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Next exam on November 14 – 15, 2026:
All concepts

Complex Numbers and Arithmetic

复数及其四则运算

Addition, subtraction, multiplication and division of complex numbers, conjugates and the modulus; division works by realising the denominator.

2 real-exam questions · 2 sittings28 in mock papers40 practice

Formulas

  • Imaginary Unit

    Definition
    i2=−1\mathrm{i}^{2}=-1

  • Modulus of a Complex Number

    Definition
    ∣z∣=∣a+bi∣=a2+b2|z|=|a+b\mathrm{i}|=\sqrt{a^{2}+b^{2}}

    Applies when：a, b ∈ R

    a, b
    real and imaginary parts of z = a+bi (a, b ∈ R)

  • Complex Conjugate

    Definition
    a+bi‾=a−bi,zz‾=∣z∣2\overline{a+b\mathrm{i}}=a-b\mathrm{i},\quad z\overline{z}=|z|^{2}

  • Complex Division

    Rule
    a+bic+di=(a+bi)(c−di)c2+d2\frac{a+b\mathrm{i}}{c+d\mathrm{i}}=\frac{(a+b\mathrm{i})(c-d\mathrm{i})}{c^{2}+d^{2}}

    Applies when：c+di ≠ 0

How the real exam tests it

Questions from real CSCA papers that test this concept. Work each one out before you open the answer.

  1. CSCA Math April 2026 Past Paper
    (z−2i)(2−i)=5( z - 2 i ) ( 2 - i ) = 5, then z=z =
    • A．2+3i2 + 3 i
    • B．2−3i2 - 3 i
    • C．3−2i3 - 2 i
    • D．3+2i3 + 2 i
    Show answer & solution

    Answer: A

    z−2i=52−i=5(2+i)5=2+iz - 2 i = \frac { 5 } { 2 - i } = \frac { 5 ( 2 + i ) } { 5 } = 2 + i. So z=2+i+2i=2+3iz = 2 + i + 2 i = 2 + 3 i.
    Take the full CSCA Math April 2026 Past Paper — timed & graded
  2. CSCA Math June 2026 Past Paper
    Given a complex number z=−1+iz = -1 + i, where ii is the imaginary unit, then z+1z+2+z+2z+1=\frac{z+1}{z+2} + \frac{z+2}{z+1} =
    • A．32−12i\frac{3}{2} - \frac{1}{2}i
    • B．32+12i\frac{3}{2} + \frac{1}{2}i
    • C．12−32i\frac{1}{2} - \frac{3}{2}i
    • D．1−i1 - i
    Show answer & solution

    Answer: A

    z+1=iz + 1 = i and z+2=1+iz + 2 = 1 + i. Then z+1z+2=i1+i=i(1−i)(1+i)(1−i)=1+i2=12+12i\frac{z+1}{z+2} = \frac{i}{1+i} = \frac{i(1-i)}{(1+i)(1-i)} = \frac{1+i}{2} = \frac{1}{2} + \frac{1}{2}i, and z+2z+1=1+ii=(1+i)(−i)i⋅(−i)=1−i\frac{z+2}{z+1} = \frac{1+i}{i} = \frac{(1+i)(-i)}{i \cdot (-i)} = 1 - i. Adding the two gives 32−12i\frac{3}{2} - \frac{1}{2}i. Choose (A).
    Take the full CSCA Math June 2026 Past Paper — timed & graded

Real-exam evidence

Recorded past-paper appearances are listed below; online verification is labelled separately.

2 questions across 2 sittings0 of 2 sittings available to verify free online
  • CSCA Math June 2026 Past Paper1 questionreading page not yet public
  • CSCA Math April 2026 Past Paper1 questionreading page not yet public

28 questions across 14 mock papers — our own papers, so a coverage signal rather than real-exam evidence

40 related practice questions availablePractise free: Complex Number

Sitting evidence comes from the CSCA past papers on record; questions whose original paper is missing are excluded, and sprint papers are deliberately not counted.

Glossary

complex number