How to solve complex numbers questions in A Level Math Paper 3

Complex numbers are the topic students most often leave until last in Paper 3, and also the one where well-organised practice pays off quickly. The questions fall into a handful of types. This guide follows the order of the P3 complex numbers revision workshop on our channel and describes the types and the points where marks are usually lost.

Know the three forms, and use radians

A complex number can be written in three forms, and the second and third are not in the formula sheet, so they must be memorised:

• Cartesian form: x + iy, with a real part and an imaginary part.
• Modulus-argument form: r(cos θ + i sin θ).
• Exponential form: r e^iθ.

Sums, differences and equations that mix real and imaginary parts are easiest in cartesian form. Products, quotients and powers are easiest in exponential form. Use radians throughout unless the question uses degrees. A negative sign in front of the sine term means a negative argument.

The basics that cost marks

• i² = −1, and the final answer should be in the form x + iy with no i left in a denominator.
• To divide, multiply top and bottom by the conjugate of the denominator. If the denominator is only a multiple of i, multiplying by i is enough.
• When a quadratic has no real roots, show the step where the negative number under the square root becomes an i. The working carries marks.

Question type 1: Equations with real and imaginary parts

Two complex numbers are equal when their real parts are equal and their imaginary parts are equal. Write z as x + iy, expand, and equate the two parts. This gives two equations in x and y. When an equation contains both z and its conjugate, substitute x + iy for one and x − iy for the other, then compare parts. Read the question for conditions such as “the real part is positive” that tell you which answer to keep.

Question type 2: Roots of polynomials

If a polynomial has real coefficients, complex roots come in conjugate pairs. This does not hold if the equation itself contains i. The number of roots equals the degree. Given one complex root, you have two ways to find the rest:

• Write the conjugate as a second root, multiply the two factors into a quadratic with real coefficients (the product of a conjugate pair is a difference of squares), then divide it into the polynomial.
• Use the sum and product of the roots, which is usually quicker. The sum of the roots is minus the second coefficient over the leading coefficient. The product is the constant term over the leading coefficient, with a plus sign for an even degree and a minus sign for an odd degree.

Question type 3: Modulus and argument

The modulus is the distance from the origin. For the argument, find the basic angle from the positive values of the two parts, then place it using the quadrant: positive in the first and second, negative in the third and fourth. The argument must lie between −π and π. If it falls outside, add or subtract 2π.

• For a product, the moduli multiply and the arguments add. For a quotient, the moduli divide and the arguments subtract.
• For a power, use exponential form, and bring the argument back into range.
• In “by considering arguments” questions, keep the inverse tangent values exact and use the product or quotient property to link them.

Question type 4: Square roots of a complex number

In cartesian form, let x + iy squared equal the number, then equate real and imaginary parts. Substituting one equation into the other gives a quartic that is a disguised quadratic. Because x is real, reject any negative value of x². The two square roots are negatives of each other, not conjugates. In exponential form, raise to the power of one half to get one root. The other is the same distance from the origin and half a turn away, so add π to the argument if it is negative, or subtract π if it is positive. A square root is not the same as a root of an equation.

Question type 5: Loci on an Argand diagram

This type appears most often in the hardest questions. First put the condition in standard form, with the expression in the form z minus a fixed number. Take a minus sign out if needed. Every locus then reduces to one of three pictures:

• The modulus of z minus a equal to r is a circle with centre a and radius r.
• The modulus of z minus a equal to the modulus of z minus b is the perpendicular bisector of the two points.
• The argument of z minus a equal to a fixed angle is a half-line starting at a, with the angle measured from a horizontal line through a.

Inequalities shade the region. For two distances, shade the side of the bisector closest to the point that z must be nearer to. Draw the diagram accurately: the same scale on both axes, a compass for circles and a protractor for arguments. A line at 45 degrees (π/4) goes the same distance across as up, so it must pass through the matching grid points. A strict inequality uses a dashed line.

For the greatest or least value of the modulus, use the shortest distance to a line, which is the perpendicular, and for a circle use the line through the centre: the distance to the centre minus the radius for the least, and plus the radius for the greatest. For the greatest or least argument of a point on a circle, draw the two tangents from the origin and use the right-angled triangles they form with the centre.

Habits that protect the marks

• Sketch before you calculate. An Argand diagram shows the quadrant, the shape of a locus and the answer to many questions at a glance.
• State the reason. For “write down the other root”, the reason is the conjugate root property for real coefficients.
• Show every line of working in complex number questions. Quadratic roots with a bare answer risk losing the method marks.
• Check by substitution. Putting a root back into the original equation catches sign errors.
• Keep exact values for moduli and arguments, such as surds and multiples of pi, unless the question asks for decimals.