Each worksheet prints cleanly on A4 or Letter. Answer keys are on a separate page.

The Lumenary · Grades 9–12

Complex numbers

Name: ______________________

Date: ______________________

You will learn to

Add, multiply and divide complex numbers $a + bi$ , and picture them as points and arrows in the complex plane
Compute the modulus, argument and conjugate of a complex number, and convert between Cartesian and polar form
Prove that multiplying complex numbers multiplies their moduli and adds their arguments, and prove De Moivre's theorem by induction
Find the $n$ -th roots of unity and of other complex numbers, and solve quadratic equations with negative discriminant
Explain why $\C$ cannot be ordered compatibly with its arithmetic, and why $\sqrt{ab} = \sqrt a \, \sqrt b$ fails for negative $a$ and $b$

1.
Compute $(4 - 3i)(2 + 5i)$ and enter its real part.
2.
What is $i^{2027}$ ?
- A. $-i$
- B. $i$
- C. $-1$
- D. $1$
3.
Compute $\left|(3 + 4i)(5 - 12i)\right|$ — ideally without multiplying out first.
4.
Which is the polar form of $z = -1 + i\sqrt 3$ ?
- A. $2\operatorname{cis}\tfrac{2\pi}{3}$
- B. $2\operatorname{cis}\left(-\tfrac{\pi}{3}\right)$
- C. $4\operatorname{cis}\tfrac{2\pi}{3}$
- D. $2\operatorname{cis}\tfrac{\pi}{3}$
5.
Solve $z^2 - 4z + 13 = 0$ .
- A. $z = 2 \pm 3i$
- B. $z = -2 \pm 3i$
- C. $z = 2 \pm 6i$
- D.There are no solutions
6.
Which of the following are sixth roots of unity, i.e. solutions of $z^6 = 1$ ? (Choose all that apply.)
- A. $-1$
- B. $\tfrac12 + \tfrac{\sqrt 3}{2} i$
- C. $i$
- D. $-\tfrac12 - \tfrac{\sqrt 3}{2} i$
- E. $1 + i$
7.
What is wrong with this "proof" that $1 = -1$ ? $1 = \sqrt 1 = \sqrt{(-1)(-1)} = \sqrt{-1}\,\sqrt{-1} = i \cdot i = -1.$
- A.The step $\sqrt{(-1)(-1)} = \sqrt{-1}\,\sqrt{-1}$
- B.The step $1 = \sqrt 1$
- C.The step $i \cdot i = -1$
- D.The step $\sqrt 1 = \sqrt{(-1)(-1)}$
8.
In the Lean module, cis θ means $\cos\theta + i\sin\theta$ and n : ℕ. Which Lean statement is De Moivre's theorem?
- A.cis θ ^ n = cis (n * θ)
- B.cis (θ ^ n) = cis θ ^ n
- C.cis θ * n = cis (n * θ)
- D.cis (θ + n) = cis θ * cis n
9.
Write nth_roots(w, n) that returns a list of the $n$ complex $n$ -th roots of a non-zero complex number w, using polar form: if $w = R\operatorname{cis}\varphi$ , the roots are $R^{1/n}\operatorname{cis}\frac{\varphi + 2\pi k}{n}$ for $k = 0, \dots, n-1$ . Python's cmath.polar(w) returns $(R, \varphi)$ and cmath.rect(r, theta) returns $r\operatorname{cis}\theta$ .
Computational — Practice

Answer key — Complex numbers

1.
$(4 - 3i)(2 + 5i) = 8 + 20i - 6i - 15i^2 = 8 + 14i + 15 = 23 + 14i$ . The real part is $23$ . (Formula check: $ac - bd = 4 \cdot 2 - (-3)(5) = 23$ .)
2.
Since $i^4 = 1$ , only the remainder of the exponent modulo $4$ matters. $2027 = 4 \cdot 506 + 3$ , so $i^{2027} = (i^4)^{506} \cdot i^3 = -i$ . Geometrically: $2027$ quarter turns is $506$ full turns plus three quarter turns.
3.
$|3 + 4i| = 5$ and $|5 - 12i| = 13$ , so $|(3 + 4i)(5 - 12i)| = 5 \cdot 13 = 65$ . Check: $(3 + 4i)(5 - 12i) = 15 - 36i + 20i - 48i^2 = 63 - 16i$ , and $63^2 + 16^2 = 3969 + 256 = 4225 = 65^2$ .
4.
$|z| = 2$ , and $z/2 = -\tfrac12 + \tfrac{\sqrt 3}{2} i = \cos\tfrac{2\pi}3 + i\sin\tfrac{2\pi}3$ . So $z = 2\operatorname{cis}\tfrac{2\pi}3$ . (Indeed $z$ is twice a primitive cube root of unity, so $z^3 = 8$ .)
5.
$\Delta = (-4)^2 - 4 \cdot 13 = -36 < 0$ , so $z = \tfrac{4 \pm i\sqrt{36}}{2} = 2 \pm 3i$ . Check: $(2 + 3i)^2 - 4(2 + 3i) + 13 = (4 + 12i - 9) - 8 - 12i + 13 = 0$ . The two roots are conjugates, and Vieta holds: sum $4$ , product $4 + 9 = 13$ .
6.
The sixth roots of unity are $\operatorname{cis}\tfrac{2\pi k}{6}$ for $k = 0, \dots, 5$ : $1, \tfrac12 \pm \tfrac{\sqrt3}2 i, -\tfrac12 \pm \tfrac{\sqrt3}2 i, -1$ . So $-1$ , $\tfrac12 + \tfrac{\sqrt3}2 i$ and $-\tfrac12 - \tfrac{\sqrt3}2 i$ are among them; $i$ (angle $\tfrac\pi2$ ) and $1 + i$ (modulus $\sqrt2$ ) are not.
7.
The error is $\sqrt{(-1)(-1)} = \sqrt{-1}\sqrt{-1}$ : the product rule for square roots holds for non-negative reals only. Every non-zero complex number has two square roots, and no consistent choice of "the" square root on all of $\C$ makes $\sqrt{ab} = \sqrt a\sqrt b$ true.
8.
De Moivre's theorem is cis θ ^ n = cis (n * θ). Option (b) is false, option (c) confuses $n$ -th powers with multiplication by $n$ , and option (d) is a true instance of Theorem 2 rather than De Moivre.
9.
```
import cmath
import math

def nth_roots(w, n):
    R, phi = cmath.polar(w)
    return [cmath.rect(R ** (1 / n), (phi + 2 * math.pi * k) / n) for k in range(n)]
```
Each root has modulus $R^{1/n}$ , and by De Moivre its $n$ -th power has angle $\varphi + 2\pi k$ , i.e. the same direction as $w$ . The $n$ angles $\frac{\varphi + 2\pi k}{n}$ differ by multiples of $\frac{2\pi}{n}$ , so the roots are distinct and form a regular $n$ -gon. The checks use a tolerance because floating-point arithmetic rounds.