Conjugates, division and equating parts Problem set B Worked solutions

Answer each part.

1. (a) Write down the conjugate of \(z = -3 + 8i\).

   • \(z^* = -3 - 8i\) (change the sign of the imaginary part)

2. (b) \(z = 9 - 12i\). Find \(zz^*\).

   • \(zz^* = 9^2 + (-12)^2 = 225\)

3. (c) \(z = 5 - 6i\). Find \(z^*\) and \(zz^*\).

   • \(z^* = 5 + 6i\)
   • \(zz^* = 5^2 + (-6)^2 = 61\)

Let \(z = x + iy\), where \(x\) and \(y\) are real numbers.

1. (a) Show that \(z - z^*\) is purely imaginary.

   • \(z - z^* = (x + iy) - (x - iy) = 2iy\)
   • 2iy has no real part.

2. (b) Show that \(z^2 - (z^*)^2\) is purely imaginary.

   • \(z^2 = x^2 + 2ixy - y^2,\quad (z^*)^2 = x^2 - 2ixy - y^2\)
   • \(z^2 - (z^*)^2 = 4ixy\)
   • This has no real part.

Write each quotient in the form \(x + iy\). Give exact fractions where needed.

1. (a) \(\dfrac{11 - 10i}{4 + i}\)

   • \(\dfrac{11 - 10i}{4 + i} = \dfrac{(11 - 10i)(4 - i)}{(4 + i)(4 - i)}\)
   • \(= \dfrac{44 - 11i - 40i + 10i^2}{4^2 + 1^2} = \dfrac{34 - 51i}{17}\)
   • \(= 2 - 3i\)

2. (b) \(\dfrac{5 - 2i}{3 + 2i}\)

   • \(\dfrac{5 - 2i}{3 + 2i} = \dfrac{(5 - 2i)(3 - 2i)}{(3 + 2i)(3 - 2i)}\)
   • \(= \dfrac{15 - 10i - 6i + 4i^2}{3^2 + 2^2} = \dfrac{11 - 16i}{13}\)
   • \(= \frac{11}{13} - \frac{16}{13}i\)

3. (c) \(\dfrac{2}{1 + 4i}\)

   • \(\dfrac{2}{1 + 4i} = \dfrac{(2)(1 - 4i)}{(1 + 4i)(1 - 4i)}\)
   • \(= \dfrac{2 - 8i}{1^2 + 4^2} = \dfrac{2 - 8i}{17}\)
   • \(= \frac{2}{17} - \frac{8}{17}i\)

Find the complex number \(z\) that satisfies each equation.

1. (a) \(2z + z^* = 12 - 5i\)

   • \(z = x + iy,\ z^* = x - iy\)
   • \(2(x + iy) + (x - iy) = 3x + iy\)
   • \(3x = 12,\quad y = -5\)
   • \(x = 4,\ y = -5,\ z = 4 - 5i\)

2. (b) \(z - 2z^* = -6 + 9i\)

   • \(z = x + iy,\ z^* = x - iy\)
   • \((x + iy) - 2(x - iy) = -x + 3iy\)
   • \(-x = -6,\quad 3y = 9\)
   • \(x = 6,\ y = 3,\ z = 6 + 3i\)

\(a\) and \(b\) are real numbers. Find \(a\) and \(b\) in each equation.

1. (a) \((a + bi)(2 + i) = -6 + 7i\)

   • \((a + bi)(2 + i) = (2a - b) + (a + 2b)i\)
   • \(2a - b = -6,\quad a + 2b = 7\)
   • From the calculator (simultaneous equations): \(a = -1,\ b = 4\)

2. (b) \((a + bi)(3 - i) = 7 + i\)

   • \((a + bi)(3 - i) = (3a + b) + (-a + 3b)i\)
   • \(3a + b = 7,\quad -a + 3b = 1\)
   • From the calculator (simultaneous equations): \(a = 2,\ b = 1\)

In a control system the output signal is \(y = 5 + 10i\) and the input signal is \(u = 2 - i\), both in volts. The system ratio is \(R = \dfrac{y}{u}\).

1. (a) Find the ratio \(R\) in the form \(x + iy\).

   • \(\dfrac{5 + 10i}{2 - i} = \dfrac{(5 + 10i)(2 + i)}{(2 - i)(2 + i)}\)
   • \(= \dfrac{10 + 5i + 20i + 10i^2}{2^2 + (-1)^2} = \dfrac{25i}{5}\)
   • \(= 5i\)

2. (b) Show that \(uu^*\) is real and find its value.

   • \(uu^* = (2 - i)(2 + i) = 2^2 + 1^2 = 5\)
   • A number times its conjugate is real.

Find the complex number \(z\) such that \(z(1 - 2i) + z^* = 4 - 6i\), where \(z^*\) is the conjugate of \(z\).

• \(z(1 - 2i) + z^* = (x + iy)(1 - 2i) + (x - iy)\)
• \(= (2x + 2y) + (-2x)i\)
• \(2x + 2y = 4,\quad -2x = -6\)
• From the calculator (simultaneous equations): \(x = 3,\ y = -1\)
• \(z = 3 - i\)

\(z = 3 - i\).

1. (a) Find \(zz^*\) and \(z + z^*\).

   • \(zz^* = 3^2 + (-1)^2 = 10\)
   • \(z + z^* = 2(3) = 6\)

2. (b) Hence show that \(\dfrac{1}{z} + \dfrac{1}{z^*}\) is real, and find its value.

   • \(\dfrac{1}{z} + \dfrac{1}{z^*} = \dfrac{z^* + z}{zz^*}\)
   • \(= \dfrac{6}{10} = \frac{3}{5}\)
   • The result is a real number.