# Examples of holomorphic functions

Since we’re going to be studying holomorphic functions, it’s probably worth seeing a few examples. I’m going to give a bunch of functions, and then prove that they’re holomorphic. I suppose the easiest would be the constant functions. Pick any ${c\in{\mathbb C}}$, and define ${f:{\mathbb C}\rightarrow{\mathbb C}}$ by ${f(z)=c}$. It’s clear that ${f}$ is continuous. It’s also a super easy computation to see that ${f'(z)=0}$ for every ${z\in{\mathbb C}}$, so ${f'}$ exists everywhere, meaning ${f}$ is holomorphic.

Well, this computation was easy, but as a general rule, explicitly computing the derivative, and checking for continuity is tedious. Instead, I’m going to make use of the theorem we proved last time:

Theorem 1 Let ${f:U\rightarrow{\mathbb C}}$ where ${U}$ is an open subset of ${{\mathbb C}}$. Write ${f=u+iv}$. Then ${f\in{\mathcal H}(U)}$ if and only if all of the partial derivatives ${u_x}$, ${u_y}$, ${v_x}$, and ${v_y}$ exist, and satisfy

$\displaystyle u_x=v_y,u_y=-v_x$

For the constant function, we can also see that all of the partials exist, and that they satisfy the Cauchy-Riemann equations. Indeed, if ${c=a+bi}$, ${f=u+iv}$ by ${u(z)=a}$ and ${v(z)=b}$. Then ${u_x=0=v_y}$ and ${u_y=0=-v_x}$.

Let’s move on to something slightly more interesting. All polynomials are also holomorphic functions. The derivative is exactly what you would expect, and it’s easy to see that the partial derivatives all exist. We should check the Cauchy Riemann equations too, but to save space, I’ll leave that as an exercise.

Here’s another function which is holomorphic, but not a polynomial: Let’s try the function ${e^z}$. We know what this means when ${z}$ is a real number, but more generally, what should it mean. If ${z=x+iy}$, I define ${e^z=e^{x+iy}=e^x\cos y+ie^x\sin y}$. This jives with our normal definition of ${e^x}$ for ${x\in{\mathbb R}}$, because the two definitions coincide when the imaginary part of ${z}$ is zero. Why is this holomorphic? Well, we’ve multiplied and added a bunch of functions which we know have well-defined partials, so the result has well-defined partials as well. We can just check that ${e^z}$ satisfies the Cauchy-Riemann equations. This is pretty straightforward. First I break up ${e^z}$ into it’s real and imaginary parts: ${e^z=u(z)+iv(z)}$ where ${u(x,y)=e^x\cos y}$ and ${v(x,y)=e^x\sin y}$ (remember that I’m using ${u(x+iy)}$ and ${u(x,y)}$ interchangeably). Then

$\displaystyle \begin{array}{cc} u_x(z)=e^x\cos & v_x(z)=e^x\sin y\\ u_y(z)=-e^x\sin y & v_y(z)=e^x\cos y \end{array}$

which clearly satisfies the Cauchy-Riemann equations.

What about ${1/z}$? Is that holomorphic? Well, we need to be precise. Over all of ${{\mathbb C}}$, certainly not. It’s not even defined at zero. But remember that for holomorphic functions, I get to specify the domain. So on the domain ${{\mathbb C}-\{0\}}$, it turns out that ${1/z}$ is holomorphic. Again, taking ${z=x+iy}$, The function ${f:z\mapsto 1/z}$, can be written as

$\displaystyle f(z)=f(x,y)=\frac{x-iy}{x^2+y^2}=\displaystyle\frac{x}{|z|^2}-i\cdot\frac{y}{|z|^2},$

which has all partials defined except when ${x=y=0}$ (which is outside our domain). As usual, we write ${u(z)=x/(x^2+y^2)}$ and ${v(z)=-y/(x^2+y^2)}$, and check:

$\displaystyle \begin{array}{cc} u_x(z)=\displaystyle\frac{y^2-x^2}{(x^2+y^2)^2} & v_x(z)=\displaystyle\frac{2xy}{(x^2+y^2)^2}\\ u_y(z)=-\displaystyle\frac{2xy}{(x^2+y^2)^2} & v_y(z)=\displaystyle\frac{y^2-x^2}{(x^2+y^2)^2} \end{array}.$

Thus, ${f\in{\mathcal H}({\mathbb C}-\{0\})}$.

No list of examples would be complete without a juxtaposed non-example, so consider the conjugation function. That is, let ${f(z)=\bar z}$. This function is not holomorphic on any open set. Yes, it’s continuous. Yes, it’s partial derivatives exist in every direction. But alas, it does not satisfy the Cauchy-Riemann equations, so it’s complex derivatives do not exist anywhere. We can write ${f=u+iv}$, where ${u(x,y)=x}$ and ${v(x,y)=-y}$. Taking partial derivatives, we can see that ${u_y=0=-v_x}$, but ${u_x=1}$ and ${v_y=-1}$. I can’t make ${u_x}$ be equal to ${v_y}$ no matter what values I plug in, so ${f}$ is not holomorphic anywhere.