This note covers complex numbers written as \(a+bi\), as \(r(\cos\theta+i\sin\theta)\) and as \(re^{i\theta}\), and shows how to add, multiply, divide and raise them to powers. Complex numbers extend the real numbers so that every quadratic equation has a solution, and they describe rotation and oscillation neatly.
Key ideas
The imaginary unit. Define \(i\) by \(i^{2}=-1\). A complex number is \(z=a+bi\) where \(a\) is the real part, \(\operatorname{Re}(z)\), and \(b\) is the imaginary part, \(\operatorname{Im}(z)\). Both \(a\) and \(b\) are real numbers. Two complex numbers are equal only when their real parts are equal and their imaginary parts are equal.
Arithmetic in Cartesian form.
- Add or subtract by combining real parts and imaginary parts separately.
- Multiply by expanding the brackets and replacing \(i^{2}\) with \(-1\).
- The conjugate of \(z=a+bi\) is \(z^{*}=a-bi\). Then \(zz^{*}=a^{2}+b^{2}\), which is real.
- To divide, multiply the top and bottom by the conjugate of the denominator. This makes the denominator real.
The Argand diagram. Plot \(z=a+bi\) as the point \((a,b)\), with the real axis horizontal and the imaginary axis vertical. The modulus \(|z|=r=\sqrt{a^{2}+b^{2}}\) is the distance from the origin. The argument \(\arg z=\theta\) is the angle from the positive real axis, measured anticlockwise. In these notes \(-\pi\lt\theta\le\pi\).
Find \(\theta\) from \(\tan\theta=\dfrac{b}{a}\), but always sketch the point first. A calculator gives an angle in the wrong quadrant when \(a\lt0\). In that case add or subtract \(\pi\) as needed.
Polar and exponential form. Since \(a=r\cos\theta\) and \(b=r\sin\theta\),
\[ z=r(\cos\theta+i\sin\theta)=r\operatorname{cis}\theta=re^{i\theta} \]The last version comes from Euler's formula \(e^{i\theta}=\cos\theta+i\sin\theta\). Always use radians for \(\theta\) in the exponential form. For example \(e^{i\pi}=-1\).
Multiplication, division and powers in polar form. If \(z_1=r_1e^{i\theta_1}\) and \(z_2=r_2e^{i\theta_2}\), then
\[ z_1z_2=r_1r_2e^{i(\theta_1+\theta_2)},\qquad \frac{z_1}{z_2}=\frac{r_1}{r_2}e^{i(\theta_1-\theta_2)},\qquad z^{n}=r^{n}e^{in\theta} \]So multiplying multiplies the moduli and adds the arguments, which rotates and scales the point. The rule for powers is de Moivre's theorem. If an argument ends up outside \(-\pi\lt\theta\le\pi\), add or subtract multiples of \(2\pi\) to bring it back.
Quadratics with complex roots. If \(b^{2}-4ac\lt0\), the quadratic formula gives a pair of roots that are conjugates of each other. For \(x^{2}+2x+5=0\), the roots are \(x=\dfrac{-2\pm\sqrt{-16}}{2}=-1\pm2i\).
Your GDC can work in complex mode to check results, and can convert between forms. Use it to check your working but show the steps.
Meaning on the Argand diagram. Adding a complex number moves a point by a fixed amount, just like adding vectors. Multiplying by a complex number of modulus 1 rotates the point about the origin without changing its distance. For example, multiplying by \(i=e^{i\pi/2}\) turns a point a quarter-turn anticlockwise. Multiplying by a real number \(k\gt0\) stretches the point away from the origin by a factor \(k\). A general product does both at once, a rotation by the sum of arguments and a scaling by the product of moduli.
Useful properties of the conjugate. The conjugate reflects the point in the real axis. It follows that \(z+z^{*}=2a\) is real, \(z-z^{*}=2bi\) is purely imaginary and \(|z^{*}|=|z|\). Also \(zz^{*}=|z|^{2}\), which is the reason the conjugate removes \(i\) from a denominator.
Powers of \(i\). The powers of \(i\) repeat every four steps: \(i^{1}=i\), \(i^{2}=-1\), \(i^{3}=-i\), \(i^{4}=1\). To find \(i^{n}\), divide \(n\) by 4 and use the remainder. For instance \(i^{27}\) has remainder 3 when 27 is divided by 4, so \(i^{27}=-i\).
Exact values to know. Polar form is easiest to convert when the argument is a standard angle. Learn \(\cos\) and \(\sin\) of \(0\), \(\dfrac{\pi}{6}\), \(\dfrac{\pi}{4}\), \(\dfrac{\pi}{3}\) and \(\dfrac{\pi}{2}\), because they let you convert back to \(a+bi\) exactly. When an exact answer is not required, round to 3 s.f.
Choosing a form. Use Cartesian form to add and subtract. Use polar or exponential form to multiply, divide or take powers. Convert back to Cartesian form at the end if the question asks for \(a+bi\).
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(i^{2}=-1\) | Defines the imaginary unit. |
| \(z=a+bi\) | Cartesian form, with real part \(a\) and imaginary part \(b\). |
| \(z^{*}=a-bi\) | Complex conjugate. Used for division. |
| \(|z|=\sqrt{a^{2}+b^{2}}\) | Modulus, the distance from the origin on the Argand diagram. |
| \(\theta=\arg z\), \(\tan\theta=\dfrac{b}{a}\) | Argument, with the quadrant checked by a sketch. |
| \(z=r(\cos\theta+i\sin\theta)=re^{i\theta}\) | Polar and exponential forms. |
| \(z_1z_2=r_1r_2e^{i(\theta_1+\theta_2)}\) | Product in polar form. |
| \(z^{n}=r^{n}e^{in\theta}\) | Power of a complex number (de Moivre). |
Worked examples
Example 1 Given \(z=3+4i\) and \(w=2-i\), find \(zw\), and find \(\dfrac{5+2i}{1-3i}\) in the form \(a+bi\).
Product: \(zw=(3+4i)(2-i)=6-3i+8i-4i^{2}\).
Replace \(i^{2}\) with \(-1\): \(6+5i+4=10+5i\).
Quotient: multiply top and bottom by the conjugate of \(1-3i\), which is \(1+3i\).
Numerator: \((5+2i)(1+3i)=5+15i+2i+6i^{2}=-1+17i\).
Denominator: \((1-3i)(1+3i)=1+9=10\).
So the result is \(\dfrac{-1+17i}{10}=-0.1+1.7i\).
Answer: \(zw=10+5i\); \(\dfrac{5+2i}{1-3i}=-0.1+1.7i\).
Example 2 Write \(z=-3+3i\) in polar form and in exponential form.
Modulus: \(r=\sqrt{(-3)^{2}+3^{2}}=\sqrt{18}=3\sqrt2\approx4.24\).
The point \((-3,3)\) is in the second quadrant. The reference angle is \(\tan^{-1}\left(\dfrac{3}{3}\right)=\dfrac{\pi}{4}\).
In the second quadrant, \(\theta=\pi-\dfrac{\pi}{4}=\dfrac{3\pi}{4}\approx2.36\).
Polar form: \(3\sqrt2\left(\cos\dfrac{3\pi}{4}+i\sin\dfrac{3\pi}{4}\right)\).
Answer: \(z=3\sqrt2\,e^{3\pi i/4}\) (modulus \(4.24\), argument \(2.36\) radians to 3 s.f.).
Example 3 Let \(z_1=4e^{i\pi/3}\) and \(z_2=2e^{i\pi/6}\). Find \(z_1z_2\), \(\dfrac{z_1}{z_2}\) and \(z_1^{3}\).
Product: moduli \(4\times2=8\), arguments \(\dfrac{\pi}{3}+\dfrac{\pi}{6}=\dfrac{\pi}{2}\). So \(z_1z_2=8e^{i\pi/2}=8i\).
Quotient: moduli \(4\div2=2\), arguments \(\dfrac{\pi}{3}-\dfrac{\pi}{6}=\dfrac{\pi}{6}\). So \(\dfrac{z_1}{z_2}=2e^{i\pi/6}\), which is \(\sqrt3+i\) in Cartesian form.
Power: \(z_1^{3}=4^{3}e^{i\pi}=64e^{i\pi}=-64\).
Answer: \(z_1z_2=8i\); \(\dfrac{z_1}{z_2}=2e^{i\pi/6}\); \(z_1^{3}=-64\).
Common errors
- Common error: forgetting that \(i^{2}=-1\) when expanding, so that a term \(-4i^{2}\) is left as \(-4\) instead of \(+4\).
- Common error: using the calculator value of \(\tan^{-1}(b/a)\) without checking the quadrant, which gives the wrong argument when \(a\lt0\).
- Common error: mixing degrees and radians in the exponential form \(re^{i\theta}\).
- Common error: adding the moduli when multiplying. Moduli are multiplied and arguments are added.
- Common error: including \(i\) in the imaginary part. For \(3+4i\), \(\operatorname{Im}(z)=4\), not \(4i\).
Exam tip: sketch the point on an Argand diagram before finding an argument. A two-second sketch prevents a wrong-quadrant answer.
Check yourself
- Find \(|5-12i|\) and the conjugate of \(5-12i\).
Answer
\(|5-12i|=\sqrt{25+144}=13\). The conjugate is \(5+12i\). - Calculate \((1+i)^{8}\).
Answer
\(1+i=\sqrt2e^{i\pi/4}\). So \((1+i)^{8}=(\sqrt2)^{8}e^{2\pi i}=16\times1=16\). - Solve \(x^{2}+2x+5=0\).
Answer
\(x=\dfrac{-2\pm\sqrt{4-20}}{2}=\dfrac{-2\pm4i}{2}=-1\pm2i\). - Write \(2e^{i\pi/2}\) in the form \(a+bi\).
Answer
\(2(\cos\frac{\pi}{2}+i\sin\frac{\pi}{2})=2(0+i)=2i\).