Mastering the four operations, conjugation, and modulus — the computational foundation of complex analysis
Complex addition and subtraction work component-wise, just like vector addition:
If $z_1 = a + bi$ and $z_2 = c + di$, then:
Let $z_1 = 3 + 4i$ and $z_2 = 1 - 2i$. Find $z_1 + z_2$ and $z_1 - z_2$.
In the complex plane, $z_1 + z_2$ is found using the parallelogram rule, identical to adding 2D vectors. This is why $|z_1 + z_2| \leq |z_1| + |z_2|$ (triangle inequality).
Multiply complex numbers using the distributive property (FOIL), remembering that $i^2 = -1$:
Compute $(2 + 3i)(4 - i)$.
Using FOIL:
Compute $(1 + i)^2$ and $(1 + i)^4$.
Starting from a complex number, repeated squaring can land back on a real number!
$|z_1 z_2| = |z_1| \cdot |z_2|$. Multiplication scales lengths and adds angles (explored further in Section 7.2).
The conjugate is the single most important tool for simplifying complex expressions.
If $z = a + bi$, then $\bar{z} = a - bi$. The conjugate flips the sign of the imaginary part.
| Property | Formula |
|---|---|
| Sum | $\overline{z_1 + z_2} = \bar{z}_1 + \bar{z}_2$ |
| Product | $\overline{z_1 z_2} = \bar{z}_1 \bar{z}_2$ |
| Quotient | $\overline{\left(\frac{z_1}{z_2}\right)} = \frac{\bar{z}_1}{\bar{z}_2}$ |
| Product with conjugate | $z \bar{z} = |z|^2 = a^2 + b^2$ |
| Real part extraction | $\text{Re}(z) = \frac{z + \bar{z}}{2}$ |
| Imaginary part extraction | $\text{Im}(z) = \frac{z - \bar{z}}{2i}$ |
| Double conjugate | $\overline{\bar{z}} = z$ |
Students often write $\text{Im}(3 + 4i) = 4i$. This is wrong! The imaginary part is a real number: $\text{Im}(3 + 4i) = 4$, not $4i$.
To divide complex numbers, multiply numerator and denominator by the conjugate of the denominator. This makes the denominator real.
Compute $\displaystyle\frac{3 + 4i}{1 - 2i}$.
Multiply top and bottom by the conjugate of the denominator, $1 + 2i$:
Numerator: $(3+4i)(1+2i) = 3 + 6i + 4i + 8i^2 = 3 + 10i - 8 = -5 + 10i$
Denominator: $(1)^2 + (2)^2 = 5$
Simplify $\displaystyle\frac{i}{2+i} + \frac{2}{1-i}$.
Rationalize each fraction:
Add: $\displaystyle\frac{1+2i}{5} + 1 + i = \frac{1+2i + 5 + 5i}{5} = \frac{6 + 7i}{5}$
The modulus generalizes absolute value to the complex plane.
$|z| = |a + bi| = \sqrt{a^2 + b^2}$. The distance between $z_1$ and $z_2$ is $|z_1 - z_2|$.
Find $|3 - 4i|$ and the distance between $z_1 = 1 + i$ and $z_2 = 4 + 5i$.
For any complex numbers $z_1, z_2$:
Step 1: Add componentwise
Addition treats $1$ and $i$ as independent directions, exactly like vectors:
$$z + w = (3 + 1) + (2 - 4)i = 4 - 2i$$Step 2: Multiply by FOIL, then use $i^2 = -1$
$$zw = (3 + 2i)(1 - 4i) = 3 - 12i + 2i - 8i^2$$Since $i^2 = -1$, the term $-8i^2$ becomes $+8$:
$$zw = 3 + 8 - 10i = 11 - 10i$$Step 3: Conjugate first, then multiply
The conjugate flips the sign of the imaginary part: $\bar{z} = 3 - 2i$.
$$\bar{z}w = (3 - 2i)(1 - 4i) = 3 - 12i - 2i + 8i^2 = 3 - 8 - 14i$$Final answers:
$$\boxed{z + w = 4 - 2i, \qquad zw = 11 - 10i, \qquad \bar{z}w = -5 - 14i}$$Step 1: Multiply top and bottom by the conjugate of the denominator
The conjugate of $2 - 3i$ is $2 + 3i$. Multiplying by $\frac{2+3i}{2+3i}$ changes nothing but clears the $i$ out of the denominator:
$$\frac{4 + 7i}{2 - 3i} \cdot \frac{2 + 3i}{2 + 3i}$$Step 2: Expand the numerator
$$(4 + 7i)(2 + 3i) = 8 + 12i + 14i + 21i^2 = 8 - 21 + 26i = -13 + 26i$$Step 3: The denominator is always real
$$(2 - 3i)(2 + 3i) = 2^2 + 3^2 = 13$$This is $|2 - 3i|^2$ — the whole point of the conjugate trick.
Step 4: Divide through
$$\frac{-13 + 26i}{13} = -1 + 2i$$ $$\boxed{\frac{4 + 7i}{2 - 3i} = -1 + 2i}$$Step 1: Multiply a concrete number by $i$
Take $z = 3 + i$, the point $(3, 1)$ in the Argand plane. Then
$$iz = i(3 + i) = 3i + i^2 = -1 + 3i$$which is the point $(-1, 3)$.
Step 2: The length did not change
$$|z| = \sqrt{3^2 + 1^2} = \sqrt{10}, \qquad |iz| = \sqrt{(-1)^2 + 3^2} = \sqrt{10}$$Step 3: The two vectors are perpendicular
Treating them as vectors in $\mathbb{R}^2$, the dot product is
$$(3)(-1) + (1)(3) = -3 + 3 = 0$$so the angle between $z$ and $iz$ is exactly $90^\circ$.
Step 4: Generalise
For any $z = a + bi$,
$$i(a + bi) = -b + ai$$which is precisely the map $(a, b) \mapsto (-b, a)$ — the standard counter-clockwise quarter turn.
$$\boxed{\text{Multiplying by } i \text{ rotates a point } 90^\circ \text{ counter-clockwise about the origin.}}$$Step 1: Write $z = a + bi$ and square it
$$z^2 = (a + bi)^2 = a^2 - b^2 + 2abi$$Step 2: Match real and imaginary parts
Two complex numbers are equal exactly when both parts agree:
$$a^2 - b^2 = 5, \qquad 2ab = -12 \;\Longrightarrow\; b = -\frac{6}{a}$$Step 3: Substitute and solve the resulting quadratic in $a^2$
$$a^2 - \frac{36}{a^2} = 5 \;\Longrightarrow\; a^4 - 5a^2 - 36 = 0 \;\Longrightarrow\; (a^2 - 9)(a^2 + 4) = 0$$Because $a$ is real, we must discard $a^2 = -4$ and keep $a^2 = 9$, so $a = \pm 3$.
Step 4: Recover $b$ from $b = -6/a$
If $a = 3$ then $b = -2$; if $a = -3$ then $b = 2$.
$$\boxed{z = 3 - 2i \quad \text{or} \quad z = -3 + 2i}$$These problems are similar in style and difficulty to past exam questions. Click each problem to reveal the step-by-step solution.
Step 1: Compute $z\bar{w}$
Here $\bar{w} = 3 + i$, so
$$z\bar{w} = (2 + 5i)(3 + i) = 6 + 2i + 15i + 5i^2 = 6 - 5 + 17i = 1 + 17i$$Step 2: Reuse that product for the quotient
The conjugate trick says $\dfrac{z}{w} = \dfrac{z\bar{w}}{w\bar{w}} = \dfrac{z\bar{w}}{|w|^2}$, and $|w|^2 = 3^2 + 1^2 = 10$:
$$\frac{z}{w} = \frac{1 + 17i}{10} = \frac{1}{10} + \frac{17}{10}i$$Step 3: Take the modulus
$$\left|\frac{z}{w}\right| = \frac{\sqrt{1^2 + 17^2}}{10} = \frac{\sqrt{290}}{10}$$Step 4: Cross-check with the multiplicative property
Since $|z/w| = |z|/|w|$,
$$\frac{|z|}{|w|} = \frac{\sqrt{4 + 25}}{\sqrt{9 + 1}} = \frac{\sqrt{29}}{\sqrt{10}} = \frac{\sqrt{290}}{10} \quad \checkmark$$ $$\boxed{z\bar{w} = 1 + 17i, \qquad \frac{z}{w} = \frac{1}{10} + \frac{17}{10}i, \qquad \left|\frac{z}{w}\right| = \frac{\sqrt{290}}{10}}$$Step 1: Rationalise using the conjugate $1 + i$
$$\frac{1+i}{1-i} \cdot \frac{1+i}{1+i} = \frac{(1+i)^2}{(1-i)(1+i)}$$Step 2: Expand numerator and denominator
$$(1+i)^2 = 1 + 2i + i^2 = 2i, \qquad (1-i)(1+i) = 1^2 + 1^2 = 2$$ $$\frac{1+i}{1-i} = \frac{2i}{2} = i$$Step 3: Use the four-step cycle of powers of $i$
Since $i^1 = i,\; i^2 = -1,\; i^3 = -i,\; i^4 = 1$, only the remainder of the exponent mod $4$ matters:
$$2026 = 4(506) + 2 \;\Longrightarrow\; i^{2026} = i^2 = -1$$ $$\boxed{\frac{1+i}{1-i} = i \qquad \text{and} \qquad \left(\frac{1+i}{1-i}\right)^{2026} = -1}$$This is exactly the arithmetic you will do in Chapter 2 and again in Section 7.3, where $r^2 + 4r + 13 = 0$ is the characteristic equation of $y'' + 4y' + 13y = 0$.
Step 1: Square the root
$$(-2 + 3i)^2 = 4 - 12i + 9i^2 = 4 - 9 - 12i = -5 - 12i$$Step 2: Multiply by the linear coefficient
$$4(-2 + 3i) = -8 + 12i$$Step 3: Add everything, including the constant
$$(-5 - 12i) + (-8 + 12i) + 13 = (-5 - 8 + 13) + (-12 + 12)i = 0 + 0i \quad \checkmark$$Step 4: Why the conjugate is automatically a root
Conjugation respects sums and products, so applying a bar to $r^2 + 4r + 13 = 0$ gives
$$\overline{r^2 + 4r + 13} = \bar{r}^2 + 4\bar{r} + 13 = \bar{0} = 0$$because the coefficients $1, 4, 13$ are real and therefore unchanged by conjugation.
$$\boxed{r = -2 \pm 3i \text{ are the roots; complex roots of a real polynomial always come in conjugate pairs.}}$$Try these problems to test your understanding. Click "Show Solution" to check your work.
Let $z = 2 - 5i$ and $w = -1 + 3i$. Compute $z \cdot w$ and $|z \cdot w|$.
Verify: $|z| = \sqrt{4+25} = \sqrt{29}$, $|w| = \sqrt{1+9} = \sqrt{10}$, and $|z||w| = \sqrt{29}\sqrt{10} = \sqrt{290}$. Checks out!
Find all complex numbers $z$ such that $z^2 = -8 + 6i$.
Let $z = a + bi$. Then $z^2 = a^2 - b^2 + 2abi = -8 + 6i$.
Matching real and imaginary parts: $a^2 - b^2 = -8$ and $2ab = 6$, so $b = 3/a$.
Substituting: $a^2 - 9/a^2 = -8$, so $a^4 + 8a^2 - 9 = 0$. This factors as $(a^2+9)(a^2-1) = 0$.
Since $a$ is real, $a^2 = 1$, so $a = \pm 1$ and $b = \pm 3$.
Show that $\displaystyle\frac{1}{z} = \frac{\bar{z}}{|z|^2}$ for any $z \neq 0$.
Let $z = a + bi$. Then:
This is exactly the division rule with numerator $1$.
Test your understanding with these multiple-choice questions. Click a choice to see if you're correct.
Answer: (C) $11 - 10i$. FOIL gives $3 - 12i + 2i - 8i^2$. The trap is $-8i^2$: since $i^2 = -1$ this becomes $+8$, so the real part is $3 + 8 = 11$ and the imaginary part is $-12 + 2 = -10$.
Answer: (A) $-1 + 2i$. Multiply top and bottom by $2 + 3i$: the numerator becomes $-13 + 26i$ and the denominator becomes $2^2 + 3^2 = 13$. Dividing gives $-1 + 2i$. Check: $(-1+2i)(2-3i) = 4 + 7i$.
Answer: (D) $-1$. Powers of $i$ repeat with period $4$. Since $2026 = 4(506) + 2$, we get $i^{2026} = i^2 = -1$.
Answer: (B) $13$. The modulus is $\sqrt{(-5)^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$. Note the real part is squared, so its minus sign disappears — the modulus is a distance and can never be negative.
Answer: (A) $3 + 2i$. $i(2 - 3i) = 2i - 3i^2 = 3 + 2i$. Geometrically $(2,-3)$ has been rotated $90^\circ$ counter-clockwise to $(3,2)$; the modulus stays $\sqrt{13}$ because $|i| = 1$.
Answer: (C) $a^2 + b^2$. $(a+bi)(a-bi) = a^2 - (bi)^2 = a^2 + b^2 = |z|^2$, which is real and non-negative. Option (B) is $z + \bar{z} = 2a$, and option (D) is the real part of $z^2$.
Now that you can add, subtract, multiply, and divide complex numbers, the next section introduces the polar form $z = re^{i\theta}$. This representation turns multiplication into adding angles and multiplying lengths — a massive simplification. It also leads to Euler's formula, one of the most beautiful equations in mathematics.