretty important, I'm starting with it.

Here's the definition: The function $f$ is holomorphic at the point $z_0 \in \Omega$ if the quotient $$\frac{f(z_0+h)-f(z_0)}{h}$$ converges to a limit when $h \to 0$.

Then we are given two examples. First of the function $f(z)=z$ being holomorphic on any open set in $\Bbb C$, and $f'(z) = 1$. And second of the function $f(z)=\bar z$ being not holomorphic.

While I think I understand first example, I cannot see how the second will be different, how the fact of the complex conjugate changes the set up? Basically, why $f(z)=\bar z$ is not holomorphic by definition? I am aware of the Cauchy-Riemann equations, but I want to see the proof based on the definition (especially that the fact is claimed at this point in the book).

I tried to search for an answer but didn't find anything particularly related. I also think I might miss some very basic concept. With all that said I will really appreciate your input.

• I had this same concern. Instead of just testing the CR equations, try using them from first principles and setting up the limit definitions with $\overline{z}$ specifically. It helped me, anyway. – The Count Apr 17 '17 at 17:29
• @TheCount Thanks for the comment! I'm not sure if I completely understoo it, but from what I did :) I'd say that the goal was really to look at this problem as if I had no idea about CR equations. – awesome_Leni Apr 17 '17 at 20:07
• that's exactly what I meant. I hope it helped! – The Count Apr 17 '17 at 23:06

A good rule of thumb for complex analysis is that, if something has gone wrong, $0$ is almost certainly to blame.

Here is the difference quotient for $\overline{z}$ at $0$ taken along two different paths in the complex plane:

Tending to $0$ along the positive imaginary axis produces:

$\lim_{h\rightarrow0}\frac{\overline{\left(0+ih\right)}-\overline{0}}{ih}=\lim_{h\rightarrow0}\frac{-ih}{ih}=-1$

On the other hand, tending to $0$ along the positive real axis produces:

$\lim_{h\rightarrow0}\frac{\overline{\left(0+h\right)}-\overline{0}}{h}=\lim_{h\rightarrow0}\frac{h}{h}=1$

which is completely different. Since the limits do not agree, the limit of this difference quotient as $h$ tends to $0$ in $\mathbb{C}$ does not exist. Thus, the conjugate function is not complex differentiable at $0$.

Generally, when working "from first principles" as it were, one has to rely on the two-dimensionality of $\mathbb{C}$ (along with its algebraic/"arithmetic" properties) to show that things are different than the case of single real-variable analysis. Indeed, it is exactly because the condition of holomorphy requires these two-dimensional limits to exist that holomorphic functions are so marvelously well-behaved (in stark contrast to functions of multiple real variables).

• This is exactly the kind of explanation I was looking for, thank you very much! – awesome_Leni Apr 17 '17 at 18:56
• Happy to hear so, and am glad to have been helpful. :) – MCS Apr 17 '17 at 19:17
• Indeed I missed the basic thing: for some reason I was considering $f(\bar z_0 + h)$ and not $f(\overline {{z_0 + h}})$ – awesome_Leni Apr 17 '17 at 20:01
• I really like this answer, and I will save it in my mental back pocket. – The Count Apr 18 '17 at 0:39

Long comment about problems like that ("prove that ... isn't complex differentiable without using C-R").

Let be $c = a + bi$. It's easy to see that the condition $$\exists \lim_{z\to c}\frac{f(z) - f(c)}{z - c}$$ is equivalent to "$f$ considered as function of two real variables is $\Bbb R$-differentiable at $(a,b)$ and the differential is $\Bbb C$-linear". And differential being $\Bbb C$-linear $\iff$ C-R equations. Then, there are two possibilities for $f$ not being $\Bbb C$-differentiable at some point $c$:

• $f$ considered as function of two real variables isn't $\Bbb R$-differentiable at $(a,b)$: the $\Bbb R$-differential does not exists.

• $f$ considered as function of two real variables is $\Bbb R$-differentiable at $(a,b)$ but fails to verify the C-R equations: the $\Bbb R$-differential does exists but isn't $\Bbb C$-linear.

The examples usually seen (conjugation, modulus...) are $\Bbb R$-differentiable at (almost) every point but they do not satisfy C-R.

But what are the partial derivatives? Limits in a subspace. When we check that $\lim_{z\to c}\frac{f(z) - f(c)}{z - c}$ does not exists finding that the limits in different subspaces are different (like in the other answer), we are actually checking C-R.

Bottom line: posing the problems of this type without clarification can mislead the students.