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How do complex numbers extend the real number system and let us solve every polynomial?

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

Cartesian and polar forms of complex numbers, modulus-argument arithmetic, De Moivre's theorem and roots, for TCE Mathematics Specialised Unit 3.

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

The conjugate of z=x+iyz = x+iy is zˉ=xiy\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)(cid)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(cosθ+isinθ)z = r(\cos\theta + i\sin\theta), often abbreviated rcisθr\operatorname{cis}\theta, and in exponential form z=reiθz = re^{i\theta} via Euler's formula eiθ=cosθ+isinθe^{i\theta} = \cos\theta + i\sin\theta.

De Moivre's theorem

For any integer nn,

(rcisθ)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=Rcisϕw = R\operatorname{cis}\phi are

zk=R1/ncis ⁣(ϕ+2πkn),k=0,1,,n1. 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}.

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: r1eiθ1r2eiθ2=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 420=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 practice questions

Practice questions written in the style of TASC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

TCE 20217 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 tanπ12\tan\dfrac{\pi}{12}.
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(a) Cartesian: multiply top and bottom by the conjugate 3i\sqrt{3} - i. The numerator is (1+i)(3i)=3i+3ii2=(3+1)+(31)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)(3i)=3+1=4(\sqrt{3} + i)(\sqrt{3} - i) = 3 + 1 = 4. So w=(3+1)+(31)i4w = \dfrac{(\sqrt{3} + 1) + (\sqrt{3} - 1)i}{4}. Polar: w=1+i3+i=22|w| = \dfrac{|1 + i|}{|\sqrt{3} + i|} = \dfrac{\sqrt{2}}{2}, and arg(w)=arg(1+i)arg(3+i)=π4π6=π12\arg(w) = \arg(1 + i) - \arg(\sqrt{3} + i) = \dfrac{\pi}{4} - \dfrac{\pi}{6} = \dfrac{\pi}{12}. So w=22cisπ12w = \dfrac{\sqrt{2}}{2}\operatorname{cis}\dfrac{\pi}{12}. (4 marks)

(b) Comparing the two forms, tanπ12=Im(w)Re(w)=313+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 31\sqrt{3} - 1: (31)2(3+1)(31)=323+131=4232=23\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 tanπ12=23\tan\dfrac{\pi}{12} = 2 - \sqrt{3}. (3 marks)

TCE 20244 marksIf arg(z5)=2π3\arg(z - 5) = \dfrac{2\pi}{3}, find the minimum possible value of z|z|.
Show worked answer →

The condition arg(z5)=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 π2π3=π3\pi - \dfrac{2\pi}{3} = \dfrac{\pi}{3}. The perpendicular (shortest) distance is therefore 5sinπ3=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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