Lesson 1: Complex Numbers, the Plane, and Basic Topology

Arithmetic and geometry of complex numbers

Identify $z=x+iy$ with the point $(x,y)\in\mathbb{R}^2$. Addition is exactly vector addition: $z_1+z_2=(2+i)+(-1+3i)=1+4i$, tip-to-tail, the same parallelogram law from vector calculus.

Multiplication is the operation that isn't just $\mathbb{R}^2$ — it's defined by declaring $i^2=-1$ and expanding like any other binomial: $(1+i)(1+i)=2i$, and $(2+i)(-1+3i)=-5+5i$.

The geometric rule. Rather than stating "moduli multiply, angles add," it's worth deriving it from a single example. Compute $|1+i|=\sqrt2$ and $|2i|=2$, and notice $2=\sqrt2\cdot\sqrt2$. Then check angles: $\arg(1+i)=45°$, $\arg(2i)=90°$, and $45°+45°=90°$. The pattern generalizes: $$|z_1z_2|=|z_1||z_2|, \qquad \arg(z_1z_2)=\arg(z_1)+\arg(z_2)$$ This is the same fact that made $z^2$ double the angle in Lesson 0.

Modulus, conjugate, and $z\bar z=|z|^2$. Take $z=3+4i$, a clean Pythagorean triple. $|z|=5$, $\bar z=3-4i$, and $$z\bar z=(3+4i)(3-4i)=9+16=25=|z|^2$$ via difference-of-squares. The reason this always works: conjugation flips the sign of $iy$, so the cross terms in the product cancel, leaving exactly $x^2+y^2$.

Division follows directly: $\dfrac1z=\dfrac{\bar z}{|z|^2}$, so $\dfrac{1}{1+i}=\dfrac{1-i}{2}=\dfrac12-\dfrac12 i$.

Key takeaways: $z=x+iy\leftrightarrow(x,y)$; addition = vector addition (tip-to-tail). $|z_1z_2|=|z_1||z_2|$; $\arg(z_1z_2)=\arg(z_1)+\arg(z_2)$. $z\bar z=|z|^2 \implies \dfrac1z=\dfrac{\bar z}{|z|^2}$.

Polar form and the argument

Lesson 0 used $z=re^{i\theta}$ to explain why $z^2$ doubles the angle; here it's made precise. Write $z=re^{i\theta}=r(\cos\theta+i\sin\theta)$, where $r=|z|$ and $\theta=\arg(z)$.

Converting $1+i=\sqrt2\,e^{i\pi/4}$ is the clean warm-up case ($45°$). A less obliging example: $-1+i\sqrt3$ has $r=2$, $\theta=2\pi/3$, found via the second-quadrant reference angle.

Multivaluedness of $\arg$. Take $z=i$. Both $\pi/2$ and $\pi/2+2\pi=5\pi/2$ (or equivalently $-3\pi/2$) are valid arguments: they point in the same direction, differing by a full revolution. (Note that $-\pi/2$ is not valid here: that's the direction of $-i$, not $i$.) The principal argument $\operatorname{Arg}(z)\in(-\pi,\pi]$ picks out one representative. This multivaluedness is the seed of branch cuts, which show up in Lesson 3.

Euler's formula, $e^{i\theta}=\cos\theta+i\sin\theta$, is stated here without proof; Lesson 3 derives it properly from the power series definition of $e^z$. It gives $e^{i\pi}=-1$ directly, since $\cos\pi=-1$ and $\sin\pi=0$.

De Moivre / the payoff. $z^n=r^ne^{in\theta}$. In polar form, $(1+i)^8$ is immediate: $r^8=2^4=16$, $8\theta=2\pi$, so $(1+i)^8=16e^{i2\pi}=16$, far less painful than a binomial expansion.

Key takeaways: $z=re^{i\theta}$, $r=|z|$, $\theta=\arg(z)$ (mod $2\pi$); $\operatorname{Arg}\in(-\pi,\pi]$. Euler: $e^{i\theta}=\cos\theta+i\sin\theta$; $e^{i\pi}=-1$. De Moivre: $z^n=r^ne^{in\theta}$.

Roots of unity

How many solutions does $z^3=1$ have over $\mathbb{C}$? The Fundamental Theorem of Algebra (proved properly in Lesson 5) guarantees exactly 3, counted with multiplicity, for a degree-3 polynomial, a fact worth invoking here even before it's proven.

Finding them is a "solve in polar form, match mod $2\pi$" argument that's genuinely worth doing as an explicit six-step process the first time through, since the compressed version tends to hide the two places it's easy to get confused:

  1. Write $z=re^{i\theta}$ and $1=1\cdot e^{i0}$ in polar form.
  2. Cube: $z^3=r^3e^{i3\theta}$.
  3. Match moduli: $r^3=1\implies r=1$.
  4. Match angles mod $2\pi$: $e^{i3\theta}=e^{i0}\implies 3\theta=2\pi k,\ k\in\mathbb{Z}$. This is the crux: $e^{i\alpha}=e^{i\beta}$ iff $\alpha-\beta$ is a multiple of $2\pi$, so the equation only pins down $3\theta$ up to an added $2\pi k$, not $3\theta=0$ outright.
  5. Solve: $\theta=\dfrac{2\pi k}{3}$.
  6. Enumerate $k=0,1,2,3,\dots$: $k=3$ gives $\theta=2\pi\equiv0$, repeating $k=0$ — so there are exactly 3 distinct roots.

(The other easy confusion: it's $\theta=\arg(z)$, the thing solved for, that gets plugged back into $z=e^{i\theta}$, not $3\theta=\arg(z^3)$, since those are angles of different numbers.)

The roots: $1$, $e^{i2\pi/3}=-\tfrac12+\tfrac{\sqrt3}{2}i$, $e^{i4\pi/3}=-\tfrac12-\tfrac{\sqrt3}{2}i$, three points spaced $120°$ apart on the unit circle, forming an equilateral triangle.

All three roots are powers of the primitive root $\omega=e^{i2\pi/3}$: $\omega^0,\omega^1,\omega^2$. And since a degree-3 polynomial has at most 3 roots, there are no more to find.

The same six steps generalize directly. For $z^3=8i$, write $8i=8e^{i\pi/2}$ and run the identical argument: $r=2$ (the real cube root of 8) and $\theta=\frac{\pi}{6}+\frac{2\pi k}{3}$, giving roots at $30°,150°,270°$, the same equilateral triangle, scaled to radius 2 and rotated by $30°$.

Key takeaways: $n$th roots of unity: $e^{2\pi ik/n}$, $k=0,\dots,n-1$, a regular $n$-gon on the unit circle. All roots are powers of the primitive root $\omega=e^{2\pi i/n}$. General case $z^n=w$ ($w=\rho e^{i\phi}$): same polygon, radius $\rho^{1/n}$, rotated by $\phi/n$.

Basic topology of the plane

Everything so far has been points and curves. The moment calculus enters the picture (Lesson 2 onward), it needs vocabulary for where — what kind of set a domain is. This connects directly to vector calculus: domains of vector fields, open regions, the "nice domains" theorems like Green's theorem require.

Open disk. $D(z_0,r)=\lbrace z:|z-z_0|<r\rbrace$. Testing $D(0,1)$ against $z=0.9$, $z=i$, $z=1$: $0.9$ is interior, while $i$ and $1$ sit exactly on the boundary circle ($|z|=1$, not $<1$), which is exactly why the disk is called "open": the boundary is excluded.

Open set. Every point has some $\varepsilon$-disk around it fully contained in the set. Three test cases:

Closed sets are defined as sets whose complement is open. $\bar D(0,1)$ is closed: its complement $\lbrace|z|>1\rbrace$ is open, by the same argument as the open-disk case, just outside the circle.

Open and closed aren't opposites. The half-open annulus $A=\lbrace0<|z|\le1\rbrace$ (includes the outer boundary, excludes the origin) is neither:

Connectedness and regions. Informally, a set is connected if it's one piece — you can travel between any two points without leaving the set. (Contrast $D(0,1)$ with two disjoint disks.) A region (or domain) is defined as open and connected. The punctured disk $D(0,1)\setminus\lbrace0\rbrace$ is still connected (route around the missing point) and still open, so it qualifies as a region, relevant later when residues show up on punctured regions.

Proving $D(z_0,r)$ is open. For $z_1\in D(z_0,r)$, let $\varepsilon=r-|z_1-z_0|>0$, the leftover room between $z_1$ and the boundary. For any $w\in D(z_1,\varepsilon)$, the goal is $|w-z_0|<r$. Write $$|w-z_0|=|(w-z_1)+(z_1-z_0)|\le|w-z_1|+|z_1-z_0|$$ by the triangle inequality, and then $$|w-z_0|<\varepsilon+(r-\varepsilon)=r.$$ So every point of $D(z_0,r)$ has room to spare: the disk is open.

Key takeaways: Open set: every point has some $\varepsilon$-disk contained in the set. Closed set: complement is open. Open and closed are not opposites: $A=\lbrace0<|z|\le1\rbrace$ is neither. Region/domain = open + connected. Triangle inequality: $|a+b|\le|a|+|b|$. $D(z_0,r)$ is open: given $z_1\in D(z_0,r)$, take $\varepsilon=r-|z_1-z_0|$; the triangle inequality gives $|w-z_0|\le|w-z_1|+|z_1-z_0|<\varepsilon+(r-\varepsilon)=r$.