Complex Numbers - Cartesian (AI HL)
A complex number written as \(a+bi\) behaves almost like an ordinary algebraic expression, except that \(i^2=-1\) whenever it appears - which is what makes multiplying and dividing complex numbers by hand a little different from real arithmetic. This page covers Cartesian-form arithmetic and quadratics with complex roots, with worked examples and the mistakes that cost marks. It's part of the broader Complex Numbers topic.
27 questions on this sub-topic.
The two formulas
Covered under IB syllabus reference AHL1.12: the number \(i\) such that \(i^2=-1\), Cartesian form, real and imaginary parts, conjugates, and complex numbers as solutions to real quadratics with \(b^2-4ac<0\).
Cartesian form
\(z=a+bi\)
In the formula booklet. Add, subtract and multiply like ordinary binomials, replacing \(i^2\) with \(-1\) wherever it appears; to divide, multiply top and bottom by the conjugate \(a-bi\).
Complex roots of a quadratic
\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]
In the formula booklet. When \(b^2-4ac<0\), the square root is of a negative number, giving a conjugate pair of complex roots \(a\pm bi\).
Need the full syllabus wording and formula-booklet reference table? See Complex Numbers.
Worked examples
Let \(z_1 = 3 + 2i\) and \(z_2 = 1 - 4i.\) Find, in the form \(a + bi\):
(a) \(z_1 z_2\)
(b) \(\dfrac{z_1}{z_2}.\)
Worked solution
(a) Find \(z_1 z_2\). \((3+2i)(1-4i) = 3 - 12i + 2i - 8i^2 = 3 - 10i + 8\) M1
\(= 11 - 10i.\) A1
(b) Find \(\dfrac{z_1}{z_2}\). Multiply by the conjugate: \(\dfrac{(3+2i)(1+4i)}{(1-4i)(1+4i)} = \dfrac{-5 + 14i}{17}\) M1
\(= -\tfrac{5}{17} + \tfrac{14}{17}i.\) A1
A cubic with real coefficients has roots \(2\) and \(3 - i\).
(a)(i) State the third root.
(a)(ii) State its multiplicity.
(b) Find the cubic in expanded form \(z^3 + bz^2 + cz + d\).
Worked solution
(a) Real coefficients force conjugate pairs, so the third root is \(3 + i.\) M1 A1
(b) \((z-2)(z^2 - 6z + 10)\) M1
\(= z^3 - 8z^2 + 22z - 20.\) A1
Common mistakes
- Forgetting complex roots come in conjugate pairs. If a real-coefficient polynomial has one complex root \(a+bi\), the other root is automatically \(a-bi\) - there's no need to solve for it separately, as in the cubic example above.
- Losing the minus sign when \(i^2\) appears. \(i^2=-1\), so a term like \(2i\times 3i\) becomes \(-6\), not \(6i^2\) left unevaluated or \(6\) with the sign dropped - a very easy slip when expanding quickly.
- Multiplying only the numerator by the conjugate when dividing. To rationalise \(\dfrac{z_1}{z_2}\), both the numerator and denominator must be multiplied by the conjugate of \(z_2\), otherwise the value of the fraction changes.
Ready to practise properly?
27 Cartesian complex-number questions, marked instantly like the real exam.
Quick answers
What is the Cartesian form of a complex number?
\(z=a+bi\), where \(a\) is the real part, \(b\) is the imaginary part, and \(i^2=-1\). It's the \(a+bi\) form you use for adding, subtracting, multiplying and dividing complex numbers by hand.
Why do complex roots of a quadratic always come in pairs?
When a quadratic (or any polynomial) has real coefficients and \(b^2-4ac<0\), the quadratic formula produces two roots that are complex conjugates of each other, \(a+bi\) and \(a-bi\), because the \(\pm\) in front of the square root of a negative number flips only the sign of the imaginary part.