Powers and de Moivre's formula Problem set B One question at a time

Work by hand. Give exact values.

1. (a) \(z = 2e^{i\frac{\pi}{15}}\). Find \(z^{5}\) in the form \(x + iy\).

   • \(z^5 = \left(2e^{i\frac{\pi}{15}}\right)^5 = 2^{5}e^{i \times 5 \times \frac{\pi}{15}} = 32e^{i\frac{\pi}{3}}\)
   • \(z^5 = 32e^{i\frac{\pi}{3}} = 32\left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right) = 16+16 \sqrt{3}i\)

2. (b) \(z = 4e^{i\frac{\pi}{4}}\). Find \(z^{3}\) in the form \(x + iy\).

   • \(z^3 = \left(4e^{i\frac{\pi}{4}}\right)^3 = 4^{3}e^{i \times 3 \times \frac{\pi}{4}} = 64e^{i\frac{3\pi}{4}}\)
   • \(z^3 = 64e^{i\frac{3\pi}{4}} = 64\left(\cos\frac{3\pi}{4} + i\sin\frac{3\pi}{4}\right) = -32 \sqrt{2}+32 \sqrt{2}i\)

3. (c) \(z = \sqrt{3}e^{i\frac{7\pi}{6}}\). Find \(z^{4}\) in the form \(x + iy\).

   • \(z^4 = \left(\sqrt{3}e^{i\frac{7\pi}{6}}\right)^4 = (\sqrt{3})^{4}e^{i \times 4 \times \frac{7\pi}{6}} = 9e^{i\frac{14\pi}{3}}\)
   • \(\frac{14\pi}{3} - 4\pi = \frac{2\pi}{3}\)
   • \(z^4 = 9e^{i\frac{2\pi}{3}} = 9\left(\cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3}\right) = -\frac{9}{2}+\frac{9 \sqrt{3}}{2}i\)

Find the modulus and argument first, with \(0 \le \arg z < 2\pi\). Give exact values.

1. (a) \(z = - \sqrt{3}+i\). Write \(z^{3}\) in the form \(x + iy\).

   • \(|z| = \sqrt{(- \sqrt{3})^2 + (1)^2} = 2\)
   • \(\text{The point is in the second quadrant. The acute angle is } \arctan\frac{1}{\sqrt{3}} = \frac{\pi}{6}, \text{ so } \arg z = \frac{5\pi}{6}\)
   • \(z = 2e^{i\frac{5\pi}{6}}\)
   • \(z^3 = \left(2e^{i\frac{5\pi}{6}}\right)^3 = 2^{3}e^{i \times 3 \times \frac{5\pi}{6}} = 8e^{i\frac{5\pi}{2}}\)
   • \(\frac{5\pi}{2} - 2\pi = \frac{\pi}{2}\)
   • \(z^3 = 8e^{i\frac{\pi}{2}} = 8\left(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\right) = 8i\)

2. (b) \(z = -2-2i\). Write \(z^{6}\) in the form \(x + iy\).

   • \(|z| = \sqrt{(-2)^2 + (-2)^2} = 2 \sqrt{2}\)
   • \(\text{The point is in the third quadrant. The acute angle is } \arctan\frac{2}{2} = \frac{\pi}{4}, \text{ so } \arg z = \frac{5\pi}{4}\)
   • \(z = 2 \sqrt{2}e^{i\frac{5\pi}{4}}\)
   • \(z^6 = \left(2 \sqrt{2}e^{i\frac{5\pi}{4}}\right)^6 = (2 \sqrt{2})^{6}e^{i \times 6 \times \frac{5\pi}{4}} = 512e^{i\frac{15\pi}{2}}\)
   • \(\frac{15\pi}{2} - 6\pi = \frac{3\pi}{2}\)
   • \(z^6 = 512e^{i\frac{3\pi}{2}} = 512\left(\cos\frac{3\pi}{2} + i\sin\frac{3\pi}{2}\right) = -512i\)

3. (c) \(z = 2-2 \sqrt{3}i\). Write \(z^{2}\) in the form \(x + iy\).

   • \(|z| = \sqrt{2^2 + (- 2 \sqrt{3})^2} = 4\)
   • \(\text{The point is in the fourth quadrant. The acute angle is } \arctan\frac{2 \sqrt{3}}{2} = \frac{\pi}{3}, \text{ so } \arg z = \frac{5\pi}{3}\)
   • \(z = 4e^{i\frac{5\pi}{3}}\)
   • \(z^2 = \left(4e^{i\frac{5\pi}{3}}\right)^2 = 4^{2}e^{i \times 2 \times \frac{5\pi}{3}} = 16e^{i\frac{10\pi}{3}}\)
   • \(\frac{10\pi}{3} - 2\pi = \frac{4\pi}{3}\)
   • \(z^2 = 16e^{i\frac{4\pi}{3}} = 16\left(\cos\frac{4\pi}{3} + i\sin\frac{4\pi}{3}\right) = -8-8 \sqrt{3}i\)

A geometric sequence has first term \(u_1 = 16\) and common ratio \(r = -0.45 + 0.6i\).

1. (a) Find \(u_2\) and \(u_3\) in the form \(x + iy\).

   • \(u_2 = 16(-0.45 + 0.6i) = -7.2 + 9.6i\)
   • \(u_3 = (-7.2 + 9.6i)(-0.45 + 0.6i) = -2.52 - 8.64i\)

2. (b) Show that \(|u_1|, |u_2|, |u_3|, \dots\) is also a geometric sequence, and write down its common ratio.

   • \(|u_1| = 16,\ |u_2| = 12,\ |u_3| = 9\).
   • Each modulus is 0.75 times the one before, since \(|u_{n+1}| = |u_n||r|\) and \(|r| = \sqrt{(-0.45)^2 + 0.6^2} = 0.75\).

3. (c) Find the sum of the infinite sequence \(|u_1|, |u_2|, |u_3|, \dots\).

   • \(|r| < 1\), so \(S_\infty = \dfrac{16}{1 - 0.75} = 64\).

In a video animation, a point on a shrinking, turning logo starts at \(1\) on the complex plane, measured in centimetres from the centre of the screen. At each frame the point's position is multiplied by \(m = 0.8e^{i\frac{\pi}{9}}\). After \(n\) frames its position is \(m^n\).

1. (a) Find the position of the point after 3 frames in the form \(x + iy\). Give \(x\) and \(y\) to 3 significant figures.

   • \(m^3 = 0.8^3e^{i \times 3 \times \frac{\pi}{9}} = 0.512e^{i\frac{\pi}{3}}\)
   • \(x = 0.512\cos\frac{\pi}{3} = 0.256,\quad y = 0.512\sin\frac{\pi}{3} = 0.4434…\)
   • \(m^3 \approx 0.256 + 0.443i\)

2. (b) Find the distance of the point from the centre of the screen after 3 frames. Give your answer to 3 significant figures.

   • \(|m^3| = 0.8^3 = 0.512\)
   • The distance from the centre is 0.512 cm.

\(z = 1 - i\).

1. (a) Write \(z\) in exponential form, with \(0 \le \arg z < 2\pi\).

   • \(|z| = \sqrt{1 + 1} = \sqrt{2}. \text{ The point is in the fourth quadrant, so } \arg z = 2\pi - \frac{\pi}{4} = \frac{7\pi}{4}\)
   • \(z = \sqrt{2}e^{i\frac{7\pi}{4}}\)

2. (b) Find the smallest positive integer \(n\) for which \(z^n\) is real and negative, and write down \(z^n\).

   • \(z^n = (\sqrt{2})^ne^{i\frac{7n\pi}{4}}\)
   • \(\text{Real and negative: } \frac{7n\pi}{4} = \pi, 3\pi, 5\pi, 7\pi, \ldots. \text{ } n = 1, 2, 3 \text{ give } \frac{7\pi}{4}, \frac{7\pi}{2}, \frac{21\pi}{4}, \text{ and } n = 4 \text{ gives } 7\pi\)
   • \(z^4 = (\sqrt{2})^4e^{i7\pi} = 4 \times (-1) = -4\)