Complex Numbers - Cartesian Form (AA HL)
This page focuses on complex numbers written \(a+bi\): adding, multiplying and dividing them, taking conjugates, and solving quadratics whose discriminant is negative. Polar form and De Moivre's theorem build on these skills but are covered separately. It's part of the broader Complex Numbers topic.
25 questions on this sub-topic.
Key facts
Covered under IB syllabus reference AHL1.12: the number \(i\), where \(i^2=-1\); Cartesian form \(z=a+bi\); the terms real part, imaginary part, conjugate, modulus and argument; the complex plane (Argand diagram).
Cartesian form and conjugate
\(z=a+bi,\ \bar z=a-bi\)
Not in the formula booklet - this is prior knowledge you're expected to already have from SL work with \(i\).
Complex roots of a quadratic
If \(z^2+pz+q=0\) has real coefficients and one root \(a+bi\), the other root is \(a-bi\).
Non-real roots of a polynomial with real coefficients always occur in conjugate pairs - useful for finding a second root without repeating the whole calculation.
Need the full syllabus wording and formula-booklet reference table? See Complex Numbers. For calculator-specific tips, see the parent topic's GDC guidance.
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
Solve \(z^2 - 4z + 13 = 0\), giving the roots in the form \(a + bi.\)
Worked solution
\(z = \dfrac{4 \pm \sqrt{16 - 52}}{2}.\) M1 \(= \dfrac{4 \pm \sqrt{-36}}{2}.\) A1
\(\sqrt{-36} = 6i\) M1 \(z = \dfrac{4 \pm 6i}{2}\) A1 \(= 2 \pm 3i.\) A1
An AC circuit has voltage \(V = 50 + 0i\) V and impedance \(Z = 6 + 8i\ \Omega\). Using \(V = IZ\), find the current \(I\) in the form \(a + bi\) and its magnitude.
Worked solution
\(I = \dfrac{50}{6 + 8i}\) M1
\(= \dfrac{50(6 - 8i)}{100}\) A1
\(= \dfrac{50(6 - 8i)}{100}\) A1
\(= 3 - 4i\) A. A1
\(|I| = \sqrt{9 + 16}\) M1
\(= 5\) A. A1
Common mistakes
- Forgetting \(i^2=-1\) when expanding brackets. \((3+2i)(1-4i)\) has an \(i^2\) term buried in the expansion - if you leave it as \(-8i^2\) instead of \(+8\), the whole real part comes out wrong.
- Dividing without multiplying by the conjugate. You can't simplify \(\dfrac{z_1}{z_2}\) by cancelling term by term - multiply numerator and denominator by the conjugate of the denominator so the bottom becomes a real number.
- Only writing down one root of a real-coefficient quadratic. If the discriminant is negative, the two roots are automatically a conjugate pair - once you have \(2+3i\), the second root \(2-3i\) follows immediately, no extra working needed.
Ready to practise properly?
26 Cartesian complex-number questions, marked instantly like the real exam.
Quick answers
What is i in a complex number?
\(i\) is defined by \(i^2 = -1\). A complex number in Cartesian form is written \(z = a + bi\), where \(a\) is the real part and \(b\) is the imaginary part.
Why do complex roots of a quadratic always come in pairs?
If a polynomial has real coefficients, any complex root \(a+bi\) must be paired with its conjugate \(a-bi\), since the coefficients being real forces the imaginary parts to cancel symmetrically.