Complex Differentiability

Complex Differentiability

$\text{C}$-Differentiability

Suppose $U\subset \text{C}$ is open, a function $f:U\to \text{C}$ is $\text{C}$-differentiable at $z_{0\in }U$ if ${\lim }_{\Delta z\to 0}\frac{f(z_0+\Delta z)-f(z_0)}{\Delta z}$exists, in which case we call the limit $f^{\prime }(z_0)$. Equivalently, $f$ is $\text{C}$-differentiable if there exists a complex number $f^{\prime }(z_0)$ such that ${\lim }_{\Delta z\to 0}\frac{|f(z_0+\Delta z)-f(z_0)-f^{\prime }(z_0)\Delta z|}{|\Delta z|}=0$

e.g. $z\mapsto \bar{z}$ is $\mathbb{R}$-differentiable but not $\text{C}$-differentiable.

The Cauchy-Riemann Equations

Cauchy-Riemann Equations

Let $U\in \text{C}$ be open, let $f:U\to \text{C}$, and let $z_0=x_0+i{y}_0\in U$. Write $f(x+iy)=u(x,y)+i\cdot v(x,y)$. Then the Cauchy-Riemann equations are $\frac{\partial u}{\partial x}(x_0,y_0)=\frac{\partial v}{\partial y}(x_0,y_0)=a,\frac{\partial v}{\partial x}(x_0,y_0)=-\frac{\partial u}{\partial y}(x_0,y_0)=b$

Dolbeaut Operators

The Dolbeaut operators for some complex functions are defined as: \begin{align}\frac{\partial f}{\partial z}:=\frac12\left(\frac{\partial f}{\partial x}-i\frac{\partial f}{\partial y}\right)\\\frac{\partial f}{\partial\bar{z}}:=\frac12\left(\frac{\partial f}{\partial x}+i\frac{\partial f}{\partial y}\right)\end{align}

e.g. $\begin{array}{rl}\frac{\partial {\bar{z}}^2}{\partial \bar{z}}& =\frac{\partial }{\partial \bar{z}}((x^2-y^2)+i(-2xy))\\ & =\frac{1}{2}((2x-2iy)+i(-2y-2ix))\\ & =2x-2iy=2\bar{z}\end{array}$This agrees with how we think differentiating with respect to $\bar{z}$ should work, and indeed the same goes for applying $\frac{\partial }{\partial z}$ and $\frac{\partial }{\partial \bar{z}}$ to polynomials in $z$ and $\bar{z}$. We will see for one thing that $z\mapsto {\bar{z}}^n$ is $\text{C}$-differentiable only at $z=0$ by the following theorem:

Theorem

Let $U\in \text{C}$ be open, let $f:U\to \text{C}$, and let $z_0=x_0+i{y}_0\in U$. Write $f(x+iy)=u(x,y)+i\cdot v(x,y)$. Then the following are equivalent:

• $f$ is $\text{C}$-differentiable at $z_0$ and $f^{\prime }(z_0)=a+ib$.
• $f$ is $\mathbb{R}$-differentiable at $z_0$ and the Jacobian $J_f(x_0,y_0)=\left(\begin{array}{cc}a& -b\\ b& a\end{array}\right)$.
• $f$ is $\mathbb{R}$-differentiable at $z_0$, and we have the Cauchy-Riemann equations hold.
• $f$ is $\mathbb{R}$-differentiable at $z_0$, and $\frac{\partial f}{\partial \overline{z}}=0$. In this case $f^{\prime }(z_0)=\frac{\partial f}{\partial z}(z_0)$.

Proof The last three arguments are equivalent simply by definition of Jacobian. To see the first two are equivalent, suppose $f^{\prime }(z_0)=a+ib$, we just check this is equivalent to the Jacobian in terms of multiplication:$f^{\prime }(z_0)\Delta z=(a\Delta x-b\Delta y)+i(b\Delta x+a\Delta y)$and accordingly in ${\mathbb{R}}^2$, we have $$\begin{pmatrix}a\Delta x-b\Delta y\b\Delta x+a\Delta y\end{pmatrix}=\begin{pmatrix}a&-b\b&a\end{pmatrix}\begin{pmatrix}\Delta x\\Delta y\end{pmatrix}$$$\square$

e.g. Consider $z\mapsto e^z$. Cleary $e^{x+iy}=e^x(\cos y+i\sin y)$ is $\mathbb{R}$-differentiable. Now check that the Cauchy-Riemann equation holds: $\frac{\partial (e^x\cos y)}{\partial x}=e^x\cos y=\frac{\partial (e^x\sin y)}{\partial y},\frac{\partial (e^x\sin y)}{\partial x}=e^x\sin y=-\frac{\partial (e^x\cos y)}{\partial y}$So $z\mapsto e^z$ is $\text{C}$-differentiable everywhere.

Corollary

$\text{C}$-differentiable functions are continuous.

Proof $\text{C}$-differentiability implies $\mathbb{R}$-differentiability, and $\mathbb{R}$-differentiability implies continuity. $\square$