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FP1.1 - Complex numbers

Syllabus
2019
Topic
—
Level
AS

Learning objectives

—FP1.1.1 - Definition of complex numbersDefinition of complex numbers in The meaning of conjugate, modulus, argument, real part, the form a + ib and imaginary part and equality of complex numbers should be known. rcos θ + irsin θ.—FP1.1.2 - Sum, product and quotient of complex numbersSum, product and quotient of | z z | = | z | | z | 1 2 1 2 complex numbers.; Knowledge of the result arg(z z) = arg z + arg z is not 1 2 1 2 required.—FP1.1.3 - Argand diagrams and complex operationsGeometrical representation of complex numbers in the Argand diagram.; Geometrical representation of sums, products and quotients of complex numbers.—FP1.1.4 - Complex solutions of quadratic equationsComplex solutions of quadratic equations with real coefficients.—FP1.1.5 - Finding conjugate complex rootsFinding conjugate complex roots Knowledge that if z is a root of f(z) = 0 then z * is 1 1 and a real root of a cubic equation also a root. with integer coefficients.—FP1.1.6 - Finding conjugate complex rootsFinding conjugate complex roots For example, and/or real roots of a quartic (i) f(x) = x4 − x3 − 5x2 + 7x + 10 equation with real coefficients.; Given that x = 2 + i is a root of f(x) = 0, use algebra to find the three other roots of f(x) = 0 (ii) g(x) = x4 − x3 + 6x2 + 14x − 20 Given g(1) = 0 and g(−2) = 0, use algebra to solve g(x) = 0 completely.

Describe a complex number in two forms

A complex number combines a real part and an imaginary part: z=a+ibz=a+ib, where a,b∈Ra,b\in\mathbb R and i2=−1i^2=-1. Its polar form records the same number by its distance rr from the origin and its directed angle θ\theta from the positive real axis.

z=a+ib=r(cos⁡θ+isin⁡θ),r=∣z∣=a2+b2z=a+ib=r(\cos\theta+i\sin\theta),\qquad r=|z|=\sqrt{a^2+b^2}

Feature Meaning
Re⁡z=a\operatorname{Re}z=a real part
Im⁡z=b\operatorname{Im}z=b imaginary part, without the factor ii
z∗=a−ibz^*=a-ib conjugate: reflection in the real axis
arg⁡z=θ\arg z=\theta principal argument, −π<θ≤π-\pi<\theta\le\pi for z≠0z\ne0
equality a+ib=c+ida+ib=c+id exactly when a=ca=c and b=db=d

For z=−3+33,iz=-3+3\sqrt3,i, ∣z∣=9+27=6,arg⁡z=2π3,|z|=\sqrt{9+27}=6,\qquad \arg z=\frac{2\pi}{3}, because the point lies in quadrant II. Hence z=6(cos⁡2π3+isin⁡2π3),z∗=−3−33,i.z=6\left(\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}\right),\qquad z^*=-3-3\sqrt3,i.

Use the signs of both components to choose the correct quadrant; a calculator value of an−1(b/a)an^{-1}(b/a) alone may give the wrong argument. The zero complex number has modulus 00 but no defined argument.

Calculate with complex numbers

Add and subtract complex numbers by matching real and imaginary parts. For products, expand and replace i2i^2 by −1-1. For a quotient, multiply numerator and denominator by the conjugate of the denominator so the denominator becomes real.

Operation Example
sum (2+i)+(−3+4i)=−1+5i(2+i)+(-3+4i)=-1+5i
product (2+i)(1−3i)=5−5i(2+i)(1-3i)=5-5i
quotient 2+i1−2i=(2+i)(1+2i)(1−2i)(1+2i)=i\displaystyle\frac{2+i}{1-2i}=\frac{(2+i)(1+2i)}{(1-2i)(1+2i)}=i

∣z1z2∣=∣z1∣ ∣z2∣|z_1z_2|=|z_1|\,|z_2|

The conjugate identity (c+id)(c−id)=c2+d2(c+id)(c-id)=c^2+d^2 explains why rationalising a quotient works. The modulus-product rule can find an unknown modulus without first expanding the product; for example, if ∣z1z2∣=15|z_1z_2|=15 and ∣z1∣=3|z_1|=3, then ∣z2∣=5|z_2|=5.

Keep the final answer in the requested form, usually a+iba+ib, and simplify both parts. This unit does not require memorising rg(z_1z_2)=rg z_1+rg z_2; do not use that unrequired result as a substitute for the specified algebra or geometry.

Read complex operations on an Argand diagram

An Argand diagram plots z=x+iyz=x+iy as the point (x,y)(x,y): the horizontal axis is Re⁡z\operatorname{Re}z and the vertical axis is Im⁡z\operatorname{Im}z. The vector from the origin has length ∣z∣|z| and direction arg⁡z\arg z.

Operation Geometrical effect
add ww translate by the vector representing ww
multiply by r(cos⁡θ+isin⁡θ)r(\cos\theta+i\sin\theta) enlarge by factor rr and rotate through θ\theta about the origin
divide by r(cos⁡θ+isin⁡θ)r(\cos\theta+i\sin\theta) scale by 1/r1/r and reverse that rotation

Multiplication by ii makes the geometry visible algebraically: i(x+iy)=−y+ix.i(x+iy)=-y+ix. Thus (x,y)(x,y) maps to (−y,x)(-y,x), a 90∘90^\circ anticlockwise rotation. Multiplication by 2i2i adds enlargement by factor 22; division by ii rotates 90∘90^\circ clockwise.

Complex roots with real coefficients often appear as conjugate points reflected in the real axis. Distances, perimeters and areas on an Argand diagram are ordinary coordinate geometry once each complex number has been placed at its real and imaginary coordinates.

Label the real and imaginary axes in the correct order and preserve relative scale. A conjugate changes only the sign of the imaginary coordinate; it is not reflection in the imaginary axis.

Solve a quadratic with complex roots

A quadratic with real coefficients can have complex solutions when its discriminant is negative. Use the ordinary quadratic formula and rewrite the square root of a negative number using −k=ik\sqrt{-k}=i\sqrt{k} for k>0k>0.

az2+bz+c=0⟹z=−b±b2−4ac2aaz^2+bz+c=0\quad\Longrightarrow\quad z=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

For z2−6z+13=0z^2-6z+13=0, the discriminant is (−6)2−4(1)(13)=−16.(-6)^2-4(1)(13)=-16. Therefore z=6±−162=6±4i2=3±2i.z=\frac{6\pm\sqrt{-16}}{2}=\frac{6\pm4i}{2}=3\pm2i. Substitution or the sum and product of the two roots checks the result.

Do not stop at −16\sqrt{-16} or discard it as having no solution: it has no real square root, but the complex square roots are ±4i\pm4i. With real coefficients, the non-real solutions produced by the quadratic formula form a conjugate pair.

Complete a cubic from one complex root

For a polynomial with real—in particular integer—coefficients, every non-real root occurs with its complex conjugate. In a cubic, a known non-real root therefore supplies a quadratic factor, leaving one real linear factor.

a+ib and a−ib⟹[x−(a+ib)][x−(a−ib)]=(x−a)2+b2a+ib\text{ and }a-ib\quad\Longrightarrow\quad[x-(a+ib)][x-(a-ib)]=(x-a)^2+b^2

Suppose f(x)=x3−5x2+11x−15f(x)=x^3-5x^2+11x-15 and 1+2i1+2i is a root. Then 1−2i1-2i is also a root, giving [x−(1+2i)][x−(1−2i)]=x2−2x+5.[x-(1+2i)][x-(1-2i)]=x^2-2x+5. Division gives f(x)=(x2−2x+5)(x−3),f(x)=(x^2-2x+5)(x-3), so the third root is 33.

Form the real quadratic factor from the conjugate pair; divide the cubic by it; solve the remaining linear factor; then state all three roots. Multiplying the factors back together checks that no coefficient or sign has drifted.

The conjugate-root conclusion depends on the polynomial coefficients being real. Do not conjugate a real root into a new root: its conjugate is itself. A factor is x−rx-r, so a root rr introduces the opposite sign inside its factor.

Recover every root of a quartic

A quartic with real coefficients has four roots counting multiplicity. Use every supplied root to build factors: a non-real root supplies its conjugate, while each stated real root supplies one linear factor. Divide or compare coefficients to find the remaining factor.

Given evidence Factor secured
root a+iba+ib, b≠0b\ne0 (x−a)2+b2(x-a)^2+b^2
real root rr x−rx-r
two real roots r,sr,s (x−r)(x−s)(x-r)(x-s)

Let f(x)=x4−7x3+13x2+x−20,f(x)=x^4-7x^3+13x^2+x-20, with roots 2+i2+i and −1-1. Real coefficients also give 2−i2-i, so [x−(2+i)][x−(2−i)]=x2−4x+5.[x-(2+i)][x-(2-i)]=x^2-4x+5. The root −1-1 gives x+1x+1. Dividing by (x2−4x+5)(x+1)(x^2-4x+5)(x+1) leaves x−4x-4, so the four roots are 2+i, 2−i, −1, 4.2+i,\ 2-i,\ -1,\ 4.

State all four roots and count multiplicity. Do not assume every remaining root is real unless the factorisation proves it, and do not use decimal approximations when exact algebraic roots are available. Multiply the completed factors to verify the original quartic.