Functions of a Complex Variable
\[
Z=x+\ii y,\quad x,y\in\mathbb R;\qquad f(z)=u(x,y)+\ii v(x,y),\quad u,v\in\mathbb R.
\]
Analytic function A function defined and differentiable throughout a domain \(D\) is
analytic in \(D\); e.g. \(z\), \(\sin z\).
Non-analytic function A function that is not differentiable is not analytic; e.g. \(\bar z\).
Necessary conditions: Cauchy–Riemann equations
\begin{align*}
\pd{u}{x}&=\pd{v}{y}, & \pd{u}{y}&=-\pd{v}{x} &&\text{(Cartesian form)},\\
\pd{u}{r}&=\frac1r\pd{v}{\theta}, & \pd{u}{\theta}&=-r\pd{v}{r} &&\text{(polar form)}.
\end{align*}
Example 5.3 Determine whether the following functions are analytic or non-analytic.
Solution. (a) \(f(z)=z=x+\ii y=u+\ii v\), \(u=x\), \(v=y\): \(\pd{u}{x}=\pd{v}{y}=1\), \(\pd{u}{y}=-\pd{v}{x}=0\). The C–R equations hold, so \(f\) is analytic.
(b) \(f(z)=z^2=(x+\ii y)^2=x^2-y^2+2\ii xy\), \(u=x^2-y^2\), \(v=2xy\):
\[
\pd{u}{x}=2x,\quad \pd{v}{y}=2x,\qquad \pd{u}{y}=-2y,\quad -\pd{v}{x}=-2y\;\Rightarrow\;\pd{u}{x}=\pd{v}{y}\ \&\ \pd{u}{y}=-\pd{v}{x}.
\]
The C–R equations hold, so \(f\) is analytic.
(c) \(f(z)=z^3=(x+\ii y)^3=x^3+\ii^3y^3+3x^2\ii y+3x\ii^2y^2=x^3-\ii y^3+3x^2\ii y-3xy^2\), so
\(u=x^3-3xy^2\), \(v=3x^2y-y^3\):
\[
\pd{u}{x}=3x^2-3y^2,\quad \pd{v}{y}=3x^2-3y^2;\qquad \pd{u}{y}=-6xy,\quad -\pd{v}{x}=-6xy,
\]
so the function is analytic.
In general a Taylor series \(f(z)=\sum_{n=0}^\infty a_n(z-z_0)^n\) is analytic inside its circle of
convergence, since every power \(z^n\) is analytic.
(d) \(f(z)=\sin z=\sin(x+\ii y)=\sin x\cos\ii y+\cos x\sin\ii y=\sin x\cosh y+\ii\cos x\sinh y\), so
\(u=\sin x\cosh y\), \(v=\cos x\sinh y\):
\[
\pd{u}{x}=\cos x\cosh y,\quad \pd{v}{y}=\cos x\cosh y;\qquad \pd{u}{y}=\sin x\sinh y,\quad -\pd{v}{x}=\sin x\sinh y,
\]
so the function is analytic.
(e) \(f(z)=\bar z=x-\ii y\), \(u=x\), \(v=-y\): \(\pd{u}{x}=1\), \(\pd{v}{y}=-1\), \(\pd{u}{y}=0\), \(\pd{v}{x}=0\). Since
\(\pd{u}{x}\neq\pd{v}{y}\), the function is not analytic.
The special thing about \(\bar z\)
For \(f(z)=\bar z\),
\begin{align*}
f'(z)&=\lim_{\Delta z\to0}\frac{f(z+\Delta z)-f(z)}{\Delta z}=\lim_{\Delta z\to0}\frac{\overline{z+\Delta z}-\bar z}{\Delta z}\\
&=\lim_{\Delta z\to0}\frac{\overline{\Delta z}}{\Delta z}=\lim_{\Delta z\to0}\frac{\Delta x-\ii\Delta y}{\Delta x+\ii\Delta y}.
\end{align*}
Along different paths,
\[
\lim_{\Delta x\to0}\frac{\Delta x-\ii\Delta y}{\Delta x+\ii\Delta y}=-1,\qquad
\lim_{\Delta y\to0}\frac{\Delta x-\ii\Delta y}{\Delta x+\ii\Delta y}=1,\qquad
\lim_{\Delta y\to\Delta x}\frac{\Delta x-\ii\Delta y}{\Delta x+\ii\Delta y}=\frac{1-\ii}{1+\ii}.
\]
The limits differ, so the derivative does not exist: a function that genuinely depends on \(\bar z\) is
non-analytic.
Example 5.4 \(f(z)=(-x+\ii y+7)^7\).
Solution. \(f(z)=-(x-\ii y-7)^7=-(\bar z-7)^7\). The function is non-analytic.
Example 5.5 \(f(z)=z\bar z\).
Solution. \(f(z)=z\bar z=(x+\ii y)(x-\ii y)=(x^2+y^2)+0\ii\), so \(u=x^2+y^2\), \(v=0\):
\(\pd{u}{x}=2x\), \(\pd{v}{y}=0\), \(\pd{u}{y}=2y\), \(\pd{v}{x}=0\). The C–R equations hold only at the origin, so \(f\) is
differentiable there alone and analytic nowhere.
Example 5.6 \(f(z)=\ee^z\). This is an analytic function.
Solution. \(f(z)=\ee^z=\ee^{(x+\ii y)}=\ee^x\ee^{\ii y}=\ee^x[\cos y+\ii\sin y]=\ee^x\cos y+\ii\ee^x\sin y\), so
\(u=\ee^x\cos y\), \(v=\ee^x\sin y\):
\[
\pd{u}{x}=\ee^x\cos y,\quad \pd{v}{y}=\ee^x\cos y,\qquad \pd{u}{y}=-\ee^x\sin y,\quad -\pd{v}{x}=-\ee^x\sin y.
\]
So this is an analytic function.
Example 5.7 \(f(z)=\ln z\).
Solution. With \(z=r\ee^{\ii\theta}\), \(f(z)=\ln r\ee^{\ii\theta}=\ln r+\ln\ee^{\ii\theta}=\ln r+\ii\theta\), so \(u=\ln r\), \(v=\theta\):
\[
\pd{u}{r}=\frac1r,\ \pd{v}{\theta}=1\;\Rightarrow\;\pd{u}{r}=\frac1r\pd{v}{\theta};\qquad
\pd{u}{\theta}=0,\ \pd{v}{r}=0\;\Rightarrow\;\pd{u}{\theta}=-r\pd{v}{r}.
\]
So \(\ln z\) is analytic except at \(z=0\) and on its branch cut.
Finding \(u\) or \(v\) when either of them is known
Example 5.8 If the real part \(u\) of an analytic function is given, find the imaginary part \(v\).
Solution. (a) \(u=x^2-y^2\):
\begin{align*}
\pd{u}{x}&=\pd{v}{y}=2x\;\Rightarrow\;\partial v=2x\,\partial y\;\Rightarrow\;v=2xy+g(x),\tag{1}\\
\pd{u}{y}&=-\pd{v}{x}=-2y\;\Rightarrow\;\partial v=2y\,\partial x\;\Rightarrow\;v=2yx+h(y).\tag{2}
\end{align*}
From (1) and (2), the imaginary part is \(v=2xy\).
(b) \(u=x^2-y^2-y\):
\begin{align*}
\pd{u}{x}&=\pd{v}{y}=2x\;\Rightarrow\;v=2xy+g(x), &(1)\\
\pd{u}{y}&=-\pd{v}{x}=-2y-1\;\Rightarrow\;\pd{v}{x}=2y+1\;\Rightarrow\;v=\int(2y+1)\,\partial x=2yx+x+h(y). &(2)
\end{align*}
From (1) and (2), the imaginary part is \(v=2xy+x\).
(c) \(u=x+\frac12(x^2-y^2)\):
\[
\pd{u}{x}=1+x=\pd{v}{y}\;\Rightarrow\;v=y+xy+g(x),\qquad
\pd{u}{y}=-y=-\pd{v}{x}\;\Rightarrow\;\pd{v}{x}=y\;\Rightarrow\;v=yx+h(y).
\]
So \(v=xy+y\).
(d) \(u=\log\abs z\): with \(z=r\ee^{\ii\theta}\), \(\abs z=\abs{r(\cos\theta+\ii\sin\theta)}=\sqrt{r^2(\cos^2\theta+\sin^2\theta)}=r\), so \(u=\log r\):
\[
\pd{u}{r}=\frac1r\pd{v}{\theta}\Rightarrow\frac1r=\frac1r\pd{v}{\theta}\Rightarrow v=\theta+g(r),\qquad
\pd{u}{\theta}=-r\pd{v}{r}\Rightarrow0=\pd{v}{r}\Rightarrow v=c.
\]
So \(v=\theta\).
(e) \(u=x^3-3xy^2\):
\begin{align*}
\pd{u}{x}&=\pd{v}{y}\;\Rightarrow\;3x^2-3y^2=\pd{v}{y}\;\Rightarrow\;v=3x^2y-\frac{3y^3}3=3x^2y-y^3+g(x), &(1)\\
\pd{u}{y}&=-\pd{v}{x}\;\Rightarrow\;-6xy=-\pd{v}{x}\;\Rightarrow\;v=\frac{6x^2y}2=3x^2y+h(y). &(2)
\end{align*}
From (1) and (2), \(v=3x^2y-y^3\).
(f) \(u=\ee^{-x}\sin2y\): here \(\nabla^2u=\ee^{-x}\sin2y-4\ee^{-x}\sin2y=-3\ee^{-x}\sin2y\neq0\), so \(u\) is not
harmonic and no analytic function has it as its real part. Indeed the two Cauchy–Riemann equations are
inconsistent:
\begin{gather*}
\pd{v}{y}=\pd{u}{x}=-\ee^{-x}\sin2y\;\Rightarrow\;v=\ee^{-x}\frac{\cos2y}2+g(x),\\
\pd{v}{x}=-\pd{u}{y}=-2\ee^{-x}\cos2y\;\Rightarrow\;v=2\ee^{-x}\cos2y+h(y),
\end{gather*}
and no choice of \(g\), \(h\) reconciles them. (Compare \(u=\ee^{-x}\sin y\), for which \(v=\ee^{-x}\cos y\).)
(g) \(u=\cos x\cosh y\):
\begin{align*}
\pd{u}{x}&=-\sin x\cosh y=\pd{v}{y}\;\Rightarrow\;v=-\sin x\sinh y+g(x), &(1)\\
\pd{u}{y}&=\cos x\sinh y=-\pd{v}{x}\;\Rightarrow\;v=-\sin x\sinh y+h(y). &(2)
\end{align*}
So \(v=-\sin x\sinh y\).
(h) \(u=\log\sqrt{x^2+y^2}=\log r\) \(\bigl[r=\sqrt{x^2+y^2}\bigr]\):
\(\pd{u}{r}=\frac1r=\frac1r\pd{v}{\theta}\Rightarrow v=\theta+g(r)\); \(\pd{u}{\theta}=-r\pd{v}{r}\Rightarrow v=c\). So
\(v=\theta=\tan^{-1}\bigl(\frac yx\bigr)\).
(i) \(u=2x(1-y)\):
\begin{align*}
\pd{u}{x}&=(2-2y)=\pd{v}{y}\;\Rightarrow\;v=2y-y^2+g(x), &(1)\\
\pd{u}{y}&=-2x=-\pd{v}{x}\;\Rightarrow\;v=\frac{2x^2}2+h(y). &(2)
\end{align*}
So \(v=x^2-y^2+2y\).
(j) \(u=\bigl(r-\frac1r\bigr)\sin\theta\):
\[
\pd{u}{r}=\frac1r\pd{v}{\theta}\;\Rightarrow\;\sin\theta+\frac{\sin\theta}{r^2}=\frac1r\pd{v}{\theta}\;\Rightarrow\;\pd{v}{\theta}=r\sin\theta+\frac{\sin\theta}r,
\]
\[
v=\int r\sin\theta\,\partial\theta+\int\frac{\sin\theta}r\partial\theta=-r\cos\theta-\frac{\cos\theta}r+g(r);\tag{1}
\]
\[
\pd{u}{\theta}=-r\pd{v}{r}\;\Rightarrow\;r\cos\theta-\frac{\cos\theta}r=-r\pd{v}{r}\;\Rightarrow\;\pd{v}{r}=\frac{\cos\theta}{r^2}-\cos\theta,
\]
\[
v=\int\frac{\cos\theta}{r^2}\partial r-\int\cos\theta\,\partial r=-\frac{\cos\theta}r-r\cos\theta+h(\theta).\tag{2}
\]
So \(v=-\dfrac{\cos\theta}r-r\cos\theta\).
