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 $f(z)=r^2{e}^{i2\theta }$. The function squares the modulus and 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, remarkably, 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 — analytic functions — turn out to have remarkable properties: infinitely differentiable, equal to their Taylor series, rigid in ways no real function is. The rest of Vector Calculus and Complex Analysis’s Part III develops these.