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Complex numbers - TCE Mathematics Specialised (Tasmania)

Syllabus dot point

“Represent complex numbers in Cartesian and polar form, perform arithmetic, and apply De Moivre's theorem.”

TCESpecialist MathematicsUnit 39 min read

Quick answer

A complex number is z=x+iyz = x + iy where i2=−1i^2 = -1. You must move fluently between Cartesian form x+iyx+iy and polar form r(cos⁡θ+isin⁡θ)r(\cos\theta + i\sin\theta), do arithmetic in whichever form is easier, and use De Moivre's theorem to find powers and roots.

What this dot point is asking

Complex numbers extend the real numbers by introducing the imaginary unit ii, defined by i2=−1i^2 = -1. Once we allow ii, every polynomial equation has a solution, which is the content of the Fundamental Theorem of Algebra. This dot point is about being completely comfortable with the two standard representations and knowing when to use each.

Cartesian form

In Cartesian (rectangular) form we write z=x+iyz = x + iy, where x=Re⁡(z)x = \operatorname{Re}(z) is the real part and y=Im⁡(z)y = \operatorname{Im}(z) is the imaginary part. Addition and subtraction are done component by component:

(a+ib)+(c+id)=(a+c)+i(b+d). (a+ib) + (c+id) = (a+c) + i(b+d).

Multiplication uses the distributive law together with i2=−1i^2 = -1:

(a+ib)(c+id)=(ac−bd)+i(ad+bc). (a+ib)(c+id) = (ac - bd) + i(ad + bc).

The conjugate of z=x+iyz = x+iy is zˉ=x−iy\bar z = x - iy. It is the key tool for division, because zzˉ=x2+y2z\bar z = x^2 + y^2 is always a non-negative real number. To divide, multiply top and bottom by the conjugate of the denominator:

a+ibc+id=(a+ib)(c−id)c2+d2. \frac{a+ib}{c+id} = \frac{(a+ib)(c-id)}{c^2 + d^2}.

Modulus and argument

The modulus is ∣z∣=r=x2+y2|z| = r = \sqrt{x^2 + y^2}, the distance from the origin to zz on the Argand diagram. The argument arg⁡(z)=θ\arg(z) = \theta is the angle measured anticlockwise from the positive real axis. The principal argument is taken in (−π,π](-\pi, \pi]. Be careful: θ=arctan⁡(y/x)\theta = \arctan(y/x) only directly when zz is in the first or fourth quadrant; otherwise adjust by ±π\pm\pi using the signs of xx and yy.

Polar and exponential form

In polar form z=r(cos⁡θ+isin⁡θ)z = r(\cos\theta + i\sin\theta), often abbreviated rcis⁡θr\operatorname{cis}\theta, and in exponential form z=reiθz = re^{i\theta} via Euler's formula eiθ=cos⁡θ+isin⁡θe^{i\theta} = \cos\theta + i\sin\theta.

Polar multiplication and division

z1z2=r1r2cis⁡(θ1+θ2),z1z2=r1r2cis⁡(θ1−θ2). z_1 z_2 = r_1 r_2 \operatorname{cis}(\theta_1 + \theta_2), \qquad \frac{z_1}{z_2} = \frac{r_1}{r_2}\operatorname{cis}(\theta_1 - \theta_2).

Multiply the moduli and add the arguments; divide the moduli and subtract the arguments.

De Moivre's theorem

For any integer nn,

(rcis⁡θ)n=rncis⁡(nθ). \big(r\operatorname{cis}\theta\big)^n = r^n \operatorname{cis}(n\theta).

This makes powers trivial in polar form and is the route to finding roots. The nn solutions of zn=wz^n = w where w=Rcis⁡ϕw = R\operatorname{cis}\phi are

zk=R1/ncis⁡ ⁣(ϕ+2πkn),k=0,1,…,n−1. z_k = R^{1/n}\operatorname{cis}\!\left(\frac{\phi + 2\pi k}{n}\right), \qquad k = 0, 1, \dots, n-1.

These nn roots lie equally spaced on a circle of radius R1/nR^{1/n}.

Cube roots of −8-8

Find all solutions of z3=−8z^3 = -8.

Step 1: Write the right-hand side in polar form

Before applying De Moivre's theorem we need the right-hand side expressed as Rcis⁡ϕR\operatorname{cis}\phi, because the root formula takes the modulus and argument of that number directly. Here −8-8 lies on the negative real axis, so it has modulus 88 and argument π\pi:

−8=8cis⁡π. -8 = 8\operatorname{cis}\pi.

Step 2: Apply the nn-th root formula

The cube roots of Rcis⁡ϕR\operatorname{cis}\phi all have modulus R1/3R^{1/3} and arguments that step through equally spaced values. With R=8R = 8 and ϕ=π\phi = \pi, the modulus of each root is 81/3=28^{1/3} = 2 and the three arguments are:

θk=π+2πk3,k=0,1,2. \theta_k = \frac{\pi + 2\pi k}{3}, \qquad k = 0, 1, 2.

Step 3: Evaluate each root

Substituting k=0,1,2k = 0, 1, 2 in turn gives the three distinct roots.

For k=0k=0: θ0=π3\theta_0 = \dfrac{\pi}{3}, giving z0=2cis⁡π3=2(12+i32)=1+i3z_0 = 2\operatorname{cis}\dfrac{\pi}{3} = 2\big(\tfrac12 + i\tfrac{\sqrt3}{2}\big) = 1 + i\sqrt3.

For k=1k=1: θ1=π+2π3=π\theta_1 = \dfrac{\pi + 2\pi}{3} = \pi, giving z1=2cis⁡π=−2z_1 = 2\operatorname{cis}\pi = -2.

For k=2k=2: θ2=π+4π3=5π3\theta_2 = \dfrac{\pi + 4\pi}{3} = \dfrac{5\pi}{3}, giving z2=2cis⁡5π3=1−i3z_2 = 2\operatorname{cis}\dfrac{5\pi}{3} = 1 - i\sqrt3.

Final answer: The three cube roots are −2,  1+i3,  1−i3-2,\; 1 + i\sqrt3,\; 1 - i\sqrt3. As a check, the two complex roots are conjugates (expected since the coefficients are real), and they are spaced 120∘120^\circ apart on the circle of radius 22.

Argument in the wrong quadrant

A very common error is to compute arg⁡(z)\arg(z) as arctan⁡(y/x)\arctan(y/x) without checking the quadrant. For z=−1−iz = -1 - i, a calculator gives arctan⁡ ⁣(−1−1)=arctan⁡(1)=π4\arctan\!\big(\tfrac{-1}{-1}\big) = \arctan(1) = \tfrac{\pi}{4}, but zz is in the third quadrant, so the principal argument is π4−π=−3π4\tfrac{\pi}{4} - \pi = -\tfrac{3\pi}{4}. Always sketch zz on the Argand diagram first and read off the correct quadrant before quoting an argument.

Converting and multiplying in polar form

Express z=−1+i3z = -1 + i\sqrt{3} in polar form, then use it to evaluate z(1+i)z(1 + i).

Step 1: Find the modulus and argument of zz

To convert to polar form we need both r=∣z∣r = |z| and θ=arg⁡(z)\theta = \arg(z). The modulus comes from Pythagoras and the argument requires us to identify the correct quadrant before applying arctangent.

The modulus is ∣z∣=(−1)2+(3)2=4=2|z| = \sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{4} = 2. The point (−1,3)(-1, \sqrt{3}) is in the second quadrant. The reference angle is arctan⁡31=π3\arctan\dfrac{\sqrt{3}}{1} = \dfrac{\pi}{3}, so the principal argument is π−π3=2π3\pi - \dfrac{\pi}{3} = \dfrac{2\pi}{3}. Hence

z=2cis⁡2π3. z = 2\operatorname{cis}\dfrac{2\pi}{3}.

Step 2: Write 1+i1 + i in polar form

We also convert the second factor, since multiplying in polar form is much simpler than expanding in Cartesian form: we just multiply the moduli and add the arguments.

1+i=2cis⁡π4. 1 + i = \sqrt{2}\operatorname{cis}\dfrac{\pi}{4}.

Step 3: Multiply using the polar multiplication rule

Multiplying in polar form multiplies moduli and adds arguments, so there is no bracket expansion needed at all.

z(1+i)=22cis⁡ ⁣(2π3+π4)=22cis⁡11π12. z(1 + i) = 2\sqrt{2}\operatorname{cis}\!\left(\dfrac{2\pi}{3} + \dfrac{\pi}{4}\right) = 2\sqrt{2}\operatorname{cis}\dfrac{11\pi}{12}.

Final answer: z(1+i)=22cis⁡11π12z(1+i) = 2\sqrt{2}\operatorname{cis}\dfrac{11\pi}{12}. Converting back to Cartesian form (if required) uses cos⁡11π12\cos\dfrac{11\pi}{12} and sin⁡11π12\sin\dfrac{11\pi}{12}, but the polar answer is exact and compact, which is exactly why polar form is preferred for products.

Choosing the right form

Use Cartesian form for addition and subtraction, and for stating exact real and imaginary parts. Use polar or exponential form for multiplication, division, powers and roots. The exponential form z=reiθz = re^{i\theta} is the same information as polar form, and it makes the multiplication and division rules obvious because exponents add and subtract: r1eiθ1⋅r2eiθ2=r1r2ei(θ1+θ2)r_1 e^{i\theta_1}\cdot r_2 e^{i\theta_2} = r_1 r_2 e^{i(\theta_1 + \theta_2)}. Switching at the right moment is what separates fast, clean solutions from long algebraic struggles, so practise converting both ways until it is automatic.

The fundamental theorem in practice

Allowing ii guarantees solutions to equations that have none over the reals. For instance z2+2z+5=0z^2 + 2z + 5 = 0 has discriminant 4−20=−16<04 - 20 = -16 < 0, so over the reals it has no solution. Using −16=4i\sqrt{-16} = 4i, the quadratic formula gives z=−2±4i2=−1±2iz = \dfrac{-2 \pm 4i}{2} = -1 \pm 2i, a conjugate pair. This is the bridge to factorising polynomials over the complex field, where every quadratic factor with a negative discriminant splits into two complex linear factors.

Exam-style questions

Questions in the style of TASC exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice7 marks
(a) Express w=1+i3+iw = \dfrac{1 + i}{\sqrt{3} + i} in both Cartesian and polar form. (b) Hence, or otherwise, deduce an exact value for tan⁡π12\tan\dfrac{\pi}{12}.
Show worked answer →

(a) Cartesian: multiply top and bottom by the conjugate 3−i\sqrt{3} - i. The numerator is (1+i)(3−i)=3−i+3 i−i2=(3+1)+(3−1)i(1 + i)(\sqrt{3} - i) = \sqrt{3} - i + \sqrt{3}\,i - i^2 = (\sqrt{3} + 1) + (\sqrt{3} - 1)i, and the denominator is (3+i)(3−i)=3+1=4(\sqrt{3} + i)(\sqrt{3} - i) = 3 + 1 = 4. So w=(3+1)+(3−1)i4w = \dfrac{(\sqrt{3} + 1) + (\sqrt{3} - 1)i}{4}. Polar: ∣w∣=∣1+i∣∣3+i∣=22|w| = \dfrac{|1 + i|}{|\sqrt{3} + i|} = \dfrac{\sqrt{2}}{2}, and arg⁡(w)=arg⁡(1+i)−arg⁡(3+i)=π4−π6=π12\arg(w) = \arg(1 + i) - \arg(\sqrt{3} + i) = \dfrac{\pi}{4} - \dfrac{\pi}{6} = \dfrac{\pi}{12}. So w=22cis⁡π12w = \dfrac{\sqrt{2}}{2}\operatorname{cis}\dfrac{\pi}{12}. (4 marks)

(b) Comparing the two forms, tan⁡π12=Im⁡(w)Re⁡(w)=3−13+1\tan\dfrac{\pi}{12} = \dfrac{\operatorname{Im}(w)}{\operatorname{Re}(w)} = \dfrac{\sqrt{3} - 1}{\sqrt{3} + 1}. Rationalise by multiplying top and bottom by 3−1\sqrt{3} - 1: (3−1)2(3+1)(3−1)=3−23+13−1=4−232=2−3\dfrac{(\sqrt{3} - 1)^2}{(\sqrt{3} + 1)(\sqrt{3} - 1)} = \dfrac{3 - 2\sqrt{3} + 1}{3 - 1} = \dfrac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3}. So tan⁡π12=2−3\tan\dfrac{\pi}{12} = 2 - \sqrt{3}. (3 marks)

Practice4 marks
If arg⁡(z−5)=2π3\arg(z - 5) = \dfrac{2\pi}{3}, find the minimum possible value of ∣z∣|z|.
Show worked answer →

The condition arg⁡(z−5)=2π3\arg(z - 5) = \dfrac{2\pi}{3} describes a ray starting at the point 55, that is (5,0)(5, 0), pointing in the direction making angle 2π3\dfrac{2\pi}{3} with the positive real axis.

Here ∣z∣|z| is the distance from the origin to a point on this ray, and the minimum distance is the perpendicular distance from the origin to the line carrying the ray.

The segment from (5,0)(5, 0) to the origin points in the direction of angle π\pi and has length 55. The angle between this segment and the ray is π−2π3=π3\pi - \dfrac{2\pi}{3} = \dfrac{\pi}{3}. The perpendicular (shortest) distance is therefore 5sin⁡π3=5325\sin\dfrac{\pi}{3} = \dfrac{5\sqrt{3}}{2}.

Checking that the foot of the perpendicular lies on the ray (not its backward extension): the projection of the vector from the start point to the origin onto the ray direction is positive, so the foot is on the ray. Hence the minimum value of ∣z∣|z| is 532\dfrac{5\sqrt{3}}{2}.

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