Modulus-argument form Problem set B One question at a time

Write each number in modulus-argument form \(r\,\text{cis}\,\theta\), with \(0 \le \theta < 2\pi\). Sketch each point first and give exact values.

1. (a) \(3 \sqrt{3} + 3i\)

   • \(r = \sqrt{(3 \sqrt{3})^2 + (3)^2} = 6\)
   • \(\tan\alpha = \dfrac{3}{3 \sqrt{3}}\), so \(\alpha = \dfrac{\pi}{6}\). The point is in the first quadrant.
   • \(\theta = \alpha = \dfrac{\pi}{6}\)

2. (b) \(- 2 \sqrt{3} + 2i\)

   • \(r = \sqrt{(- 2 \sqrt{3})^2 + (2)^2} = 4\)
   • \(\tan\alpha = \dfrac{2}{2 \sqrt{3}}\), so \(\alpha = \dfrac{\pi}{6}\). The point is in the second quadrant.
   • \(\theta = \pi - \alpha = \dfrac{5\pi}{6}\)

3. (c) \(-3 - 3i\)

   • \(r = \sqrt{(-3)^2 + (-3)^2} = 3 \sqrt{2}\)
   • \(\tan\alpha = \dfrac{3}{3}\), so \(\alpha = \dfrac{\pi}{4}\). The point is in the third quadrant.
   • \(\theta = \pi + \alpha = \dfrac{5\pi}{4}\)

4. (d) \(2 - 2 \sqrt{3}i\)

   • \(r = \sqrt{(2)^2 + (- 2 \sqrt{3})^2} = 4\)
   • \(\tan\alpha = \dfrac{2 \sqrt{3}}{2}\), so \(\alpha = \dfrac{\pi}{3}\). The point is in the fourth quadrant.
   • \(\theta = 2\pi - \alpha = \dfrac{5\pi}{3}\)

Write down \(r\) and \(\theta\) for each number, to 3 significant figures, with \(-\pi < \theta \le \pi\). Use the calculator in radian mode.

1. (a) \(15 - 8i\)

   • \(r = \sqrt{(15)^2 + (-8)^2} = 17\)
   • \(\alpha = \tan^{-1}\left(\dfrac{8}{15}\right) = 0.490\). The point is in the fourth quadrant.
   • \(\theta = -\alpha = -0.490\)

2. (b) \(-5 + 3i\)

   • \(r = \sqrt{(-5)^2 + (3)^2} = 5.83\)
   • \(\alpha = \tan^{-1}\left(\dfrac{3}{5}\right) = 0.540\). The point is in the second quadrant.
   • \(\theta = \pi - \alpha = 2.60\)

Write each complex number in the form \(x + iy\). Give exact values for the first two and 3 significant figures for the third.

1. (a) \(10\,\text{cis}\,\dfrac{\pi}{6}\)

   • \(x = 10\cos \dfrac{\pi}{6} = 5 \sqrt{3}\)
   • \(y = 10\sin \dfrac{\pi}{6} = 5\)
   • \(z = 5 \sqrt{3} + 5i\)

2. (b) \(4\,\text{cis}\,\dfrac{3\pi}{4}\)

   • \(x = 4\cos \dfrac{3\pi}{4} = - 2 \sqrt{2}\)
   • \(y = 4\sin \dfrac{3\pi}{4} = 2 \sqrt{2}\)
   • \(z = - 2 \sqrt{2} + 2 \sqrt{2}i\)

3. (c) \(9\,\text{cis}(-1.10)\)

   • From the calculator (radian mode): \(x = 9\cos(-1.10) = 4.08\)
   • \(y = 9\sin(-1.10) = -8.02\)
   • \(z = 4.08 - 8.02i\)

Write the conjugate of each number in the form \(r\,\text{cis}\,\theta\) with \(-\pi < \theta \le \pi\). Sketch \(z\) and \(z^*\) for the first.

1. (a) \(9\,\text{cis}\left(\dfrac{\pi}{8}\right)\)

   • \(z^* = 9\,\text{cis}\left(-\dfrac{\pi}{8}\right)\)
   • The argument is already in \(-\pi < \theta \le \pi\).

2. (b) \(2\,\text{cis}\left(-\dfrac{3\pi}{4}\right)\)

   • \(z^* = 2\,\text{cis}\left(\dfrac{3\pi}{4}\right)\)
   • The argument is already in \(-\pi < \theta \le \pi\).

3. (c) \(6\,\text{cis}\left(\dfrac{5\pi}{4}\right)\)

   • \(z^* = 6\,\text{cis}\left(-\dfrac{5\pi}{4}\right)\)
   • \(-\dfrac{5\pi}{4} + 2\pi = \dfrac{3\pi}{4}\), so \(z^* = 6\,\text{cis}\left(\dfrac{3\pi}{4}\right)\) with \(-\pi < \theta \le \pi\).

Take the pilot as the origin of an Argand diagram, with east as the real axis and north as the imaginary axis. Distances are in metres. A drone is 50 m from the pilot on a bearing of 135 degrees.

1. (a) Find the argument of the drone's position, in radians and measured anticlockwise from east, with \(-\pi < \theta \le \pi\).

   • A bearing is measured clockwise from north. The argument is measured anticlockwise from east.
   • \(90^\circ - 135^\circ = -45^\circ = -\dfrac{\pi}{4}\) radians

2. (b) Write the drone's position as a complex number in the form \(x + iy\), to 3 significant figures.

   • \(x = 50\cos\left(-\dfrac{\pi}{4}\right) = 35.4\)
   • \(y = 50\sin\left(-\dfrac{\pi}{4}\right) = -35.4\)
   • The drone is at \(35.4 - 35.4i\).

3. (c) A tree is at \(-8 + 6i\). Find its distance from the pilot and its argument, to 3 significant figures, with \(-\pi < \theta \le \pi\).

   • \(r = \sqrt{(-8)^2 + 6^2} = 10\)
   • The point is in the second quadrant: \(\theta = \pi - \tan^{-1}\left(\dfrac{6}{8}\right) = 2.50\)

A student writes \(z = -5\,\text{cis}\,\dfrac{\pi}{4}\). The modulus of a complex number cannot be negative. Write \(z\) in the form \(r\,\text{cis}\,\theta\) with \(r > 0\) and \(0 \le \theta < 2\pi\).

• \(-1\) times a number reflects it through the origin: the modulus stays 5 and the angle turns through \(\pi\).
• \(\dfrac{\pi}{4} + \pi = \dfrac{5\pi}{4}\)
• \(z = 5\,\text{cis}\,\dfrac{5\pi}{4}\)

Write \(z = 4\left(\cos\dfrac{2\pi}{7} - i\sin\dfrac{2\pi}{7}\right)\) in the form \(r\,\text{cis}\,\theta\).

• \(\cos(-\theta) = \cos\theta\) and \(\sin(-\theta) = -\sin\theta\).
• So \(\cos\dfrac{2\pi}{7} - i\sin\dfrac{2\pi}{7} = \cos\left(-\dfrac{2\pi}{7}\right) + i\sin\left(-\dfrac{2\pi}{7}\right)\).
• \(z = 4\,\text{cis}\left(-\dfrac{2\pi}{7}\right)\). This is the conjugate of \(4\,\text{cis}\,\dfrac{2\pi}{7}\).

The complex number \(z\) has modulus 10 and argument \(\dfrac{5\pi}{6}\).

1. (a) Write \(z\) in the form \(x + iy\).

   • \(x = 10\cos\dfrac{5\pi}{6} = -5\sqrt{3}\), \(y = 10\sin\dfrac{5\pi}{6} = 5\)
   • \(z = -5\sqrt{3} + 5i\)

2. (b) Write \(z^*\) in the form \(x + iy\) and in the form \(r\,\text{cis}\,\theta\) with \(-\pi < \theta \le \pi\).

   • Change the sign of the imaginary part: \(z^* = -5\sqrt{3} - 5i\).
   • In polar form the argument changes sign: \(z^* = 10\,\text{cis}\left(-\dfrac{5\pi}{6}\right)\).

3. (c) Find the distance between the points that represent \(z\) and \(z^*\).

   • The points are the same distance above and below the real axis.
   • Distance \(= 2 \times 5 = 10\)