Products, quotients and powers in cis form Problem set A Worked solutions

Give each argument in the range \(0 \le \theta < 2\pi\).

1. (a) \(z = 2\,\text{cis}\left(\frac{\pi}{7}\right)\) and \(w = 6\,\text{cis}\left(\frac{2\pi}{7}\right)\). Find \(zw\).

   • \(|zw| = 2 \times 6 = 12\)
   • \(\arg(zw) = \frac{\pi}{7} + \frac{2\pi}{7} = \frac{3\pi}{7}\)
   • \(zw = 12\,\text{cis}\left(\frac{3\pi}{7}\right)\)

2. (b) \(z = 5\,\text{cis}\left(\frac{3\pi}{4}\right)\) and \(w = 3\,\text{cis}\left(\frac{\pi}{2}\right)\). Find \(zw\).

   • \(|zw| = 5 \times 3 = 15\)
   • \(\arg(zw) = \frac{3\pi}{4} + \frac{\pi}{2} = \frac{5\pi}{4}\)
   • \(zw = 15\,\text{cis}\left(\frac{5\pi}{4}\right)\)

3. (c) \(z = 4\,\text{cis}\left(\frac{5\pi}{3}\right)\) and \(w = 2.5\,\text{cis}\left(\frac{2\pi}{3}\right)\). Find \(zw\).

   • \(|zw| = 4 \times 2.5 = 10\)
   • \(\arg(zw) = \frac{5\pi}{3} + \frac{2\pi}{3} = \frac{7\pi}{3}\)
   • \(\frac{7\pi}{3} - 2\pi = \frac{\pi}{3}\)
   • \(zw = 10\,\text{cis}\left(\frac{\pi}{3}\right)\)

Give each answer in the form \(r\,\text{cis}\,\theta\) with \(0 \le \theta < 2\pi\).

1. (a) \(z = 12\,\text{cis}\left(\frac{5\pi}{6}\right)\) and \(w = 3\,\text{cis}\left(\frac{\pi}{3}\right)\). Find \(\dfrac{z}{w}\).

   • \(\left|\frac{z}{w}\right| = \frac{12}{3} = 4\)
   • \(\arg\left(\frac{z}{w}\right) = \frac{5\pi}{6} - \frac{\pi}{3} = \frac{\pi}{2}\)
   • \(\frac{z}{w} = 4\,\text{cis}\left(\frac{\pi}{2}\right)\)

2. (b) \(z = 7\,\text{cis}\left(\frac{\pi}{4}\right)\) and \(w = 2\,\text{cis}\left(\frac{2\pi}{3}\right)\). Find \(\dfrac{z}{w}\).

   • \(\left|\frac{z}{w}\right| = \frac{7}{2} = 3.5\)
   • \(\arg\left(\frac{z}{w}\right) = \frac{\pi}{4} - \frac{2\pi}{3} = -\frac{5\pi}{12}\)
   • \(-\frac{5\pi}{12} + 2\pi = \frac{19\pi}{12}\)
   • \(\frac{z}{w} = 3.5\,\text{cis}\left(\frac{19\pi}{12}\right)\)

Give each answer in the form \(r\,\text{cis}\,\theta\) with \(0 \le \theta < 2\pi\).

1. (a) \(z = 2\,\text{cis}\left(\frac{\pi}{9}\right)\). Find \(z^{3}\).

   • \(|z^3| = 2^3 = 8\)
   • \(\arg(z^3) = 3 \times \frac{\pi}{9} = \frac{\pi}{3}\)
   • \(z^3 = 8\,\text{cis}\left(\frac{\pi}{3}\right)\)

2. (b) \(z = 3\,\text{cis}\left(\frac{5\pi}{6}\right)\). Find \(z^{4}\).

   • \(|z^4| = 3^4 = 81\)
   • \(\arg(z^4) = 4 \times \frac{5\pi}{6} = \frac{10\pi}{3}\)
   • \(\frac{10\pi}{3} - 2\pi = \frac{4\pi}{3}\)
   • \(z^4 = 81\,\text{cis}\left(\frac{4\pi}{3}\right)\)

Work by hand. Give exact values.

1. (a) \(z = 9\,\text{cis}\left(\frac{\pi}{3}\right)\) and \(w = 3\,\text{cis}\left(\frac{\pi}{6}\right)\). Write \(\dfrac{z^2}{w^3}\) in the form \(x + iy\), giving exact values.

   • \(|z^2| = 9^2 = 81,\quad \arg(z^2) = 2 \times \frac{\pi}{3} = \frac{2\pi}{3}\)
   • \(|w^3| = 3^3 = 27,\quad \arg(w^3) = 3 \times \frac{\pi}{6} = \frac{\pi}{2}\)
   • \(\left|\frac{z^2}{w^3}\right| = \frac{81}{27} = 3,\quad \arg = \frac{2\pi}{3} - \frac{\pi}{2} = \frac{\pi}{6}\)
   • \(\frac{z^2}{w^3} = 3\,\text{cis}\left(\frac{\pi}{6}\right) = \frac{3 \sqrt{3}}{2}+\frac{3}{2}i\)

2. (b) \(z = 2\,\text{cis}\left(\frac{\pi}{4}\right)\) and \(w = 2\,\text{cis}\left(\frac{\pi}{8}\right)\). Write \(\dfrac{z^3}{w^2}\) in the form \(x + iy\), giving exact values.

   • \(|z^3| = 2^3 = 8,\quad \arg(z^3) = 3 \times \frac{\pi}{4} = \frac{3\pi}{4}\)
   • \(|w^2| = 2^2 = 4,\quad \arg(w^2) = 2 \times \frac{\pi}{8} = \frac{\pi}{4}\)
   • \(\left|\frac{z^3}{w^2}\right| = \frac{8}{4} = 2,\quad \arg = \frac{3\pi}{4} - \frac{\pi}{4} = \frac{\pi}{2}\)
   • \(\frac{z^3}{w^2} = 2\,\text{cis}\left(\frac{\pi}{2}\right) = 2i\)

An alternating voltage is stored as the complex number \(V = 12\,\text{cis}\left(\frac{\pi}{6}\right)\), in volts, with the phase in radians. A filter multiplies the signal by \(F = 0.5\,\text{cis}\left(-\frac{\pi}{4}\right)\). Give each phase in the range \(0 \le \theta < 2\pi\).

1. (a) Write the signal that leaves one filter, \(FV\), in the form \(r\,\text{cis}\,\theta\).

   • \(|FV| = 0.5 \times 12 = 6\)
   • \(\arg(FV) = -\frac{\pi}{4} + \frac{\pi}{6} = -\frac{\pi}{12}\)
   • \(-\frac{\pi}{12} + 2\pi = \frac{23\pi}{12}\)
   • \(FV = 6\,\text{cis}\left(\frac{23\pi}{12}\right)\)

2. (b) The signal passes through two identical filters in a row. Write the output \(VF^2\) in the form \(r\,\text{cis}\,\theta\) and state its amplitude and phase.

   • \(|F^2| = 0.5^2 = 0.25,\quad \arg(F^2) = 2 \times \left(-\frac{\pi}{4}\right) = -\frac{\pi}{2}\)
   • \(|VF^2| = 0.25 \times 12 = 3\)
   • \(\arg = \frac{\pi}{6} + \left(-\frac{\pi}{2}\right) = -\frac{\pi}{3}\)
   • \(-\frac{\pi}{3} + 2\pi = \frac{5\pi}{3}\)
   • \(VF^2 = 3\,\text{cis}\left(\frac{5\pi}{3}\right)\)

Answer each part.

1. (a) \(z = 2\,\text{cis}\left(\frac{\pi}{8}\right)\). Find the smallest positive integer \(n\) for which \(z^n\) is real and negative, and write down \(z^n\).

   • \(\arg(z^n) = \frac{n\pi}{8}\)
   • Real and negative: \(\arg(z^n) = \pi\), \(3\pi\), … so \(\frac{n\pi}{8} = \pi\) and \(n = 8\)
   • \(|z^8| = 2^8 = 256, \text{ so } z^8 = 256\,\text{cis}\,\pi = -256\)

2. (b) \(zw = 20\,\text{cis}\left(\frac{5\pi}{6}\right)\) and \(\dfrac{z}{w} = 5\,\text{cis}\left(\frac{\pi}{6}\right)\). Find \(z\) and \(w\) in the form \(r\,\text{cis}\,\theta\).

   • \(|zw|\times\left|\frac{z}{w}\right| = |z|^2 = 20 \times 5 = 100, \text{ so } |z| = 10 \text{ and } |w| = 20 \div 10 = 2\)
   • \(\arg z + \arg w = \frac{5\pi}{6}, \quad \arg z - \arg w = \frac{\pi}{6}\)
   • \(\text{Add: } 2\arg z = \pi, \text{ so } \arg z = \frac{\pi}{2}. \text{ Then } \arg w = \frac{5\pi}{6} - \frac{\pi}{2} = \frac{\pi}{3}\)