Step 33: Complex Numbers

The number i and arithmetic in Cartesian form

The quadratic \( z^2 + 6z + 25 = 0 \) has discriminant \( 36-100 = -64 \). With real numbers only, the story ends there: no real number has a negative square. Complex numbers continue it. We invent one new number, \( i \), with

and keep every other rule of algebra exactly as before. Then \( (8i)^2 = 64i^2 = -64 \), so \( \sqrt{-64} = 8i \), and the quadratic formula gives \( z = \dfrac{-6 \pm 8i}{2} = -3 \pm 4i \). Powers of \( i \) repeat in a cycle of four, \( i, -1, -i, 1 \), so \( i^{n} \) depends only on the remainder when \( n \) is divided by 4.

Two complex numbers are equal only when their real parts are equal and their imaginary parts are equal. So one complex equation gives two real equations, which is the most useful trick in this step.

The form \( x + iy \) is Cartesian form. Add and subtract part by part. Multiply by expanding brackets, then replace \( i^2 \) by \( -1 \): for example \( (2 + 5i)(3-i) = 6-2i + 15i-5i^2 = 11 + 13i \).

Division needs one more idea. The conjugate of \( z = x + iy \) is \( z^* = x-iy \), and

which is real. So to divide, multiply the top and bottom of the fraction by the conjugate of the bottom. The bottom becomes real and the answer splits into parts.

Conjugate pairs of roots

The roots \( -3 \pm 4i \) of the opening quadratic differ only in the sign of \( i \). That always happens when the coefficients are real.

The reason is that conjugating respects arithmetic, \( (z + w)^* = z^* + w^* \) and \( (zw)^* = z^*w^* \), while a real coefficient is its own conjugate. Conjugate every term of \( p(\alpha) = 0 \) and you get \( p(\alpha^*) = 0 \).

A conjugate pair \( p \pm qi \) multiplies out to a real quadratic factor: the sum of the roots is \( 2p \) and the product is \( p^2 + q^2 \), so

The Argand diagram, modulus and argument

A complex number is fixed by two real numbers, so draw \( x + iy \) as the point \( (x, y) \). The horizontal axis is the real axis and the vertical axis is the imaginary axis; this plane is the Argand diagram. Each piece of arithmetic becomes a picture. The conjugate is a reflection in the real axis. Addition works exactly like adding vectors (see Vectors and Lines in 3D), so \( z + w \) completes a parallelogram, and \( z-w \) is the step from \( w \) to \( z \).

Two facts follow straight from the picture: \( |z|^2 = zz^* \), and \( |z-w| \) is the distance between the points \( z \) and \( w \). To find an argument, sketch the point, find the acute angle \( \alpha = \tan^{-1}\left|\tfrac{y}{x}\right| \) to the real axis, then adjust by quadrant.

Modulus–argument form: multiplying and dividing

From the right-angled triangle, \( x = r\cos\theta \) and \( y = r\sin\theta \). So every non-zero complex number can be written in modulus–argument form (polar form):

Cartesian form is best for adding. Polar form is best for multiplying, because the compound-angle formulae give

Moduli multiply and arguments add. So multiplying by \( w \) is a stretch by \( |w| \) and an anticlockwise rotation by \( \arg w \). Multiplying by \( i \) is a quarter turn: \( i(2 + 3i) = -3 + 2i \).

The form must have a plus sign in the middle, the same angle in both places, and \( r \) must be positive. So \( 4\left(\cos\tfrac{\pi}{7}-i\sin\tfrac{\pi}{7}\right) \) must be rewritten as \( 4\left[\cos\left(-\tfrac{\pi}{7}\right) + i\sin\left(-\tfrac{\pi}{7}\right)\right] \).

Repeated multiplication explains a classic economics result. In Samuelson’s multiplier–accelerator model, national income depends on the two previous years: \( Y_t = c(1 + v)Y_{t-1}-cvY_{t-2} + G \), where \( c \) is the marginal propensity to consume and \( v \) the accelerator. Take \( c = 0.75 \) and \( v = 1 \). The deviation \( d_t \) of income from equilibrium satisfies \( d_t = 1.5d_{t-1}-0.75d_{t-2} \), and it is built from powers \( \lambda^t \) where \( \lambda^2-1.5\lambda + 0.75 = 0 \). The discriminant is \( 2.25-3 = -0.75 < 0 \), so

with \( |\lambda| = \sqrt{0.75} = \tfrac{\sqrt{3}}{2} \approx 0.866 \) and \( \arg\lambda = \pm\tfrac{\pi}{6} \).

Each year rotates by \( \tfrac{\pi}{6} \) and shrinks by 0.866. Twelve years make a full turn, so income swings in a 12-year cycle, and each cycle is \( 0.866^{12} = 0.75^6 \approx 0.18 \) times the size of the one before. Complex roots mean cycles; a modulus below 1 means they die out, and above 1 that they grow. If a shock leaves income $10bn above equilibrium in year 0 and $7.5bn above in year 1, the deviation is exactly \( d_t = 10(0.866)^t\cos\tfrac{\pi t}{6} \).

Square roots, loci and regions (9709 P3 3.9)

Loci are on Cambridge 9709 Paper 3 but not IB AA HL; the square-root method suits both. To find the square roots of \( a + bi \), look for \( x + iy \) with \( (x + iy)^2 = a + bi \). Expanding and equating parts gives \( x^2-y^2 = a \) and \( 2xy = b \). A useful third equation comes from moduli: \( x^2 + y^2 = \sqrt{a^2 + b^2} \).

A locus is the set of points satisfying a condition. Read \( |z-a| \) as “distance from \( a \)” and \( \arg(z-a) \) as “direction from \( a \)”, and three shapes appear.

Replace \( = \) by an inequality and each locus becomes a region: \( |z-a| \le r \) is a disc, \( |z-a| \le |z-b| \) is the half-plane on \( a \)’s side of the bisector, and \( \alpha \le \arg(z-a) \le \beta \) is a wedge. Draw included boundaries solid and excluded ones (from a strict \( < \)) dashed.

IB HL only: exponential form, De Moivre and roots of unity

This section is for IB AA HL (AHL 1.13 to 1.15). Cambridge 9709 does not include exponential form, De Moivre’s theorem or roots of unity, so 9709 students can skip to the practice questions.

Arguments add when numbers multiply, just as indices add when powers of the same base multiply. Euler’s formula makes the link exact:

Substituting \( i\theta \) into the series for \( e^x \) (see Taylor and Maclaurin Series) separates into the series for \( \cos\theta \) and \( i\sin\theta \). With \( \theta = \pi \) it gives \( e^{i\pi} = -1 \). In exponential form, \( re^{i\alpha} \times se^{i\beta} = rse^{i(\alpha + \beta)} \) and \( (re^{i\theta})^* = re^{-i\theta} \). Applying the product rule \( n \) times gives De Moivre’s theorem, which IB HL also asks you to prove by induction for positive integers \( n \):

or, in exponential form, \( (re^{i\theta})^n = r^ne^{in\theta} \).

To solve \( z^n = w \), write \( w = Re^{i\phi} \). Adding any multiple of \( 2\pi \) to \( \phi \) gives the same \( w \), and dividing by \( n \) spreads those choices into \( n \) different roots:

They lie on a circle of radius \( R^{1/n} \), spaced \( \tfrac{2\pi}{n} \) apart: the corners of a regular \( n \)-sided polygon.

The sum is a geometric series with ratio \( \omega \ne 1 \): \( \dfrac{\omega^n-1}{\omega-1} = 0 \). The cube roots of unity are \( 1 \) and \( -\tfrac{1}{2} \pm \tfrac{\sqrt{3}}{2}i \).

Practise: 6 questions

Try each one on paper first, then check the worked answer.

Check yourself

The short version

• \( i^2 = -1 \). Multiply by expanding; divide by multiplying top and bottom by the conjugate of the bottom, since \( zz^* = x^2 + y^2 \). One complex equation gives two real ones.
• Real polynomials have non-real roots in conjugate pairs \( p \pm qi \), each giving the real factor \( z^2-2pz + (p^2 + q^2) \).
• On the Argand diagram, \( z^* \) is a reflection and \( |z-w| \) is a distance. Sketch before finding a principal argument in \( -\pi < \theta \le \pi \).
• In modulus–argument form, multiply moduli and add arguments: a stretch and a rotation. Complex roots in an economic model mean cycles.
• Square roots: solve \( x^2-y^2 = a \), \( 2xy = b \). Loci (9709): circle, perpendicular bisector, half-line. IB HL: \( (re^{i\theta})^n = r^ne^{in\theta} \), and \( z^n = w \) has \( n \) roots evenly spaced on a circle.
• The full chapter in Mathematics Made Simple has many more worked examples and a full set of practice questions with a complete answer key.