Lecture 19 - Derivatives

Differentiable

Definition

A function $f:D\to \mathbb{C}$ is differentiable at $z\in \mathbb{C}$ if the following limit exists
${\lim }_{\Delta z\to \infty }\frac{f(z+\Delta z)-f(z)}{\Delta z}.$
We denote by $f^{\prime }(z)$.

The same thing is also defined here: Differentiable

Example

Consider $f(z)=z$. We have:
$\frac{f(z+\Delta z)-f(z)}{\Delta z}=\frac{z+\Delta z}{\Delta z}=1.$
Hence $f^{\prime }(z)=1$.

We have general results to compute derivatives, as in the real case.

Proposition

Suppose $f$ and $g$ are differentiable at $z\in \mathbb{C}$. Then:

1. $(af(z)+bg(z){)}^{\prime }=a{f}^{\prime }(z)+b{g}^{\prime }(z)$ for any $a,b\in \mathbb{C}$,
2. $(f(z)g(z){)}^{\prime }=f^{\prime }(z)g(z)+f(z)g^{\prime }(z)$,
3. Also a quotient rule.

Some terminologies which will be important to understand the rest:

• Holomorphic
• Entire

Example

Any polynomial $P(z)={\sum }_{k=0}^{n}ak{z}^k$ is an entire function (follows from the proposition).

Example

Consider $f(z)=\mid z{\mid }^2=z\bar{z}$. This is not going to be holomorphic.

We compute:
$\frac{f(z+\Delta z)-f(z)}{\Delta z}=\frac{\mid z+\Delta z{\mid }^2=\mid z{\mid }^2}{\Delta z}$

Note

We say that $z=x+iy$, hence $\Delta z=a+ib$.

$=\frac{(x+a{)}^2+(y+b{)}^2-x^2-y^2}{a+ib}=\frac{a^2+b^2+2(ax+by)}{a+ib}.$
Consider the real direction.

Write

$\Delta z=\Delta x$

${\lim }_{\underset{a\to 0}{\underbrace{\Delta x\to 0}}}\frac{f(z+\Delta x)-f(z)}{\Delta x}={\lim }_{a\to 0}\frac{a^2+2ax}{a}=2x$

Similarly for the imaginary axis

Write

$\Delta z=i\Delta y$

${\lim }_{\underset{b\to 0}{\underbrace{i\Delta y\to 0}}}\frac{f(z+i\Delta y)-f(z)}{i\Delta y}={\lim }_{b\to 0}\frac{b^2+2by}{ib}=-2iy.$
This means that $f$ is not differentiable at $z\ne 0$.

Cauchy-Riemann Equations

A complex function $f(z)$ can be seen as a function of two real variables. Hence
$f(z)=f(x+iy)=f(x,y).$
When interpreted appropriately.

For instance

$f(z)=z^2=(x+iy{)}^2=x^2-y^2+2ixy$

Suppose $f$ is differentiable, we should expect relations between $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$.

The same thing is also defined here: Cauchy-Riemann Equations

Theorem

Let $z_0=x_0+i{y}_0$.

1. Suppose $f$ is differentiable at $z_0$. Then $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ exists and have $\frac{\partial f}{\partial x}(x_0,y_0)=-i\frac{\partial f}{\partial y}(x_0,y_0).$

Visual Representation

2. Suppose $f$ is such that:
   • $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ exists at $(x_0,y_0)$,
   • they are continuous in a small disk centred at $(x_0,y_0)$.
     Then $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ satisfy the condition above and $f$ is differentiable at $z_0$.
3. In both cases, we have $f^{\prime }(z_0)=\frac{\partial f}{\partial x}(x_0,y_0)$.

Proof

For 1.

Since $f$ is differentiable at $z_0$, the following limits exists
$f^{\prime }(z_0)={\lim }_{\Delta z\to 0}\frac{f(z_0+\Delta z)-f(z_0)}{\Delta z}.$

Visual representation

$f(z)=f(x,y)$

First we take $\Delta z$ real, so $\Delta z=\Delta x$. Then
$f^{\prime }(z_0)={\lim }_{\Delta x\to 0}\frac{f(x_0+\Delta x)-f(x_0,y_0)}{\Delta x}=\frac{\partial f}{\partial x}(x_0,y_0).$

Next take $\Delta z$ to be purely imaginary ($\Delta x,\Delta y\in \mathbb{R}$), so $\Delta z=i\Delta y$
($z_0+i\Delta y=x_0+i(y_0+\Delta y)$)

Then:
$f^{\prime }(z_0)={\lim }_{\Delta y\to 0}\frac{f(x_0,y_0+\Delta y)=f(x_0,y_0)}{i\Delta y}=-i\frac{\partial f}{\partial y}(x_0,y_0).$
The two expressions must coincide, since $f$ is differentiable at $z_0$.

For 2.

It is more complicated to prove. You can find the proof in the references on Canvas for the course.

For 3.

Shown in For 1.

Decomposition

We can decompose any $f:D\to \mathbb{C}$ into its real and imaginary part. We write
$f(z)=f(x,y)=u(x,y)+iv(x,y).$
or $\mathrm{Re}f=u$ and $\mathrm{Im}f=v$

Corollary

With notation as before, we have
$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y},\frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y}.$

Proof

By the previous theorem, we have
$\frac{\partial f}{\partial x}=-i\frac{\partial f}{\partial y}.$
Write $f=u+iv$. Then
$\frac{\partial u}{\partial x}+i\frac{\partial v}{\partial x}=-i\frac{\partial u}{\partial y}+\frac{\partial v}{\partial y}.$
Comparing real and imaginary parts gives the result.

Example

Consider $f(z)=z^2$.
This is seen to be holomorphic (in at least two ways).

One way is
$f^{\prime }(z)=(zz{)}^{\prime }=z^{\prime }z+z{z}^{\prime }=zz.$

Another way is to use the Cauchy-Riemann Equations. We have
$z^2=(x+iy{)}^2=(x^2-y^2)+2ixy.$
So here
$\mathrm{Re}f=u=x^2-y^2,\mathrm{Im}f=v=2xy.$
So one checks that $u$ and $v$ satisfy the Cauchy-Riemann Equations.
Since the derivatives are continuous, we get that $f=u+iv$ is holomorphic.