This note introduces complex numbers written as \(a+bi\), where \(i\) is a number whose square is \(-1\). It covers addition, multiplication, division, conjugates, the Argand diagram and solving equations whose answers are not real. Complex numbers let you solve every quadratic equation, including those with a negative discriminant.
Key ideas
The real numbers cannot solve \(x^2=-1\). We define a new number \(i\) with \(i^2=-1\). A complex number has the form \(z=a+bi\), where \(a\) and \(b\) are real. The number \(a\) is the real part, written \(\mathrm{Re}(z)\). The number \(b\) is the imaginary part, written \(\mathrm{Im}(z)\). Note that \(\mathrm{Im}(z)\) is \(b\), not \(bi\).
Two complex numbers are equal only if their real parts are equal and their imaginary parts are equal. This is very useful: one complex equation gives two real equations.
Arithmetic. Add and subtract the real and imaginary parts separately. To multiply, expand the brackets like ordinary algebra and replace \(i^2\) by \(-1\). The powers of \(i\) repeat in a cycle of four: \(i,\,-1,\,-i,\,1\).
The conjugate. The complex conjugate of \(z=a+bi\) is \(z^*=a-bi\). Multiplying a number by its conjugate always gives a real result: \(zz^*=a^2+b^2\).
Division. To divide, multiply the top and bottom by the conjugate of the denominator. The denominator becomes real, just like rationalising a surd.
The Argand diagram. Plot \(a+bi\) as the point \((a,b)\). The horizontal axis is the real axis and the vertical axis is the imaginary axis. Adding two complex numbers is the same as adding the position vectors of the points. The conjugate is the reflection of the point in the real axis.
Modulus. The distance from the origin to the point is \(|z|=\sqrt{a^2+b^2}\). So \(zz^*=|z|^2\).
Quadratic equations. If \(b^2-4ac\lt0\), the quadratic \(az^2+bz+c=0\) has two complex solutions. Use the quadratic formula and write \(\sqrt{-k}=i\sqrt{k}\) for \(k\gt0\). When the coefficients are real, the two solutions are conjugates of each other.
Useful properties
- The conjugate of a sum is the sum of the conjugates, and the conjugate of a product is the product of the conjugates: \((z+w)^*=z^*+w^*\) and \((zw)^*=z^*w^*\).
- \(z+z^*=2a\) is real, and \(z-z^*=2bi\) is purely imaginary. So the real part is \(\dfrac{z+z^*}{2}\).
- A number is real exactly when \(z=z^*\), and purely imaginary when \(z=-z^*\) and \(z\ne0\).
- The modulus of a product is the product of the moduli: \(|zw|=|z||w|\). For example, \(|(3+4i)(1+i)|=5\times\sqrt2\).
Reading the Argand diagram
On the Argand diagram, \(3+2i\) sits 3 units to the right of the origin and 2 units up. Its conjugate \(3-2i\) sits 2 units below the axis, a mirror image. The number \(-3-2i\) is on the opposite side of the origin, because negating both parts rotates the point half a turn about the origin. The modulus is the length of the line from the origin to the point, so all numbers with the same modulus lie on a circle centred on the origin.
Subtracting two complex numbers gives the vector from one point to the other, and the modulus of that difference is the distance between the points. The sum of two complex numbers is the fourth corner of the parallelogram built from the two points and the origin. Sketching a quick Argand diagram is a good way to check that an answer is sensible, for example that a conjugate pair is symmetrical about the real axis.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(i^2=-1\) | The defining property of \(i\). Use it whenever \(i^2\) appears. |
| \(z=a+bi\) | Cartesian form, with \(a=\mathrm{Re}(z)\) and \(b=\mathrm{Im}(z)\). |
| \(z^*=a-bi\) | Complex conjugate. Reflects the point in the real axis. |
| \(zz^*=a^2+b^2\) | Product with the conjugate is real. Used for division. |
| \(|z|=\sqrt{a^2+b^2}\) | Modulus, the distance from the origin on the Argand diagram. |
| \(\dfrac{z_1}{z_2}=\dfrac{z_1z_2^*}{z_2z_2^*}\) | Division of complex numbers. |
| \(a+bi=c+di\iff a=c\text{ and }b=d\) | Equating real and imaginary parts. |
Worked examples
Example 1 Given \(z=3+2i\) and \(w=1-4i\), find (a) \(zw\) and (b) \(\dfrac{5+i}{2-3i}\).
(a) Expand: \((3+2i)(1-4i)=3-12i+2i-8i^2\).
Replace \(i^2\) by \(-1\): \(3-10i+8=11-10i\).
(b) Multiply top and bottom by the conjugate of the denominator, \(2+3i\).
Numerator: \((5+i)(2+3i)=10+15i+2i+3i^2=10+17i-3=7+17i\).
Denominator: \((2-3i)(2+3i)=4+9=13\).
So the quotient is \(\dfrac{7+17i}{13}\).
Answer: (a) \(11-10i\) (b) \(\dfrac{7}{13}+\dfrac{17}{13}i\)
Example 2 Solve \(z^2-4z+13=0\), giving your answers in the form \(a+bi\).
Use the quadratic formula with \(a=1\), \(b=-4\), \(c=13\).
\(z=\dfrac{4\pm\sqrt{16-52}}{2}=\dfrac{4\pm\sqrt{-36}}{2}\)
Since \(\sqrt{-36}=6i\): \(z=\dfrac{4\pm6i}{2}=2\pm3i\).
The two roots are conjugates, as expected for real coefficients.
Answer: \(z=2+3i\) or \(z=2-3i\)
Example 3 Find the complex number \(z\) that satisfies \(z+2z^*=6-3i\).
Let \(z=x+yi\), so \(z^*=x-yi\).
Then \(z+2z^*=x+yi+2x-2yi=3x-yi\).
The equation becomes \(3x-yi=6-3i\).
Equate real parts: \(3x=6\), so \(x=2\).
Equate imaginary parts: \(-y=-3\), so \(y=3\).
Check: \(2+3i+2(2-3i)=2+3i+4-6i=6-3i\). It works.
Answer: \(z=2+3i\)
Common errors
- Forgetting that \(i^2=-1\) when expanding, which gives a wrong real part. Always replace \(i^2\) before you collect terms.
- Writing the imaginary part as \(bi\) instead of \(b\). For \(3+2i\), \(\mathrm{Im}(z)=2\).
- Multiplying by the wrong conjugate when dividing. Use the conjugate of the denominator, and change only the sign of the imaginary part.
- Writing \(|z|=a^2+b^2\) without the square root, or including the \(i\) inside the root.
- Leaving a root with a negative sign inside it. Write \(\sqrt{-36}=6i\) rather than leaving \(\sqrt{-36}\) in the answer.
Exam tip: When a question gives an equation containing \(z\) and \(z^*\), set \(z=x+yi\) straight away and equate real and imaginary parts. It turns the problem into two ordinary simultaneous equations.
Check yourself
- Simplify \(i^{23}\).
Answer
The powers repeat every 4. \(23=4\times5+3\), so \(i^{23}=i^3=-i\). - Expand and simplify \((2-i)^2\).
Answer
\((2-i)^2=4-4i+i^2=4-4i-1=3-4i\). - Write \(\dfrac{1}{1+2i}\) in the form \(a+bi\).
Answer
Multiply by \(\dfrac{1-2i}{1-2i}\): the numerator is \(1-2i\) and the denominator is \(1+4=5\). The result is \(\dfrac15-\dfrac25i\). - Find the modulus of \(3-4i\), and the value of \((3+i)(3-i)\).
Answer
\(|3-4i|=\sqrt{9+16}=5\). Also \((3+i)(3-i)=9+1=10\), because a number times its conjugate equals its modulus squared.