Complex Numbers: conjugate roots, Argand geometry & exact tangent — TMJC 2025 H2 Math Prelim P1

(a) Values of \(a\), \(b\) and the remaining roots

Because the coefficients are real, the conjugate of the given root is also a root, so multiply the two conjugate factors to get a real quadratic factor.

Since the equation has real coefficients, \(-2 + 2\mathrm{i}\) being a root implies \(-2 - 2\mathrm{i}\) is also a root. Consider \[ \begin{aligned} [z - (-2 + 2\mathrm{i})][z - (-2 - 2\mathrm{i})] &= [(z + 2) - 2\mathrm{i}][(z + 2) + 2\mathrm{i}] \\ &= (z + 2)^2 - (2\mathrm{i})^2 \\ &= z^2 + 4z + 4 + 4 \\ &= z^2 + 4z + 8. \end{aligned} \]

Writing the cubic as \((z^2 + 4z + 8)(pz + q)\):

• Coefficient of \(z^3\): \(p = 1\).
• Constant: \(8q = -16\sqrt{2} \Rightarrow q = -2\sqrt{2}\).

So \(z^3 + az^2 + bz - 16\sqrt{2} = (z^2 + 4z + 8)(z - 2\sqrt{2})\).

• Coefficient of \(z^2\): \(a = -2\sqrt{2} + 4 = 4 - 2\sqrt{2}\).
• Coefficient of \(z\): \(b = -8\sqrt{2} + 8 = 8 - 8\sqrt{2}\).

Therefore \(a = 4 - 2\sqrt{2}\), \(b = 8 - 8\sqrt{2}\), and the other two roots are \(-2 - 2\mathrm{i}\) and \(2\sqrt{2}\).

(b) Modulus, argument and the Argand diagram

Compute the modulus and argument of each root, taking care to place each in the correct quadrant.

Let \(z_1 = -2 + 2\mathrm{i}\) (2nd quadrant): \[ \begin{aligned} |z_1| &= \sqrt{(-2)^2 + 2^2} = 2\sqrt{2},\\ \arg(z_1) &= \pi - \tan^{-1}\!\left(\tfrac{2}{2}\right) = \pi - \dfrac{\pi}{4} = \dfrac{3\pi}{4}. \end{aligned} \]

Let \(z_2 = -2 - 2\mathrm{i}\) (3rd quadrant): \[ \begin{aligned} |z_2| &= 2\sqrt{2},\\ \arg(z_2) &= -\pi + \tan^{-1}\!\left(\tfrac{2}{2}\right) = -\pi + \dfrac{\pi}{4} = -\dfrac{3\pi}{4}. \end{aligned} \]

Let \(z_3 = 2\sqrt{2}\): \(|z_3| = 2\sqrt{2}\) and \(\arg(z_3) = 0\).

On the Argand diagram, plot \(A \equiv -2 + 2\mathrm{i}\) in the second quadrant, \(B \equiv -2 - 2\mathrm{i}\) in the third quadrant, and \(C \equiv 2\sqrt{2}\) on the positive real axis. Each point lies a distance \(2\sqrt{2}\) from \(O\); \(A\) has argument \(\dfrac{3\pi}{4}\), \(B\) has argument \(-\dfrac{3\pi}{4}\), and \(C\) has argument \(0\).

Geometrical relationship: \(B\) is the reflection of \(A\) in the real axis (equivalently, \(B\) is the \(90^\circ\) anti-clockwise rotation of \(A\) about \(O\), since \(\mathrm{i}A = \mathrm{i}(-2 + 2\mathrm{i}) = -2 - 2\mathrm{i} = B\)).

(c) Proving \(\tan\dfrac{3\pi}{8} = 1 + \sqrt{2}\)

Since \(OA = OC\), the sum \(z_1 + z_3\) gives the rhombus diagonal whose argument bisects \(\angle AOC\), namely \(\dfrac{3\pi}{8}\).

Since \(OA = OC = 2\sqrt{2}\), consider the point \(D\) such that \(OADC\) is a rhombus. Then \(OD\) bisects \(\angle AOC\), so \[ \angle DOC = \tfrac{1}{2}\cdot\tfrac{3\pi}{4} = \tfrac{3\pi}{8}. \]

Let \(D \equiv z_4\) where \[ z_4 = (-2 + 2\mathrm{i}) + 2\sqrt{2} = (-2 + 2\sqrt{2}) + 2\mathrm{i}. \]

Since \(z_4\) lies in the 1st quadrant, \(\arg(z_4) = \tfrac{3\pi}{8}\): \[ \begin{aligned} \tan\dfrac{3\pi}{8} &= \dfrac{2}{-2 + 2\sqrt{2}} \\ &= \dfrac{1}{-1 + \sqrt{2}} \times \dfrac{-1 - \sqrt{2}}{-1 - \sqrt{2}} \\ &= \dfrac{-1 - \sqrt{2}}{(-1)^2 - (\sqrt{2})^2} \\ &= \dfrac{-1 - \sqrt{2}}{1 - 2} \\ &= 1 + \sqrt{2} \quad \text{(shown)}. \end{aligned} \]