Complex Numbers: modulus equations & cube roots on an Argand diagram — VJC 2025 H2 Math Prelim P1
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Question
Do not use a calculator in answering this question.
(a) Find the complex number \(z\) which satisfies the equation \[ \frac{4|z|}{15 - z^*} = 5\mathrm{i}. \]
(b) The complex number \(w\) is such that \((w - \mathrm{i})^3 = -\mathrm{i}\).
(i) Given that one possible value of \(w\) is \(2\mathrm{i}\), find the two other possible values of \(w\). Give your answers in cartesian form \(a + b\mathrm{i}\).
The points \(W_1\), \(W_2\) and \(W_3\) on the Argand diagram represent the three roots of the equation \((w - \mathrm{i})^3 = -\mathrm{i}\), and the point \(A\) represents the complex number \(k\mathrm{i}\), where \(k\) is a positive real number.
(ii) Show that the points \(W_1\), \(W_2\) and \(W_3\) lie on a circle with centre \(A\) for some value of \(k\), stating the value of \(k\).
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(a) Solve for \(z\) by comparing parts
Set \(z = a + b\mathrm{i}\), clear the denominator, then match real and imaginary parts.
\[ \begin{aligned} \frac{4|z|}{15 - z^*} &= 5\mathrm{i} \\ \frac{4\sqrt{a^2+b^2}}{15-(a-b\mathrm{i})} &= 5\mathrm{i} \\ 4\sqrt{a^2+b^2} &= 75\mathrm{i} - 5a\mathrm{i} - 5b = -5b + (75-5a)\mathrm{i} \end{aligned} \]Comparing real and imaginary parts:
\[ \begin{cases} 4\sqrt{a^2+b^2} = -5b & \quad ---(1)\\ 75 - 5a = 0 \implies a = 15 & \quad ---(2) \end{cases} \]Substituting (2) into (1):
\[ \begin{aligned} 4\sqrt{15^2+b^2} &= -5b \\ 4^2\left(15^2+b^2\right) &= 25b^2 \\ 9b^2 &= 3600 \\ b^2 &= 400 \\ b &= -20 \qquad (\because b < 0 \text{ by eqn. (1)}) \end{aligned} \] \[ \therefore z = 15 - 20\mathrm{i} \](b)(i) Factorise the cubic and solve the quadratic factor
Expand the equation into a cubic in \(w\), divide out the known root \(w = 2\mathrm{i}\), then solve the remaining quadratic.
\[ \begin{aligned} (w-\mathrm{i})^3 &= -\mathrm{i} \\ w^3 + 3w^2(-\mathrm{i}) + 3w(-\mathrm{i})^2 + (-\mathrm{i})^3 + \mathrm{i} &= 0 \\ w^3 - 3\mathrm{i}w^2 - 3w + 2\mathrm{i} &= 0 \end{aligned} \]Since \(2\mathrm{i}\) is a root,
\[ w^3 - 3\mathrm{i}w^2 - 3w + 2\mathrm{i} = (w-2\mathrm{i})(Aw^2 + Bw + C) \]By comparing coefficients, \(A = 1,\ C = -1\).
\[ C - 2\mathrm{i}B = -3 \implies B = \frac{-1+3}{2\mathrm{i}} = -\mathrm{i} \] \[ w^2 - \mathrm{i}w - 1 = 0 \] \[ \begin{aligned} w &= \frac{\mathrm{i} \pm \sqrt{(-\mathrm{i})^2 - 4(-1)}}{2} \\ w &= \frac{\sqrt{3}}{2} + \frac{1}{2}\mathrm{i} \qquad \text{or} \qquad -\frac{\sqrt{3}}{2} + \frac{1}{2}\mathrm{i} \end{aligned} \](b)(ii) Locate the centre using equal moduli
Impose \(W_1A = W_2A\) to solve for \(k\), then confirm \(W_3A\) is the same distance.
Let \(W_1\), \(W_2\), \(W_3\) and \(A\) represent \(2\mathrm{i}\), \(\dfrac{\sqrt{3}}{2}+\dfrac{1}{2}\mathrm{i}\), \(-\dfrac{\sqrt{3}}{2}+\dfrac{1}{2}\mathrm{i}\) and \(k\mathrm{i}\) respectively.
For \(W_1A = W_2A\), i.e. \(\left|\dfrac{\sqrt{3}+\mathrm{i}}{2} - k\mathrm{i}\right| = |2\mathrm{i} - k\mathrm{i}|\),
\[ \begin{aligned} \left(\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{1}{2}-k\right)^2 &= (2-k)^2 \\ \frac{3}{4} + 1 - k + k^2 &= 4 - 4k + k^2 \\ k &= 1 \end{aligned} \]So, we have \(W_1A = |2\mathrm{i} - \mathrm{i}| = 1\) and \(W_2A = \left|\dfrac{\sqrt{3}+\mathrm{i}}{2} - \mathrm{i}\right| = 1\).
Check that \(W_3A = 1\) too:
\[ \left|\frac{-\sqrt{3}+\mathrm{i}}{2} - \mathrm{i}\right| = \left|\frac{-\sqrt{3}-\mathrm{i}}{2}\right| = \sqrt{\left(\frac{-\sqrt{3}}{2}\right)^2 + \left(\frac{-1}{2}\right)^2} = 1 \]Since the distances are all equal to 1, \(W_1\), \(W_2\) and \(W_3\) lie on a circle with radius 1, centred at \(A\) representing the complex number \(\mathrm{i}\), so \(k = 1\).