Complex Numbers

What you need to know

• $i^2=-1$, and powers of $i$ cycle with period $4$: $i,\,-1,\,-i,\,1$. For $z=a+bi$: conjugate $\bar z=a-bi$, modulus $|z|=\sqrt{a^2+b^2}$, and $z\bar z=|z|^2$.
• To divide, multiply numerator and denominator by the conjugate of the denominator. Also $|zw|=|z||w|$ and $|z^n|=|z|^n$.
• Two complex numbers are equal iff real parts and imaginary parts both agree; one complex equation is two real equations.
• Polar form $z=r(\cos\theta+i\sin\theta)$: multiplying multiplies moduli and adds angles, so $z^n=r^n(\cos n\theta+i\sin n\theta)$. Handy: $(1+i)^2=2i$.
• The $n$th roots of unity are the vertices of a regular $n$-gon on the unit circle and sum to $0$; $\omega=\frac{-1+i\sqrt3}2$ satisfies $\omega^3=1$ and $\omega^2+\omega+1=0$.
• A real-coefficient polynomial with root $z$ also has root $\bar z$, and $(x-z)(x-\bar z)=x^2-2\operatorname{Re}(z)\,x+|z|^2$.

How AMC 10 tests it

• Problems 5–12: evaluate $(1+i)^{20}$ or $i^{2025}+i^{2026}$ using the cycle of $i$ or $(1+i)^2=2i$.
• Problems 8–15: find real $a,b$ with $(a+bi)^2=3+4i$, or make $\frac{a+bi}{2-i}$ real, by matching parts.
• Problems 12–18: "how many complex $z$ satisfy $z^2=\bar z$" — set $z=a+bi$ or use polar form.
• Problems 15–22: geometry: $|z-1|=|z+1|$ is a line, $|z-2i|=3$ a circle; the question asks for an area, a distance, or a count.
• Occasionally a quadratic with non-real roots; Vieta still applies.

Standard approaches

1. For powers, reduce the exponent modulo $4$ (for $i$) or square first ($(1\pm i)^2=\pm2i$).
2. For equations, write $z=a+bi$, expand, and equate real and imaginary parts.
3. For modulus conditions, use $|z|^2=a^2+b^2$ instead of square roots, and read $|z-w|$ as a distance.
4. For $z^n=1$ or repeated multiplication by a fixed number, switch to polar form and track the angle.
5. Check the final answer with $|z^n|=|z|^n$; it catches most arithmetic slips.

Worked example

Let $z$ be a complex number with $|z|=2$ whose real part is $1$. What is $z^6$?

(A) $-64$ (B) $-8$ (C) $8$ (D) $64$ (E) $128$

Write $z=1+bi$. Then $1+b^2=4$, so $b=\pm\sqrt3$. Compute
$$z^2=1-b^2+2bi=-2\pm2\sqrt3\,i,$$
$$z^3=z\cdot z^2=(1\pm\sqrt3\,i)(-2\pm2\sqrt3\,i)=-2\pm2\sqrt3\,i\mp2\sqrt3\,i+(\sqrt3\,i)(2\sqrt3\,i)=-2-6=-8 .$$
Hence $z^6=(-8)^2=64$. (In polar form $z=2(\cos60^\circ\pm i\sin60^\circ)$, so $z^6=2^6(\cos360^\circ+i\sin360^\circ)=64$; the modulus check $|z^6|=2^6$ agrees.) The answer is $\boxed{\textbf{(D)}\ 64}$.

Pitfalls

• Treating $i$ as an ordinary variable and forgetting $i^2=-1$ when expanding.
• Computing $|a+bi|$ as $a^2+b^2$ (no square root) or as $\sqrt{a^2-b^2}$.
• Dividing without the conjugate, or conjugating only the numerator.
• Forgetting the case $z=0$ when counting solutions of equations like $z^n=\bar z$.