Quadratic equations with complex coefficients and applications

The quadratic formula $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ continues to work even when $a, b, c$ are themselves complex , provided we know how to take the square root of a complex number. This subtopic ties everything together.

Square root of a complex number

We want, given $c = p + qi$, to find $w = u + vi$ such that $w^2 = c$.

$w^2 = u^2 - v^2 + 2uvi = p + qi$, so $u^2 - v^2 = p, \qquad 2uv = q.$

From the modulus: $u^2 + v^2 = |w|^2 = |c| = \sqrt{p^2 + q^2}$. Combining: $u^2 = \frac{|c| + p}{2}, \quad v^2 = \frac{|c| - p}{2}.$

Take the positive square root for $u$; then $v$ has the same sign as $q$ (since $2uv = q$). The other root of the quadratic $w^2 = c$ is $-w$.

Solving general quadratics

For $az^2 + bz + c = 0$ with $a, b, c \in \mathbb{C}$, $a \ne 0$: $z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},$ where the square root is computed in $\mathbb{C}$ by the procedure above.

The conjugate-root theorem does not apply here , coefficients are not necessarily real.

Locus and region problems

A typical Class-XI question asks: "Describe the set of $z \in \mathbb{C}$ satisfying ...". These problems mix algebra and geometry.

Common loci

Worked examples

Example 1. Find $\sqrt{3 + 4i}$.

$|3 + 4i| = 5$. $u^2 = (5 + 3)/2 = 4$, $v^2 = (5 - 3)/2 = 1$, $u, v$ same sign as $4$ (positive). So $u = 2, v = 1$ , one root is $2 + i$, the other $-2 - i$.

Check: $(2 + i)^2 = 4 + 4i + i^2 = 3 + 4i$. ✓

Example 2. Solve $z^2 - (3 + 2i)z + (5 + i) = 0$.

$\Delta = (3 + 2i)^2 - 4(5 + i) = 9 + 12i - 4 - 20 - 4i = -15 + 8i$.

$\sqrt{-15 + 8i}$: $|\Delta| = \sqrt{225 + 64} = 17$. $u^2 = (17 - 15)/2 = 1$, $v^2 = (17 + 15)/2 = 16$. $u = 1, v = 4$ ($v$ matches sign of $8$, positive). So $\sqrt{\Delta} = 1 + 4i$.

$z = \dfrac{(3 + 2i) \pm (1 + 4i)}{2} = \dfrac{4 + 6i}{2} \text{ or } \dfrac{2 - 2i}{2} = 2 + 3i \text{ or } 1 - i$.

Check: $(2 + 3i) + (1 - i) = 3 + 2i$ ✓; $(2 + 3i)(1 - i) = 2 - 2i + 3i - 3i^2 = 5 + i$ ✓.

Example 3. Describe the locus $|z - 2 + i| = 3$.

$|z - (2 - i)| = 3$: a circle centred at $z_0 = 2 - i$ (i.e. the point $(2, -1)$) with radius $3$.

Example 4. Find the region of the Argand plane where $|z - 1| \le 2$ and $\text{Re}(z) \ge 1$.

First condition: closed disc of radius $2$ centred at $(1, 0)$. Second: right half-plane $x \ge 1$. Intersection: right half of the disc.

Example 5 (harder). Find all $z$ with $z^2 = 1 + i$.

$|1 + i| = \sqrt{2}$. $u^2 = (\sqrt{2} + 1)/2$, $v^2 = (\sqrt{2} - 1)/2$. $u, v$ same sign. $u = \sqrt{(\sqrt{2} + 1)/2}$, $v = \sqrt{(\sqrt{2} - 1)/2}$. So $z = \pm \left(\sqrt{\frac{\sqrt{2} + 1}{2}} + i\sqrt{\frac{\sqrt{2} - 1}{2}}\right).$

Equivalently in polar form: $1 + i = \sqrt{2}\,\text{cis}\,\pi/4$, so $\sqrt{1 + i} = 2^{1/4}\,\text{cis}\,\pi/8$ (and its negative).

Try it yourself

1. Find $\sqrt{-5 + 12i}$.
2. Solve $z^2 - 4z + (4 - 4i) = 0$.
3. Solve $z^2 + (1 + i)z + 1 = 0$.
4. Describe the locus $|z + 1| + |z - 1| = 3$.
5. Sketch the region $1 \le |z| \le 2$.
6. Sketch the locus $\arg(z - 1) = \tfrac{\pi}{4}$.
7. Find $\sqrt{-8i}$.
8. Show that $|z|^2 - z - \bar{z} = 0$ describes a circle.
9. Solve $z^2 + z + 1 = 0$ in polar form (recognise as a cube root of unity).
10. Find all $z$ with $|z - 1| = |z - i|$.
11. Solve $iz^2 - 2z + i = 0$.
12. Find the area of the region $|z| \le 3$ and $\text{Im}(z) \ge 1$.

Pitfalls / Tricks

• The conjugate-root theorem applies only to quadratics with real coefficients.
• When taking $\sqrt{c}$ for complex $c$, there are two values $\pm w$. Specify both unless context dictates.
• Loci like $\arg(\cdot) = \theta$ define rays, not full lines.
• Insight. Every Class-XI complex-number problem reduces to one of three operations: algebra (use $i^2 = -1$), modulus arithmetic (use $z\bar{z} = |z|^2$), or geometry (draw the picture).