The Argand plane and harder problems

The Argand plane pictures complex numbers: the number x + iy is the point (x, y). The horizontal axis is the real axis and the vertical one the imaginary axis. So –3 + 4i is the point (–3, 4), in the second quadrant.

Two meanings to remember:

• |z| is the distance of the point from the origin: |–3 + 4i| = √(9 + 16) = 5.
• z̄ is the mirror image of z in the real axis: the conjugate –3 – 4i is the point (–3, –4).

Comparing parts turns a complex equation into two real ones. Solve (x + iy)(2 – i) = 7 + 4i for real x, y:

• Expand: 2x – ix + 2iy – i²y = (2x + y) + i(2y – x).
• So 2x + y = 7 and 2y – x = 4, giving x = 2, y = 3. Check: (2 + 3i)(2 – i) = 7 + 4i.

Taking moduli of both sides is another tool, and works because moduli split over products. If (x + iy)(1 – 2i) = 3 + 4i, then √(x² + y²) × √5 = 5, so x² + y² = 5, without finding x and y.

Powers of quotients: simplify first. (1 – i)/(1 + i) = (1 – i)²/((1 + i)(1 – i)) = (1 – 2i + i²)/2 = –i. So ((1 – i)/(1 + i))¹⁰⁰ = (–i)¹⁰⁰ = i¹⁰⁰ = 1.

Quadratic equations with a negative discriminant now have solutions. Use the formula x = (–b ± √D)/2a as usual, writing √(negative) with i:

• x² + 4x + 13 = 0: D = 16 – 52 = –36, so x = (–4 ± 6i)/2 = –2 ± 3i.
• 2x² + x + 1 = 0: D = 1 – 8 = –7, so x = (–1 ± √7 i)/4.
• For real coefficients, such roots always come as a conjugate pair.

A good habit: after any complex calculation, check the answer by putting it back. –2 + 3i in x² + 4x + 13 gives (4 – 12i – 9) + (–8 + 12i) + 13 = 0, as it should.