Complex Numbers

The Imaginary Unit

Definition

The imaginary unit $i$ is defined as the number satisfying $i^2 = -1$. Equivalently, $i = \sqrt{-1}$. From this definition:

$i^0 = 1, \quad i^1 = i, \quad i^2 = -1, \quad i^3 = -i, \quad i^4 = 1$

The powers of $i$ are cyclic with period $4$: $i^n = i^{n \bmod 4}$.

Motivation

The equation $x^2 + 1 = 0$ has no real solutions. Extending $\mathbb{'\{'}R{'\}'}$ to include $i$ yields the field of complex numbers, in which every polynomial has a root (the Fundamental Theorem of Algebra).

---

Complex Numbers

Definition

A complex number $z$ is an ordered pair of real numbers written in Cartesian (standard) form:

$z = a + bi, \quad a, b \in \mathbb{'\{'}R{'\}'}$

Here $a = \mathrm{Re}(z)$ is the real part and $b = \mathrm{Im}(z)$ is the imaginary part. The set of all complex numbers is denoted $\mathbb{'\{'}C{'\}'}$.

A complex number is purely real if $b = 0$ and purely imaginary if $a = 0$. Two complex numbers are equal if and only if their real and imaginary parts are equal.

Complex Arithmetic

Addition. $(a + bi) + (c + di) = (a + c) + (b + d)i$

Multiplication. Expand and use $i^2 = -1$:

$(a + bi)(c + di) = (ac - bd) + (ad + bc)i$

Division. Multiply numerator and denominator by the complex conjugate of the denominator:

$\frac{a + bi}{c + di} = \frac{(a + bi)(c - di)}{(c + di)(c - di)} = \frac{(ac + bd) + (bc - ad)i}{c^2 + d^2}$

Complex Conjugate

The complex conjugate of $z = a + bi$ is $\bar{z} = a - bi$.

Properties:

• $z\bar{z} = a^2 + b^2 = |z|^2$
• $\overline{z_1 + z_2} = \bar{z}_1 + \bar{z}_2$
• $\overline{z_1 z_2} = \bar{z}_1 \cdot \bar{z}_2$
• $\bar{\bar{z}} = z$
• $z \in \mathbb{'\{'}R{'\}'} \iff z = \bar{z}$

---

The Argand Diagram

Representation

A complex number $z = a + bi$ is represented as the point $(a, b)$ in the complex plane (Argand diagram). The horizontal axis is the real axis, and the vertical axis is the imaginary axis.

Modulus

The modulus (absolute value) of $z = a + bi$ is the distance from the origin to $(a, b)$:

$|z| = \sqrt{a^2 + b^2}$

Properties:

• $|z| \ge 0$, with equality iff $z = 0$
• $|z_1 z_2| = |z_1|\,|z_2|$
• $\left|\dfrac{z_1}{z_2}\right| = \dfrac{|z_1|}{|z_2|}$, $z_2 \ne 0$
• $|z_1 + z_2| \le |z_1| + |z_2|$ (triangle inequality)

Argument

The argument of $z = a + bi$ is the angle $\theta$ measured from the positive real axis to the line segment joining the origin to $z$, measured anticlockwise:

$\theta = \arg(z) = \arctan\!\left(\frac{b}{a}\right)$

The principal argument, denoted $\mathrm{Arg}(z)$, is restricted to $(-\pi, \pi]$.

Quadrant-aware calculation:

Condition	$\arg(z)$
$a \gt 0$	$\arctan(b/a)$
$a \lt 0, b \ge 0$	$\arctan(b/a) + \pi$
$a \lt 0, b \lt 0$	$\arctan(b/a) - \pi$
$a = 0, b \gt 0$	$\pi/2$
$a = 0, b \lt 0$	$-\pi/2$

---

Modulus-Argument (Polar) Form

Definition

Every nonzero complex number can be written as:

$z = r(\cos\theta + i\sin\theta) = r\,\mathrm{cis}\;\theta$

where $r = |z|$ and $\theta = \arg(z)$. The notation $\mathrm{cis}\;\theta$ abbreviates $\cos\theta + i\sin\theta$.

Conversion

Polar to Cartesian:

$a = r\cos\theta, \quad b = r\sin\theta$

Cartesian to Polar:

$r = \sqrt{a^2 + b^2}, \quad \theta = \arg(z) \mathrm{ (using quadrant table)}$

Multiplication and Division in Polar Form

These operations are dramatically simpler in polar form:

$z_1 z_2 = r_1 r_2\,\mathrm{cis}(\theta_1 + \theta_2)$

$\frac{z_1}{z_2} = \frac{r_1}{r_2}\,\mathrm{cis}(\theta_1 - \theta_2), \quad z_2 \ne 0$

Multiplication: multiply moduli, add arguments. Division: divide moduli, subtract arguments.

Powers in Polar Form

$z^n = r^n\,\mathrm{cis}(n\theta)$

---

De Moivre's Theorem

Statement

For any integer $n$ and any complex number $z = r\,\mathrm{cis}\;\theta$:

$(\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)$

Proof by Induction (Positive Integers)

Base case ($n = 1$): Trivially true.

Inductive step: Assume $(\cos\theta + i\sin\theta)^k = \cos(k\theta) + i\sin(k\theta)$. Then:

$(\cos\theta + i\sin\theta)^{k+1} = (\cos\theta + i\sin\theta)(\cos k\theta + i\sin k\theta)$

$= \cos\theta\cos k\theta - \sin\theta\sin k\theta + i(\cos\theta\sin k\theta + \sin\theta\cos k\theta)$

$= \cos((k+1)\theta) + i\sin((k+1)\theta)$

by the angle addition formulas.

Applications

Expanding $\cos(n\theta)$ and $\sin(n\theta)$ in terms of $\cos\theta$ and $\sin\theta$:

Using the binomial theorem on $(\cos\theta + i\sin\theta)^n$ and equating real and imaginary parts:

$\cos(n\theta) + i\sin(n\theta) = \sum_{k=0}^{n} \binom{n}{k} \cos^{n-k}\theta \cdot (i\sin\theta)^k$

Example. For $n = 3$:

$\cos 3\theta = 4\cos^3\theta - 3\cos\theta$ $\sin 3\theta = 3\sin\theta - 4\sin^3\theta$

Multiple-angle identities of any order can be derived this way.

---

Roots of Complex Numbers

$n$-th Roots

The $n$-th roots of a complex number $z = r\,\mathrm{cis}\;\theta$ are:

$w_k = r^{1/n}\,\mathrm{cis}\!\left(\frac{\theta + 2k\pi}{n}\right), \quad k = 0, 1, 2, \ldots, n - 1$

These $n$ roots are equally spaced on a circle of radius $r^{1/n}$, centred at the origin, with angular separation $\dfrac{2\pi}{n}$.

Roots of Unity

The $n$-th roots of unity are the solutions to $z^n = 1$:

$\omega_k = \mathrm{cis}\!\left(\frac{2k\pi}{n}\right), \quad k = 0, 1, \ldots, n - 1$

Properties:

• $\omega_0 = 1$ is always a root.
• The roots are symmetric about the real axis (complex conjugate pairs for non-real roots).
• The sum of all $n$-th roots of unity is $0$.
• The product of all $n$-th roots of unity is $(-1)^{n+1}$.

Example. The cube roots of unity ($n = 3$):

$\omega_0 = 1, \quad \omega_1 = \mathrm{cis}\!\left(\frac{2\pi}{3}\right) = -\frac{1}{2} + \frac{\sqrt{3}}{2}i, \quad \omega_2 = \mathrm{cis}\!\left(\frac{4\pi}{3}\right) = -\frac{1}{2} - \frac{\sqrt{3}}{2}i$

Note: $1 + \omega_1 + \omega_2 = 0$.

Sum of Roots of a Quadratic

If $\alpha$ and $\beta$ are the roots of $az^2 + bz + c = 0$ with $a \ne 0$:

$\alpha + \beta = -\frac{b}{a}, \quad \alpha\beta = \frac{c}{a}$

These follow directly from expanding $a(z - \alpha)(z - \beta) = 0$.

---

Euler's Formula

Statement

$e^{i\theta} = \cos\theta + i\sin\theta$

This unifies exponential, trigonometric, and complex number theory. Substituting $\theta = \pi$ yields Euler's identity:

$e^{i\pi} + 1 = 0$

Consequences

Complex exponential form:

$z = re^{i\theta}$

Exponential laws apply directly:

$z_1 z_2 = r_1 r_2\,e^{i(\theta_1 + \theta_2)}, \qquad z^n = r^n\,e^{in\theta}$

Derivation of trig identities. From $e^{i\theta} = \cos\theta + i\sin\theta$ and $e^{-i\theta} = \cos\theta - i\sin\theta$:

$\cos\theta = \frac{e^{i\theta} + e^{-i\theta}}{2}, \qquad \sin\theta = \frac{e^{i\theta} - e^{-i\theta}}{2i}$

---

Common Pitfalls

1. Argument quadrant errors. $\arctan(b/a)$ alone does not determine the correct quadrant. Always check the signs of $a$ and $b$ and adjust by $\pi$ when the point lies in the second or third quadrant.

2. Conjugate distribution. $\overline{z_1 + z_2} = \bar{z}_1 + \bar{z}_2$ but $\overline{z_1 z_2} \ne \bar{z}_1 \cdot \bar{z}_2$ is wrong -- it IS $\bar{z}_1 \cdot \bar{z}_2$. The error is thinking conjugation distributes over all operations differently; it distributes correctly over addition and multiplication.

3. Missing roots. An $n$-th degree equation has exactly $n$ roots (counting multiplicity) in $\mathbb{'\{'}C{'\}'}$. When finding $n$-th roots, always generate all $n$ values by varying $k$.

4. Modulus of a product. $|z_1 z_2| = |z_1|\,|z_2|$, but $|z_1 + z_2| \ne |z_1| + |z_2|$ in general (equality holds only when $z_1$ and $z_2$ have the same argument).

5. Polar form of zero. $z = 0$ has no well-defined argument and cannot be expressed in polar form $r\,\mathrm{cis}\;\theta$ since $r = 0$.

6. De Moivre's theorem scope. De Moivre's theorem as stated holds for integer exponents. For non-integer exponents, use $r^n e^{in\theta}$, but note that the result is multi-valued.

---

Practice Problems

Problem 1

Express $z = 1 - \sqrt{3}\,i$ in polar form and compute $z^5$.

Problem 2

Find all complex solutions to $z^3 = -8i$.

Problem 3

Prove that $\cos 4\theta = 8\cos^4\theta - 8\cos^2\theta + 1$ using De Moivre's theorem.

Problem 4

If $z = 2 + 3i$, express $\dfrac{1}{z}$ in the form $a + bi$.

Problem 5

Find the fourth roots of $16$ and show that their sum is $0$.

Problem 6

Solve $z^2 + (2 + 4i)z + (-1 + 6i) = 0$.

Problem 7

Prove that for any complex number $z$, $z + \bar{z} = 2\mathrm{Re}(z)$ and $z - \bar{z} = 2i\,\mathrm{Im}(z)$.

Problem 8

The complex numbers $z_1$ and $z_2$ satisfy $|z_1| = 3$, $|z_2| = 5$, and $|z_1 - z_2| = 7$. Find $|z_1 + z_2|$.

Answers to Selected Problems

Problem 1: $|z| = \sqrt{1 + 3} = 2$. Since $a = 1 \gt 0$ and $b = -\sqrt{3} \lt 0$, $\arg(z) = -\pi/3$. So $z = 2\,\mathrm{cis}(-\pi/3)$. $z^5 = 2^5\,\mathrm{cis}(-5\pi/3) = 32\,\mathrm{cis}(\pi/3) = 32\!\left(\dfrac{1}{2} + \dfrac{\sqrt{3}}{2}i\right) = 16 + 16\sqrt{3}\,i$.

Problem 2: $-8i = 8\,\mathrm{cis}(-\pi/2) = 8\,\mathrm{cis}(3\pi/2)$. $z_k = 8^{1/3}\,\mathrm{cis}\!\left(\dfrac{3\pi/2 + 2k\pi}{3}\right) = 2\,\mathrm{cis}\!\left(\dfrac{\pi}{2} + \dfrac{2k\pi}{3}\right)$ for $k = 0, 1, 2$. $z_0 = 2\,\mathrm{cis}(\pi/2) = 2i$, $z_1 = 2\,\mathrm{cis}(7\pi/6) = -\sqrt{3} - i$, $z_2 = 2\,\mathrm{cis}(11\pi/6) = \sqrt{3} - i$.

Problem 3: By De Moivre: $\cos 4\theta + i\sin 4\theta = (\cos\theta + i\sin\theta)^4$. Expanding by binomial theorem: $\cos^4\theta + 4i\cos^3\theta\sin\theta - 6\cos^2\theta\sin^2\theta - 4i\cos\theta\sin^3\theta + \sin^4\theta$. Equating real parts: $\cos 4\theta = \cos^4\theta - 6\cos^2\theta\sin^2\theta + \sin^4\theta$. Using $\sin^2\theta = 1 - \cos^2\theta$: $\cos 4\theta = \cos^4\theta - 6\cos^2\theta(1-\cos^2\theta) + (1-\cos^2\theta)^2 = 8\cos^4\theta - 8\cos^2\theta + 1$.

Problem 4: $\dfrac{1}{z} = \dfrac{\bar{z}}{z\bar{z}} = \dfrac{2 - 3i}{4 + 9} = \dfrac{2 - 3i}{13} = \dfrac{2}{13} - \dfrac{3}{13}i$.

Problem 5: $16 = 16\,\mathrm{cis}(0)$. Fourth roots: $w_k = 2\,\mathrm{cis}(k\pi/2)$ for $k = 0, 1, 2, 3$. Roots: $2, \; 2i, \; -2, \; -2i$. Sum: $2 + 2i - 2 - 2i = 0$.

Problem 8: $|z_1 + z_2|^2 + |z_1 - z_2|^2 = 2(|z_1|^2 + |z_2|^2)$ (parallelogram law). $|z_1 + z_2|^2 = 2(9 + 25) - 49 = 68 - 49 = 19$. So $|z_1 + z_2| = \sqrt{19}$.

---

Worked Examples

Worked Example: Division in Polar Form

Express $\dfrac{z_1}{z_2}$ in Cartesian form where $z_1 = 4\,\mathrm{cis}\!\left(\dfrac{2\pi}{3}\right)$ and $z_2 = 2\,\mathrm{cis}\!\left(\dfrac{\pi}{6}\right)$.

Solution

In polar form:

$\frac{z_1}{z_2} = \frac{4}{2}\,\mathrm{cis}\!\left(\frac{2\pi}{3} - \frac{\pi}{6}\right) = 2\,\mathrm{cis}\!\left(\frac{\pi}{2}\right) = 2\left(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\right) = 2(0 + i) = 2i$

Converting to Cartesian form: $0 + 2i$.

Worked Example: Solving a Quadratic with Complex Coefficients

Solve $z^2 + (1 - 3i)z + (4 + 3i) = 0$.

Solution

Using the quadratic formula $z = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:

$\Delta = (1 - 3i)^2 - 4(1)(4 + 3i) = 1 - 6i - 9 - 16 - 12i = -24 - 18i$

Find $\sqrt{-24 - 18i}$. Let $\sqrt{-24 - 18i} = a + bi$ where $a, b \in \mathbb{'\{'}R{'\}'}$. Then $a^2 - b^2 = -24$ and $2ab = -18$, so $b = -9/a$.

Substituting: $a^2 - 81/a^2 = -24 \implies a^4 + 24a^2 - 81 = 0$. Let $u = a^2$: $u^2 + 24u - 81 = 0 \implies u = \dfrac{-24 \pm \sqrt{576 + 324}}{2} = \dfrac{-24 \pm 30}{2}$. Since $u = a^2 \ge 0$: $u = 3$, so $a = \sqrt{3}$ (taking $a \gt 0$) and $b = -9/\sqrt{3} = -3\sqrt{3}$.

Thus $\sqrt{\Delta} = \sqrt{3} - 3\sqrt{3}\,i$.

$z = \frac{-(1 - 3i) \pm (\sqrt{3} - 3\sqrt{3}\,i)}{2}$

$z_1 = \dfrac{-1 + 3i + \sqrt{3} - 3\sqrt{3}\,i}{2} = \dfrac{-1 + \sqrt{3}}{2} + \dfrac{3 - 3\sqrt{3}}{2}i$

$z_2 = \dfrac{-1 + 3i - \sqrt{3} + 3\sqrt{3}\,i}{2} = \dfrac{-1 - \sqrt{3}}{2} + \dfrac{3 + 3\sqrt{3}}{2}i$

Worked Example: Using De Moivre to Derive a Trig Identity

Use De Moivre's theorem to express $\cos 5\theta$ as a polynomial in $\cos\theta$.

Solution

By De Moivre: $\cos 5\theta + i\sin 5\theta = (\cos\theta + i\sin\theta)^5$.

Expanding by the binomial theorem:

$(\cos\theta + i\sin\theta)^5 = \cos^5\theta + 5i\cos^4\theta\sin\theta - 10\cos^3\theta\sin^2\theta - 10i\cos^2\theta\sin^3\theta + 5\cos\theta\sin^4\theta + i\sin^5\theta$

Equating real parts:

$\cos 5\theta = \cos^5\theta - 10\cos^3\theta\sin^2\theta + 5\cos\theta\sin^4\theta$

Using $\sin^2\theta = 1 - \cos^2\theta$:

$\cos 5\theta = \cos^5\theta - 10\cos^3\theta(1 - \cos^2\theta) + 5\cos\theta(1 - \cos^2\theta)^2$

$= \cos^5\theta - 10\cos^3\theta + 10\cos^5\theta + 5\cos\theta - 10\cos^3\theta + 5\cos^5\theta$

$= 16\cos^5\theta - 20\cos^3\theta + 5\cos\theta$

Worked Example: Fifth Roots of a Complex Number

Find all fifth roots of $-16 + 16\sqrt{3}\,i$.

Solution

Express in polar form: $z = -16 + 16\sqrt{3}\,i$. Modulus: $|z| = \sqrt{256 + 768} = \sqrt{1024} = 32$. Since $a = -16 \lt 0$ and $b = 16\sqrt{3} \gt 0$: $\arg(z) = \arctan\!\left(\dfrac{16\sqrt{3}}{-16}\right) + \pi = \arctan(-\sqrt{3}) + \pi = -\dfrac{\pi}{3} + \pi = \dfrac{2\pi}{3}$.

So $z = 32\,\mathrm{cis}\!\left(\dfrac{2\pi}{3}\right)$.

The fifth roots are:

$w_k = 32^{1/5}\,\mathrm{cis}\!\left(\frac{2\pi/3 + 2k\pi}{5}\right) = 2\,\mathrm{cis}\!\left(\frac{2\pi + 6k\pi}{15}\right), \quad k = 0, 1, 2, 3, 4$

$w_0 = 2\,\mathrm{cis}\!\left(\dfrac{2\pi}{15}\right)$, $w_1 = 2\,\mathrm{cis}\!\left(\dfrac{8\pi}{15}\right)$, $w_2 = 2\,\mathrm{cis}\!\left(\dfrac{14\pi}{15}\right)$, $w_3 = 2\,\mathrm{cis}\!\left(\dfrac{20\pi}{15}\right) = 2\,\mathrm{cis}\!\left(\dfrac{4\pi}{3}\right) = -1 - \sqrt{3}\,i$, $w_4 = 2\,\mathrm{cis}\!\left(\dfrac{26\pi}{15}\right) = 2\,\mathrm{cis}\!\left(-\dfrac{4\pi}{15}\right)$.

These five roots lie on a circle of radius $2$, equally spaced by $\dfrac{2\pi}{5}$.

---

Additional Common Pitfalls

• Forgetting to convert to polar before using De Moivre. De Moivre's theorem requires the form $r\,\mathrm{cis}\;\theta$. Attempting to raise $a + bi$ to a power directly without conversion leads to algebraic errors.

• Square root of a complex number gives two values. Every nonzero complex number has exactly two square roots. When solving $\sqrt{\Delta}$ in the quadratic formula with complex coefficients, both signs must be considered.

• Principal argument range. $\mathrm{Arg}(z) \in (-\pi, \pi]$. If De Moivre produces an argument outside this range, add or subtract $2\pi$ to normalise. For example, $\mathrm{cis}\!\left(\dfrac{7\pi}{4}\right)$ is preferred over $\mathrm{cis}\!\left(-\dfrac{\pi}{4}\right)$ in some contexts, but both are valid.

• Conjugate pair errors in polynomials. If $a + bi$ is a root of a polynomial with real coefficients, then $a - bi$ is also a root. Forgetting this symmetry loses information about the polynomial.

• Modulus of a sum vs. sum of moduli. $|z_1 + z_2| \le |z_1| + |z_2|$ with equality only when $z_1$ and $z_2$ point in the same direction. Computing $|z_1 + z_2|$ as $|z_1| + |z_2|$ is almost always wrong.

---

Exam-Style Problems

Problem 9

Given $z = 3 + 4i$ and $w = 1 - 2i$, find $|z^2 w|$ and $\arg(z^2 w)$.

Problem 10

If $|z - 2i| = |z + 4|$ and $\arg(z) = \dfrac{\pi}{4}$, find $z$ in Cartesian form.

Problem 11

Express $\sin 3\theta$ purely in terms of $\sin\theta$ using De Moivre's theorem.

Problem 12

The complex numbers $\alpha$ and $\beta$ are roots of $z^2 - 4z + 13 = 0$. Show that $\alpha^3 + \beta^3 = -10$.

Problem 13

Find the locus of points $z$ in the complex plane satisfying $|z - 3 - 4i| = 2|z + 1 + 2i|$.

Problem 14

Find all solutions to $z^4 + z^2 + 1 = 0$ and represent them on an Argand diagram.

Problem 15

Prove that $|z_1 + z_2|^2 + |z_1 - z_2|^2 = 2(|z_1|^2 + |z_2|^2)$ for all $z_1, z_2 \in \mathbb{'\{'}C{'\}'}$ (parallelogram law).

Answers to Additional Problems

Problem 9: $|z| = 5$, $|w| = \sqrt{5}$. $|z^2 w| = |z|^2|w| = 25\sqrt{5}$. $\arg(z) = \arctan(4/3)$, $\arg(w) = \arctan(-2)$. $\arg(z^2 w) = 2\arctan(4/3) + \arctan(-2)$. Since $z^2 = (3+4i)^2 = -7 + 24i$ and $z^2 w = (-7+24i)(1-2i) = -7 + 14i + 24i + 48 = 41 + 38i$. $|z^2 w| = \sqrt{41^2 + 38^2} = \sqrt{1681 + 1444} = \sqrt{3125} = 25\sqrt{5}$. Confirmed. $\arg(z^2 w) = \arctan(38/41) \approx 0.747\;\mathrm{rad}$.

Problem 10: Let $z = x + yi$ with $x \gt 0$, $y \gt 0$ (since $\arg(z) = \pi/4$), and $y = x$. $|z - 2i|^2 = x^2 + (x-2)^2 = 2x^2 - 4x + 4$. $|z + 4|^2 = (x+4)^2 + x^2 = 2x^2 + 8x + 16$. Setting equal: $2x^2 - 4x + 4 = 2x^2 + 8x + 16 \implies -12x = 12 \implies x = -1$. But $x \gt 0$ is required. This gives no solution in the principal argument range. Rechecking: $\arg(z) = \pi/4 \implies z = t(1 + i)$ for $t \gt 0$. $|t + ti - 2i|^2 = t^2 + (t-2)^2 = 2t^2 - 4t + 4$. $|t + ti + 4|^2 = (t+4)^2 + t^2 = 2t^2 + 8t + 16$. $2t^2 - 4t + 4 = 2t^2 + 8t + 16 \implies t = -1$. No positive solution exists. The locus circle and ray do not intersect.

Problem 11: From De Moivre, equating imaginary parts: $\sin 3\theta = 3\cos^2\theta\sin\theta - \sin^3\theta = 3(1 - \sin^2\theta)\sin\theta - \sin^3\theta = 3\sin\theta - 4\sin^3\theta$.

Problem 12: $\alpha + \beta = 4$ and $\alpha\beta = 13$. $\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta) = 64 - 3(13)(4) = 64 - 156 = -92$. Note: the problem statement claims $-10$, but the correct answer is $-92$. This demonstrates the importance of always verifying claims.

Problem 13: Let $z = x + yi$. $|z - (3+4i)|^2 = (x-3)^2 + (y-4)^2$. $|z - (-1-2i)|^2 = (x+1)^2 + (y+2)^2$. Setting $(x-3)^2 + (y-4)^2 = 4[(x+1)^2 + (y+2)^2]$: $x^2 - 6x + 9 + y^2 - 8y + 16 = 4x^2 + 8x + 4 + 4y^2 + 16y + 16$. $-3x^2 - 14x - 3y^2 - 24y + 5 = 0$. $x^2 + y^2 + \dfrac{14}{3}x + 8y = \dfrac{5}{3}$. Completing squares: $\left(x + \dfrac{7}{3}\right)^2 + (y + 4)^2 = \dfrac{5}{3} + \dfrac{49}{9} + 16 = \dfrac{200}{9}$. This is a circle with centre $\left(-\dfrac{7}{3}, -4\right)$ and radius $\dfrac{10\sqrt{2}}{3}$.

Problem 14: $z^4 + z^2 + 1 = 0$. Let $u = z^2$: $u^2 + u + 1 = 0 \implies u = \dfrac{-1 \pm \sqrt{3}\,i}{2}$. So $z^2 = \mathrm{cis}\!\left(\dfrac{2\pi}{3}\right)$ or $z^2 = \mathrm{cis}\!\left(-\dfrac{2\pi}{3}\right)$. $z = \pm\,\mathrm{cis}\!\left(\dfrac{\pi}{3}\right)$ or $z = \pm\,\mathrm{cis}\!\left(-\dfrac{\pi}{3}\right)$. The four roots are $\pm\dfrac{1}{2} \pm \dfrac{\sqrt{3}}{2}i$, which are the primitive 6th roots of unity.

Problem 15: $|z_1 + z_2|^2 = (z_1 + z_2)\overline{(z_1 + z_2)} = (z_1 + z_2)(\bar{z}_1 + \bar{z}_2) = |z_1|^2 + z_1\bar{z}_2 + \bar{z}_1 z_2 + |z_2|^2$. Similarly, $|z_1 - z_2|^2 = |z_1|^2 - z_1\bar{z}_2 - \bar{z}_1 z_2 + |z_2|^2$. Adding: $|z_1 + z_2|^2 + |z_1 - z_2|^2 = 2|z_1|^2 + 2|z_2|^2 = 2(|z_1|^2 + |z_2|^2)$.

---

If You Get These Wrong, Revise:

• Trigonometric identities → Review the trigonometry section for angle addition and double-angle formulas
• Polar coordinates and argument calculation → Review ./complex-numbers (section on Argument)
• Binomial theorem expansion → Review algebra and series topics
• Polynomial roots and Vieta's formulas → Review ./complex-numbers (section on Sum of Roots)

For the A-Level Further Maths treatment of this topic, see Complex Numbers.