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AMC 12 only

Complex Numbers

Operations on a + bi, conjugates, and complex roots of quadratics — the algebraic basics you need before the deeper polar-form techniques.

A complex number has the form a+bia+bi, where aa and bb are real andi2=1i^2 = -1. Addition and subtraction work componentwise; multiplication distributes like binomials, using i2=1i^2=-1 to simplify:

(a+bi)(c+di)=(acbd)+(ad+bc)i(a+bi)(c+di) = (ac-bd) + (ad+bc)i

Conjugates and division

The conjugate of a+bia+bi is abia - bi. Multiplying a complex number by its conjugate always produces a real number:

(a+bi)(abi)=a2+b2(a+bi)(a-bi) = a^2+b^2

This is exactly how you divide complex numbers: multiply numerator and denominator by the denominator's conjugate to clear ii from the bottom.

1a+bi=abia2+b2\dfrac{1}{a+bi} = \dfrac{a-bi}{a^2+b^2}

The modulus a+bi=a2+b2|a+bi| = \sqrt{a^2+b^2} measures distance from the origin in the complex plane — it's the same quantity that appears under the square root above.

Complex roots of quadratics

When the discriminant of ax2+bx+c=0ax^2+bx+c=0 is negative, the quadratic formula still works — it just produces two complex conjugate roots, since k=ik\sqrt{-k} = i\sqrt{k} for k>0k>0. Vieta's formulas still hold exactly as before, even though the roots are no longer real.

Worked example

Problem: Solve x24x+13=0x^2 - 4x + 13 = 0 and express the roots in a+bia+bi form.

Solution: Using the quadratic formula:

x=4±16522=4±362=4±6i2=2±3ix = \dfrac{4 \pm \sqrt{16-52}}{2} = \dfrac{4 \pm \sqrt{-36}}{2} = \dfrac{4 \pm 6i}{2} = 2 \pm 3i

The roots are 2+3i2+3i and 23i2-3i — a conjugate pair, as expected whenever a quadratic with real coefficients has a negative discriminant.

Practice problems

1.Simplify (3+2i)(14i)(3+2i)(1-4i).

2.Find 3+4i|3+4i|.

3.Simplify 21+i\dfrac{2}{1+i}.

4.Solve x2+2x+5=0x^2 + 2x + 5 = 0 and express the roots in a+bia+bi form.