Conjugates, division and equating parts Problem set A Worked solutions

Answer each part.

1. (a) Write down the conjugate of \(z = 4 - 9i\).

   • \(z^* = 4 + 9i\) (change the sign of the imaginary part)

2. (b) \(z = 6 + 8i\). Find \(zz^*\).

   • \(zz^* = 6^2 + 8^2 = 100\)

3. (c) \(z = -7 + 2i\). Find \(z^*\) and \(zz^*\).

   • \(z^* = -7 - 2i\)
   • \(zz^* = (-7)^2 + 2^2 = 53\)

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

1. (a) Show that \(z + z^*\) is real.

   • \(z + z^* = (x + iy) + (x - iy) = 2x\)
   • 2x is real.

2. (b) Show that \(z^2 + (z^*)^2\) is real.

   • \(z^2 = x^2 + 2ixy - y^2,\quad (z^*)^2 = x^2 - 2ixy - y^2\)
   • \(z^2 + (z^*)^2 = 2x^2 - 2y^2 = 2(x^2 - y^2)\)
   • This is real.

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

1. (a) \(\dfrac{14 + 2i}{3 - i}\)

   • \(\dfrac{14 + 2i}{3 - i} = \dfrac{(14 + 2i)(3 + i)}{(3 - i)(3 + i)}\)
   • \(= \dfrac{42 + 14i + 6i + 2i^2}{3^2 + (-1)^2} = \dfrac{40 + 20i}{10}\)
   • \(= 4 + 2i\)

2. (b) \(\dfrac{4 + 7i}{1 + 2i}\)

   • \(\dfrac{4 + 7i}{1 + 2i} = \dfrac{(4 + 7i)(1 - 2i)}{(1 + 2i)(1 - 2i)}\)
   • \(= \dfrac{4 - 8i + 7i - 14i^2}{1^2 + 2^2} = \dfrac{18 - i}{5}\)
   • \(= \frac{18}{5} - \frac{1}{5}i\)

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

   • \(\dfrac{1}{2 - 3i} = \dfrac{(1)(2 + 3i)}{(2 - 3i)(2 + 3i)}\)
   • \(= \dfrac{2 + 3i}{2^2 + (-3)^2} = \dfrac{2 + 3i}{13}\)
   • \(= \frac{2}{13} + \frac{3}{13}i\)

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

1. (a) \(z + 2z^* = 9 - 4i\)

   • \(z = x + iy,\ z^* = x - iy\)
   • \((x + iy) + 2(x - iy) = 3x - iy\)
   • \(3x = 9,\quad -y = -4\)
   • \(x = 3,\ y = 4,\ z = 3 + 4i\)

2. (b) \(3z - z^* = -6 + 8i\)

   • \(z = x + iy,\ z^* = x - iy\)
   • \(3(x + iy) - (x - iy) = 2x + 4iy\)
   • \(2x = -6,\quad 4y = 8\)
   • \(x = -3,\ y = 2,\ z = -3 + 2i\)

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

1. (a) \((a + bi)(1 + 2i) = 7 + 4i\)

   • \((a + bi)(1 + 2i) = (a - 2b) + (2a + b)i\)
   • \(a - 2b = 7,\quad 2a + b = 4\)
   • From the calculator (simultaneous equations): \(a = 3,\ b = -2\)

2. (b) \((a + bi)(2 - i) = 5 + 5i\)

   • \((a + bi)(2 - i) = (2a + b) + (-a + 2b)i\)
   • \(2a + b = 5,\quad -a + 2b = 5\)
   • From the calculator (simultaneous equations): \(a = 1,\ b = 3\)

In an alternating current circuit, the voltage is \(V = 30 + 10i\) volts and the impedance is \(Z = 4 + 2i\) ohms. The current is \(I = \dfrac{V}{Z}\).

1. (a) Find the current \(I\) in the form \(x + iy\).

   • \(\dfrac{30 + 10i}{4 + 2i} = \dfrac{(30 + 10i)(4 - 2i)}{(4 + 2i)(4 - 2i)}\)
   • \(= \dfrac{120 - 60i + 40i - 20i^2}{4^2 + 2^2} = \dfrac{140 - 20i}{20}\)
   • \(= 7 - i\)

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

   • \(ZZ^* = (4 + 2i)(4 - 2i) = 4^2 + 2^2 = 20\)
   • A number times its conjugate is real.

Find the complex number \(z\) such that \(z(2 + i) + z^* = 5 + 3i\), where \(z^*\) is the conjugate of \(z\).

• \(z(2 + i) + z^* = (x + iy)(2 + i) + (x - iy)\)
• \(= (3x - y) + (x + y)i\)
• \(3x - y = 5,\quad x + y = 3\)
• From the calculator (simultaneous equations): \(x = 2,\ y = 1\)
• \(z = 2 + i\)

\(z = 1 + 2i\).

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

   • \(zz^* = 1^2 + 2^2 = 5\)
   • \(z + z^* = 2(1) = 2\)

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

   • \(\dfrac{1}{z} + \dfrac{1}{z^*} = \dfrac{z^* + z}{zz^*}\)
   • \(= \dfrac{2}{5}\)
   • The result is a real number.