Analytic Functions & the Cauchy-Riemann Equations
Analytic Functions & the Cauchy-Riemann Equations
A complex function $f=u+iv$ is complex-differentiable (analytic/holomorphic) at $z_0$ if $f'(z_0)=\lim_{h\to 0}(f(z_0+h)-f(z_0))/h$ exists. The limit must be the same along all paths in $\mathbb{C}$, which is an extremely strong condition captured by the Cauchy-Riemann (CR) equations $u_x=v_y$ and $u_y=-v_x$. Analytic functions are infinitely differentiable, conformal (angle-preserving), and satisfy Laplace's equation.
Cauchy-Riemann Equations
Let $f=u+iv$ where $u,v:\mathbb{R}^2\to\mathbb{R}$. $f$ is complex-differentiable at $z=x+iy$ iff $u,v$ are real-differentiable there and satisfy the Cauchy-Riemann equations: $$u_x=v_y \quad\text{and}\quad u_y=-v_x.$$ When satisfied, $f'=u_x+iv_x=v_y-iu_y$. Both $u$ and $v$ are harmonic: $\Delta u=u_{xx}+u_{yy}=0$ and $\Delta v=0$. This links complex analysis to potential theory (electrostatics, fluid flow).
Liouville's Theorem
Every bounded entire function (analytic on all of $\mathbb{C}$) is constant. Proof via Cauchy's inequality: if $f$ is entire and $|f(z)|\leq M$, then $|f'(z)|\leq M/R$ for any $R>0$ (from Cauchy's estimate on a disk of radius $R$). Letting $R\to\infty$, $f'(z)=0$ for all $z$. Corollary: Fundamental Theorem of Algebra. If $p(z)$ is a non-constant polynomial, $p$ has a root in $\mathbb{C}$. Proof: if $p(z)\neq 0$ everywhere, $1/p$ is bounded entire, hence constant — contradiction.
Example 1
Verify $f(z)=e^z$ satisfies the CR equations and is entire.
Solution: $e^z=e^x(\cos y+i\sin y)$, so $u=e^x\cos y$, $v=e^x\sin y$. $u_x=e^x\cos y=v_y$ ✓. $u_y=-e^x\sin y=-v_x$ (since $v_x=e^x\sin y$... wait: $v_x=e^x\sin y$, so $-v_x=-e^x\sin y=u_y$ ✓). CR equations hold everywhere with continuous partials, so $e^z$ is entire. $f'=(e^x\cos y)+i(e^x\sin y)=e^x e^{iy}=e^z$. So $\frac{d}{dz}e^z=e^z$ as expected.
Example 2
Show $f(z)=\bar{z}$ is nowhere differentiable.
Solution: $u=x,v=-y$. CR: $u_x=1$ but $v_y=-1\neq 1$. The CR equations fail everywhere, so $\bar{z}$ is nowhere complex-differentiable, despite being $C^\infty$ as a function $\mathbb{R}^2\to\mathbb{R}^2$. This illustrates that real differentiability is far weaker than complex differentiability.
Practice
- Show that $u(x,y)=x^3-3xy^2$ is harmonic and find its harmonic conjugate $v$.
- Prove: if $f$ is analytic and $|f|$ is constant on a domain, then $f$ is constant.
- Show $f(z)=z^2$ is analytic everywhere and compute $f'$ directly from the limit definition.
- Find all analytic functions $f=u+iv$ with $u(x,y)=e^x\cos y$.
Show Answer Key
1. $u_{xx}=6x$, $u_{yy}=-6x$, so $\Delta u=0$ (harmonic). $v_x=u_y=-6xy$, so $v=-3xy^2+g(y)$. $v_y=-3x^2+g'(y)=u_x=3x^2-3y^2$, giving $g'(y)=-3y^2$, $g(y)=-y^3+C$. Thus $v=3x^2y-y^3+C$.
2. If $|f|=c$ (constant), write $f\bar{f}=c^2$. If $c=0$, done. Otherwise $\bar{f}=c^2/f$ is also analytic (as a function of $z$). But $\bar{f}$ is anti-analytic unless constant. So $f$ is constant.
3. $f'(z)=\lim_{h\to0}\frac{(z+h)^2-z^2}{h}=\lim_{h\to0}(2z+h)=2z$. CR equations: $u=x^2-y^2$, $v=2xy$; $u_x=2x=v_y$ and $u_y=-2y=-v_x$ ✓.
4. CR: $u_x=e^x\cos y=v_y$, $u_y=-e^x\sin y=-v_x$. From $v_y=e^x\cos y$: $v=e^x\sin y+g(x)$. From $v_x=e^x\sin y+g'(x)=e^x\sin y$: $g'=0$, so $g=C$. Answer: $f(z)=e^z+iC$.