Complex number arithmeticAQA A-Level Further Maths: Revision notes

Section 1

Real and imaginary parts

The imaginary unit $i$ satisfies $i^2=-1$ . A complex number is $z=x+iy$ with $x$ and $y$ real. Then $\mathrm{Re}(z)=x$ is the real part and $\mathrm{Im}(z)=y$ is the imaginary part. The imaginary part is the real number multiplying $i$ : for $z=3-2i$ , $\mathrm{Re}(z)=3$ and $\mathrm{Im}(z)=-2$ (not $-2i$ ). Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal.

Key termsimaginary unitreal partimaginary part

Common mistake

Giving the imaginary part as $-2i$ . It is $-2$ .

Section 2

Adding and subtracting

Add or subtract the real parts and the imaginary parts separately: $(a+bi)\pm(c+di)=(a\pm c)+(b\pm d)i.$ Example: $(3-2i)+(1+4i)=4+2i$ and $(3-2i)-(1+4i)=2-6i$ .

Common mistake

Not changing the sign of both terms when subtracting a bracket: $-(1+4i)=-1-4i$ .

Section 3

Multiplying

Expand like brackets and replace $i^2$ by $-1$ : $(a+bi)(c+di)=(ac-bd)+(ad+bc)i.$ Example: $(3-2i)(1+4i)=3+12i-2i-8i^2=3+8+10i=11+10i$ .

Common mistake

Leaving $-8i^2$ as $-8$ . Since $i^2=-1$ , $-8i^2=+8$ .

Section 4

Dividing using the conjugate

The complex conjugate of $x+iy$ is $x-iy$ . The product $(x+iy)(x-iy)=x^2+y^2$ is real. To divide, multiply numerator and denominator by the conjugate of the denominator: $\frac{3-2i}{1+4i}=\frac{(3-2i)(1-4i)}{(1+4i)(1-4i)}=\frac{-5-14i}{17}=-\frac{5}{17}-\frac{14}{17}i.$ Give the answer in the form $a+bi$ , with real and imaginary parts separated.

Key termscomplex conjugate

Exam tip

Check the denominator is a real number before you finish. If $i$ remains there, you used the wrong conjugate.

Section 5

Equating real and imaginary parts

If $(2+i)(x+iy)=7+i$ with $x,y$ real, expand and compare: $(2x-y)+(x+2y)i=7+i\Rightarrow 2x-y=7,\; x+2y=1,$ so $x=3$ and $y=-1$ , giving $z=3-i$ . This agrees with dividing: $z=\frac{7+i}{2+i}=3-i$ . Use this method whenever an equation contains an unknown complex number and both sides can be written as $a+bi$ .

Exam tip

Always state that you are equating real parts and imaginary parts, and write the two equations clearly.

Section 6

Solving quadratics with real coefficients

If the discriminant $b^2-4ac<0$ , the quadratic $az^2+bz+c=0$ has no real roots, but it has two complex roots found with the formula and $\sqrt{-n}=i\sqrt{n}$ : $z=\frac{-b\pm i\sqrt{4ac-b^2}}{2a}.$ Example: $z^2+2z+5=0$ has discriminant $4-20=-16$ , so $z=\frac{-2\pm4i}{2}=-1\pm2i$ . Complex roots of a real quadratic come as a pair $p\pm qi$ .

Key termsdiscriminant

Common mistake

Forgetting to divide the whole of $\pm4i$ by $2$ : the roots are $-1\pm2i$ , not $-1\pm4i$ .

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Complex number arithmetic

The complex numbers $z$ and $w$ are given by $z=3-2i$ and $w=1+4i$ .
Find $\frac{z}{w}$ in the form $a+bi$ , where $a$ and $b$ are real.2 marks
The quadratic equation $z^2-6z+13=0$ has roots $\alpha$ and $\beta$ , where $\alpha$ has positive imaginary part.
Find the value of $\alpha^2+\beta^2$ .2 marks
The complex number $z=x+iy$ , where $x$ and $y$ are real, satisfies $(2+i)z=7+i$ .
Find $z$ in the form $x+iy$ .3 marks

See the full worksheet

Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).