A real function can be drawn as a graph, but the graph of a complex function would need four dimensions. Instead we draw two planes side by side, the $z$-plane and the $w$-plane, and watch where $f$ sends points, curves and regions. The exponential form $z = re^{i\theta}$ makes the basic maps easy to read, because it shows what happens to lengths and to angles separately.
Key idea
Three maps to know by sight:
• $w = az + b$: multiply by $a$ (turn by $\operatorname{Arg} a$, stretch by $|a|$), then shift by $b$. Lines go to lines and circles to circles.
• $w = z^2$: $re^{i\theta} \mapsto r^2e^{2i\theta}$. Lengths are squared and angles are doubled, so a sector of opening $\alpha$ at $0$ becomes a sector of opening $2\alpha$.
• $w = \frac{1}{z}$: $re^{i\theta} \mapsto \frac{1}{r}e^{-i\theta}$. Points near $0$ go far away and vice versa: the circle $|z| = r$ goes to $|w| = \frac{1}{r}$.
• $w = az + b$: multiply by $a$ (turn by $\operatorname{Arg} a$, stretch by $|a|$), then shift by $b$. Lines go to lines and circles to circles.
• $w = z^2$: $re^{i\theta} \mapsto r^2e^{2i\theta}$. Lengths are squared and angles are doubled, so a sector of opening $\alpha$ at $0$ becomes a sector of opening $2\alpha$.
• $w = \frac{1}{z}$: $re^{i\theta} \mapsto \frac{1}{r}e^{-i\theta}$. Points near $0$ go far away and vice versa: the circle $|z| = r$ goes to $|w| = \frac{1}{r}$.
Why it works
Each statement is the exponential form at work. $(re^{i\theta})^2 = r^2e^{2i\theta}$ squares the modulus and doubles the argument. For $\frac{1}{z}$: $\frac{1}{re^{i\theta}} = \frac{1}{r}e^{-i\theta}$, so the modulus is inverted and the angle changes sign. For $az + b$: with $a = |a|e^{i\varphi}$, $az = |a|r\,e^{i(\theta + \varphi)}$, a rotation by $\varphi$ and a stretch by $|a|$, and adding $b$ slides everything.
Worked examples
Worked example: A disk under a linear map
Find the image of the disk $|z - 1| < 2$ under $w = 2iz + 3$.
- A linear map sends the center to the center: $2i(1) + 3 = 3 + 2i$.
- It stretches every distance by $|2i| = 2$: the radius becomes $4$.
- The image is the disk $|w - (3 + 2i)| < 4$, turned a quarter turn (which a disk does not show).
Answer: $|w - 3 - 2i| < 4$
Worked example: A sector under squaring
Find the image of the sector $0 < \arg z < \frac{\pi}{3}$ under $w = z^2$, and the image of the point $z = 1 + i$.
- Squaring doubles angles: the sector becomes $0 < \arg w < \frac{2\pi}{3}$ (lengths $r > 0$ still give every $r^2 > 0$).
- $z = 1 + i = \sqrt{2}e^{\pi i/4}$, so $w = 2e^{\pi i/2} = 2i$. (Directly: $(1 + i)^2 = 2i$.)
Answer: $0 < \arg w < \frac{2\pi}{3}$; $2i$
Worked example: A circle under inversion
Under $w = \frac{1}{z}$, where does the circle $|z| = 3$ go, and where does $z = 3i$ go?
- $|w| = \frac{1}{|z|} = \frac{1}{3}$: the circle of radius $\frac{1}{3}$.
- $\frac{1}{3i} = \frac{-i}{3}$: the top of the big circle goes to the bottom of the small one (the angle changes sign).
Answer: $|w| = \frac{1}{3}$; $-\frac{i}{3}$
Common mistake
Expecting $z^2$ to send the first quadrant to itself. Squaring keeps real numbers real, so it is tempting to think regions keep their shape. But every angle doubles: the ray at $\frac{\pi}{4}$ goes to the ray at $\frac{\pi}{2}$, and the edge at $\frac{\pi}{2}$ goes to the negative real axis. The first quadrant fills the whole upper half-plane. Check with one point: $(1 + i)^2 = 2i$, already outside the first quadrant.