Complex function

A complex function $f$ assigns a complex number $w=f(z)$ to each complex number $z$ in a domain $D\subseteq \mathbb{C}$. Same idea as a real function, but everything happens in $\mathbb{C}$ rather than $\mathbb{R}$.

What changes from the real case is visualization. A graph $y=f(x)$ of a real function sits in 2D, with $x$ and $y$ on perpendicular axes. For $w=f(z)$, both $z$ and $w$ are 2D, so a graph would live in 4D, undrawable.

The standard workaround: draw two complex planes side by side, the $z$-plane (domain) on the left and the $w$-plane (range) on the right, and represent $f$ by the way it maps geometric configurations: points to points, curves to curves, regions to regions.

Real and imaginary parts

Any complex function decomposes as

$f(z)=u(x,y)+iv(x,y),$

where $z=x+iy$ and $u,v$ are real-valued functions of two real variables. So a complex function is, secretly, a pair of real two-variable functions. The Cauchy-Riemann equations are a pair of partial differential equations on $u$ and $v$ that determine whether $f$ is complex differentiable.

Examples

$f(z)=z^2$. Rectangular form: $(x+iy{)}^2=(x^2-y^2)+i(2xy)$, so $u=x^2-y^2$, $v=2xy$. Polar form is cleaner: if $z=r{e}^{i\theta }$ then $f(z)=r^2{e}^{i2\theta }$. Squares the modulus, doubles the argument. A circle of radius $R$ in the $z$-plane maps to a circle of radius $R^2$ in the $w$-plane, traversed twice. The upper half-plane $\mathrm{Im}(z)>0$ maps to the full plane minus the non-negative real axis.

$f(z)=1/z$. $f(r{e}^{i\theta })=(1/r)e^{-i\theta }$, inverts the modulus, negates the argument (reflection across the real axis). The unit circle maps to itself. The right half-plane $\mathrm{Re}(z)>0$ maps to itself. Undefined at $z=0$; this is a singularity, and singularities are what drive the Residue theorem.

$f(z)=\bar{z}$. $u=x$, $v=-y$. Polar: $\overline{r{e}^{i\theta }}=r{e}^{-i\theta }$. Reflection across the real axis. Continuous everywhere, but complex differentiable nowhere. See Complex conjugate for why.

$f(z)=e^z=e^x(\cos y+i\sin y)$. $u=e^x\cos y$, $v=e^x\sin y$. Modulus depends only on real part; argument depends only on imaginary part. A vertical line $\mathrm{Re}(z)=a$ maps to a circle $|w|=e^a$; a horizontal line $\mathrm{Im}(z)=b$ maps to a ray from the origin at angle $b$. The map is conformal (preserves angles) wherever the derivative is nonzero. See Complex exponential.

$f(z)=\mathrm{Re}(z)=x$. $u=x$, $v=0$. Real-valued, continuous, complex differentiable nowhere.

$f(z)=|z|$. $u=\sqrt{x^2+y^2}$, $v=0$. Continuous everywhere, complex differentiable nowhere.

Why most familiar functions fail complex differentiability

The pattern: real-variable concepts of “nice function” don’t predict complex differentiability. $\bar{z}$, $\mathrm{Re}(z)$, $|z|$ are all infinitely differentiable as functions of $(x,y)$, but each fails the Complex derivative test at every point. The condition is much more demanding than real differentiability, and most functions fail it.

The ones that pass, the analytic functions, have properties no real function has: infinitely differentiable, equal to their Taylor series, rigid. Part III of the course develops these.