Unit 23 Review Exercises

Complex Numbers

These questions mix the ideas from the whole unit. All of them are HL only AHL. Paper 1 means no calculator; Paper 2 means a GDC is allowed.

1 Paper 1 AHL Let \(z = 2 - 3i\) and \(w = 1 + i\). Find, in the form \(x + iy\),

(a) \(zw^*\) \(\qquad\) (b) \(\dfrac{z}{w}\)

(a) \(w^* = 1 - i\): \(\;(2 - 3i)(1 - i) = 2 - 2i - 3i + 3i^2 = -1 - 5i\).

(b) \(\dfrac{z}{w} = \dfrac{(2 - 3i)(1 - i)}{(1 + i)(1 - i)} = \dfrac{-1 - 5i}{2} = -\dfrac{1}{2} - \dfrac{5}{2}i\). See 23.1 and 23.2.

2 Paper 1 AHL Find the complex number \(z\) such that \(z + 3z^* = 8 + 6i\).

Let \(z = x + iy\): \(\;x + iy + 3x - 3iy = 4x - 2iy = 8 + 6i\).

So \(x = 2\) and \(-2y = 6\), \(y = -3\). \(\;z = 2 - 3i\). See 23.2.

3 Paper 1 AHL Given that \(2 - 3i\) is a root of \(z^3 + pz^2 + qz - 26 = 0\), where \(p, q \in \mathbb{R}\), find \(p\) and \(q\) and the other two roots.

Real coefficients, so \(2 + 3i\) is a root; the quadratic factor is \(z^2 - 4z + 13\). Product of the roots \(= 26\), so \(13c = 26\) and the real root is \(c = 2\).

\((z^2 - 4z + 13)(z - 2) = z^3 - 6z^2 + 21z - 26\), so \(p = -6\), \(q = 21\). The roots are \(2 \pm 3i\) and \(2\). See 23.2.

4 Paper 1 AHL Let \(z = \dfrac{-1 + \sqrt{3}\,i}{1 + i}\). By writing the top and bottom in Euler form, write \(z\) in the form \(r\mathrm{e}^{i\theta}\).

\(-1 + \sqrt{3}\,i = 2\mathrm{e}^{i\frac{2\pi}{3}}\) and \(1 + i = \sqrt{2}\,\mathrm{e}^{i\frac{\pi}{4}}\).

\(z = \dfrac{2}{\sqrt{2}}\mathrm{e}^{i\left(\frac{2\pi}{3} - \frac{\pi}{4}\right)} = \sqrt{2}\,\mathrm{e}^{i\frac{5\pi}{12}}\). See 23.3 and 23.4.

5 Paper 1 AHL Find \((-1 + i)^9\) in the form \(x + iy\).

\(-1 + i = \sqrt{2}\operatorname{cis}\dfrac{3\pi}{4}\). By De Moivre, \((-1 + i)^9 = (\sqrt{2})^9\operatorname{cis}\dfrac{27\pi}{4} = 16\sqrt{2}\operatorname{cis}\dfrac{3\pi}{4}\) (since \(\dfrac{27\pi}{4} - 6\pi = \dfrac{3\pi}{4}\)).

\(= 16\sqrt{2}\left(-\dfrac{\sqrt{2}}{2} + \dfrac{\sqrt{2}}{2}i\right) = -16 + 16i\). See 23.5.

6 Paper 1 AHL Solve \(z^3 = -27i\), giving your answers in the form \(x + iy\). Verify that the sum of the roots is \(0\).

\(-27i = 27\operatorname{cis}\left(-\dfrac{\pi}{2}\right)\), so \(z = 3\operatorname{cis}\left(-\dfrac{\pi}{6} + \dfrac{2k\pi}{3}\right)\), \(k = 0, 1, -1\):

\(3\operatorname{cis}\left(-\dfrac{\pi}{6}\right) = \dfrac{3\sqrt{3}}{2} - \dfrac{3}{2}i\), \(\;3\operatorname{cis}\dfrac{\pi}{2} = 3i\), \(\;3\operatorname{cis}\left(-\dfrac{5\pi}{6}\right) = -\dfrac{3\sqrt{3}}{2} - \dfrac{3}{2}i\).

Sum: real parts \(\dfrac{3\sqrt{3}}{2} - \dfrac{3\sqrt{3}}{2} = 0\), imaginary parts \(-\dfrac{3}{2} + 3 - \dfrac{3}{2} = 0\) ✓. See 23.6.

7 Paper 2 AHL Let \(z = 3 - 5i\).

(a) Find \(|z|\) and \(\arg z\).

(b) Hence write \(z^4\) in the form \(r\operatorname{cis}\theta\), with \(-\pi < \theta \leq \pi\).

(a) \(|z| = \sqrt{34} = 5.83\). 4th quadrant: \(\arg z = -\arctan\dfrac{5}{3} = -1.0304\ldots = -1.03\) (3 s.f.).

(b) \(|z^4| = 34^2 = 1156\). \(\;4\arg z = -4.1217\ldots\), which is less than \(-\pi\), so add \(2\pi\): \(\;2.1615\ldots\)

\(z^4 = 1156\operatorname{cis}2.16\). See 23.3 and 23.5.

8 Paper 1 AHL Let \(\omega = \mathrm{e}^{i\frac{2\pi}{5}}\).

(a) Explain why \(1 + \omega + \omega^2 + \omega^3 + \omega^4 = 0\).

(b) Hence show that \(\cos\dfrac{2\pi}{5} + \cos\dfrac{4\pi}{5} = -\dfrac{1}{2}\).

(a) These are the five fifth roots of unity, the roots of \(z^5 - 1 = 0\). The coefficient of \(z^4\) is \(0\), so their sum is \(0\).

(b) Take real parts. \(\operatorname{Re}(\omega^k) = \cos\dfrac{2k\pi}{5}\), and \(\cos\dfrac{6\pi}{5} = \cos\dfrac{4\pi}{5}\), \(\cos\dfrac{8\pi}{5} = \cos\dfrac{2\pi}{5}\). So

\[ 1 + 2\cos\frac{2\pi}{5} + 2\cos\frac{4\pi}{5} = 0 \;\Longrightarrow\; \cos\frac{2\pi}{5} + \cos\frac{4\pi}{5} = -\frac{1}{2} . \]

See 23.6.

9 Paper 1 AHL \(A\) and \(B\) represent \(-1 + i\) and \(3 + 2i\). \(ABCD\) is a square with the vertices in anticlockwise order. Find the complex numbers for \(C\) and \(D\).

\(\overrightarrow{AB} = 4 + i\). Turn anticlockwise: \(\overrightarrow{BC} = i(4 + i) = -1 + 4i\).

\(c = (3 + 2i) + (-1 + 4i) = 2 + 6i\) and \(d = (-1 + i) + (-1 + 4i) = -2 + 5i\). See 23.7.

10 Paper 1 AHL (a) Solve \(z^4 + 4z^2 + 16 = 0\), giving your answers in the form \(x + iy\).

(b) The four roots are drawn in an Argand diagram. Show that they are the vertices of a rectangle, and find its area.

(a) Quadratic in \(z^2\): \(\;z^2 = \dfrac{-4 \pm \sqrt{16 - 64}}{2} = -2 \pm 2\sqrt{3}\,i = 4\mathrm{e}^{\pm i\frac{2\pi}{3}}\).

\(z^2 = 4\mathrm{e}^{i\frac{2\pi}{3}}\) gives \(z = 2\mathrm{e}^{i\frac{\pi}{3}}\), \(2\mathrm{e}^{-i\frac{2\pi}{3}}\); the conjugate equation gives \(z = 2\mathrm{e}^{-i\frac{\pi}{3}}\), \(2\mathrm{e}^{i\frac{2\pi}{3}}\).

\[ z = 1 + \sqrt{3}\,i, \quad 1 - \sqrt{3}\,i, \quad -1 + \sqrt{3}\,i, \quad -1 - \sqrt{3}\,i . \]

(b) The points are \((\pm 1, \pm\sqrt{3})\). The figure is symmetric in both axes, with sides parallel to the axes, so it is a rectangle with width \(2\) and height \(2\sqrt{3}\). Area \(= 4\sqrt{3}\). See 23.6 and 23.7.