Algebra, Quadratics, and Complex Conjugates

Problem Statement

While the complex plane is highly geometric, working with complex numbers requires rigorous algebraic manipulation. Treating $i$ as a variable where $i^2 = -1$ allows us to add, subtract, multiply, and divide complex numbers using standard algebraic rules like the difference of two squares and index laws.

One of the most important operations is division. Just as we "rationalise the denominator" to remove surds from the bottom of a fraction, we use the complex conjugate to remove imaginary numbers from the denominator.

Furthermore, the introduction of $i$ fundamentally "fixes" the quadratic formula. By using the discriminant ($\Delta = b^2 - 4ac$), we can determine if a solution is real or imaginary, and finally solve equations that were previously considered "impossible."

Consider the complex number equation: $z^2 - 4z + 13 = 0$

(a) Use the discriminant to determine if the roots of this equation are real or imaginary.

(b) Use the quadratic formula to solve for the two complex roots, $z_1$ and $z_2$.

(c) Given your roots $z_1$ and $z_2$, express the quotient $W = \frac{z_1}{z_2}$ in the standard form $x + iy$. (Assume $z_1$ has a positive imaginary part).

Hints

• Part (a): Calculate $\Delta = b^2 - 4ac$. If $\Delta > 0$, roots are real and distinct. If $\Delta = 0$, roots are real and equal. If $\Delta < 0$, roots are complex conjugates.
• Part (b): Apply the quadratic formula $z = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$. When taking the square root of a negative number, use the fact that $\sqrt{-N} = \sqrt{-1 \times N} = i\sqrt{N}$.
• Part (c): Write the fraction $\frac{z_1}{z_2}$. To eliminate the complex number from the denominator, multiply both the numerator and the denominator by the complex conjugate of $z_2$ (which is $\bar{z}_2$). Use the difference of two squares identity: $(x + iy)(x - iy) = x^2 - (iy)^2 = x^2 - i^2y^2 = x^2 + y^2$.

Solutions

Part (a): The Discriminant

1. Identify the coefficients: $a = 1, b = -4, c = 13$.
2. Calculate $\Delta$: $\Delta = (-4)^2 - 4(1)(13)$ $\Delta = 16 - 52$ $\Delta = -36$
3. Since $\Delta < 0$, the quadratic equation has no real solutions. The roots are imaginary (specifically, a complex conjugate pair).

Part (b): Solving the Quadratic Equation

1. Apply the quadratic formula: $z = \frac{-(-4) \pm \sqrt{-36}}{2(1)}$
2. Simplify the square root using $i$: $z = \frac{4 \pm \sqrt{-1 \times 36}}{2}$ $z = \frac{4 \pm i\sqrt{36}}{2}$ $z = \frac{4 \pm 6i}{2}$
3. Divide both terms by 2: $z = 2 \pm 3i$
4. Therefore, the two roots are $z_1 = 2 + 3i$ (positive imaginary part) and $z_2 = 2 - 3i$.

Part (c): Complex Division (Rationalisation)

1. We need to evaluate $W = \frac{z_1}{z_2}$: $W = \frac{2 + 3i}{2 - 3i}$
2. Multiply numerator and denominator by the conjugate of the denominator, which is $(2 + 3i)$: $W = \frac{2 + 3i}{2 - 3i} \times \frac{2 + 3i}{2 + 3i}$
3. Expand the numerator using FOIL (First, Outer, Inner, Last): $(2 + 3i)(2 + 3i) = 4 + 6i + 6i + 9i^2$ Since $i^2 = -1$: $= 4 + 12i - 9 = -5 + 12i$
4. Expand the denominator using the difference of two squares ($x^2 + y^2$): $(2 - 3i)(2 + 3i) = 2^2 + 3^2 = 4 + 9 = 13$
5. Combine the numerator and denominator: $W = \frac{-5 + 12i}{13}$
6. Write in the standard $x + iy$ form by separating the fraction: $W = -\frac{5}{13} + \frac{12}{13}i$

Takeaways

• The Conjugate Pair Theorem: If a polynomial with real coefficients has a complex root $a + ib$, then its conjugate $a - ib$ is guaranteed to also be a root. This is why the quadratic formula always produces a $\pm$ pair for negative discriminants.
• Division is Multiplication: Just like rationalising surds ($\frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}$), dividing complex numbers requires multiplying by the conjugate to force a real denominator via the difference of two squares.
• Index Laws Apply: $i$ is just a variable obeying standard index laws, with the special cyclical property: $i^1 = i$, $i^2 = -1$, $i^3 = -i$, $i^4 = 1$.

Further Readings

• HSC Complex Numbers: https://vumaths.com/booklets/hsc-complex-numbers/
• HSC Polynomials: https://vumaths.com/booklets/hsc-polynomials/
• HSC Algebra: https://vumaths.com/booklets/hsc-collections/

Connect with me

Mastering complex algebra is your ticket to scoring a Band E4 in Extension 2. If you need more structured practice, check out the booklets on Vu's Maths Hub. I regularly post solutions to difficult complex polynomial problems on YouTube. Follow me on LinkedIn for professional updates, or read my detailed articles on the HSC curriculum over at my Substack.