A-Level Complex numbers in cartesian form and the Argand diagram: worked solution

Question

The roots z₁, z₂ of z² + αz + β = 0 satisfy |z₁| = |z₂| = √26/2 with 0 < arg(z₁) < π/2, lying on the ellipse (x/2)² + (y/3)² = 1. Given α is not real, find α and β.

Method: a point z = x + iy has |z|² = x² + y². The modulus condition and the ellipse therefore give two simultaneous equations in x² and y², which pin down the roots. Then use sum and product of roots. Step 1 - turn the modulus condition into an equation in x and y. |z₁| = √26/2, so |z₁|² = 26/4 = 13/2, giving x² + y² = 13/2 Step 2 - write out the ellipse. (x/2)² + (y/3)² = 1, that is x²/4 + y²/9 = 1 Step 3 - solve the pair simultaneously. From Step 1, y² = 13/2 − x². Substitute into Step 2 and multiply every term by 36 to clear fractions: 9x² + 4y² = 36 9x² + 4(13/2 − x²) = 36 9x² + 26 − 4x² = 36 5x² = 10 x² = 2, so x = ±√2 Then y² = 13/2 − 2 = 9/2, so y = ±3/√2 = ±(3/2)√2 Step 4 - pick out z₁. The condition 0 < arg(z₁) < π/2 puts z₁ strictly in the first quadrant, so both x and y are positive: z₁ = √2 + (3/2)√2 i Step 5 - identify z₂. For z² + αz + β = 0 the sum of the roots is −α and the product is β. z₂ is one of the three remaining points, and α must NOT be real, which rules two of them out: if z₂ = √2 − (3/2)√2 i, then z₁ + z₂ = 2√2 and α = −2√2, which is real - reject; if z₂ = −√2 − (3/2)√2 i, then z₁ + z₂ = 0 and α = 0, which is also real - reject. So z₂ = −√2 + (3/2)√2 i. Step 6 - use the sum of roots to find α. z₁ + z₂ = (√2 − √2) + ((3/2)√2 + (3/2)√2)i = 3√2 i α = −(z₁ + z₂) = −3√2 i (This is not real, as required.) Step 7 - use the product of roots to find β. β = z₁z₂ = (√2 + (3/2)√2 i)(−√2 + (3/2)√2 i) This has the form (u + v)(−u + v) = v² − u², with u = √2 and v = (3/2)√2 i: u² = 2 v² = ((3/2)√2)² i² = (9/2)(−1) = −9/2 β = v² − u² = −9/2 − 2 = −13/2