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9709 · 3.9

Complex numbers — common mistakes

Common exam mistakes on 9709 Complex numbers. Learn what loses marks, then practise the topic with Examiner’s Ink.

Exam tip 1

When a question asks for the argument, it implies the principal argument unless stated otherwise. After performing multiplication or division, your resulting argument might be outside the range (π,π](-\pi, \pi]. You must add or subtract multiples of 2π2\pi to bring it back into the principal range.

What exactly is 'i'?

ii is the imaginary unit, defined as i=1i = \sqrt{-1}. It's not a 'real' number because no real number can be squared to produce a negative result. It's a foundational concept that extends our number system.

Why are there so many forms for complex numbers (Cartesian, polar, exponential)?

Each form is useful for different operations. Cartesian form (x+iyx+iy) is best for addition and subtraction. Polar (r(cosθ+isinθ)r(\cos\theta + i\sin\theta)) and exponential (reiθre^{i\theta}) forms are far superior for multiplication, division, and finding powers and roots.

My calculator has a complex number mode. Can I just use that in the exam?

You can use it to check your answers, but you cannot rely on it for your working. Exam questions are designed to test your understanding of the methods, such as finding the modulus and argument or applying De Moivre's theorem. You must show these steps clearly to earn method marks.

How do I divide complex numbers in Cartesian form?

To divide z1z2\frac{z_1}{z_2}, you multiply the numerator and denominator by the complex conjugate of the denominator, z2z_2^*. For example, to compute 2+3i1i\frac{2+3i}{1-i}, you would calculate (2+3i)(1+i)(1i)(1+i)\frac{(2+3i)(1+i)}{(1-i)(1+i)}. This process is called 'realising the denominator' because the new denominator (1i)(1+i)=12i2=1(1)=2(1-i)(1+i) = 1^2 - i^2 = 1 - (-1) = 2 will be a real number.