Cartesian form: mixed practice Problem set A One question at a time

Find the complex number \(z\) in each case.

1. (a) \(z^* - 2z = 3 - 6i\)

   • Put \(z = x + iy\): \(x - iy - 2x - 2iy = 3 - 6i\), that is \(-x - 3iy = 3 - 6i\).
   • Real: \(-x = 3\). Imaginary: \(-3y = -6\).
   • \(x = -3,\ y = 2\), so \(z = -3 + 2i\)

2. (b) \(2iz + z^* = 4 + 5i\)

   • \(2i(x + iy) + x - iy = x - 2y + (2x - y)i = 4 + 5i\)
   • Real: \(x - 2y = 4\). Imaginary: \(2x - y = 5\).
   • \(x = 2,\ y = -1\), so \(z = 2 - i\)

Find the complex number \(z\) such that \(5z - 2 = 3iz - 1 - 21i\).

• Collect the terms in \(z\): \(5z - 3iz = 2 - 1 - 21i\), so \((5 - 3i)z = 1 - 21i\).
• \(z = \dfrac{(1 - 21i)(5 + 3i)}{(5 - 3i)(5 + 3i)} = \dfrac{5 + 3i - 105i - 63i^2}{34} = \dfrac{68 - 102i}{34}\)
• \(z = 2 - 3i\)

Solve each equation. Give exact values where you can and otherwise 3 significant figures.

1. (a) \(x^2 + 20 = 0\)

   • \(b^2 - 4ac = -80 < 0\), so the roots are complex.
   • From the calculator (complex mode): \(x = \pm 4.47i\)

2. (b) \(x^2 - 6x + 10 = 0\)

   • \(b^2 - 4ac = -4 < 0\), so the roots are complex.
   • From the calculator (complex mode): \(x = 3 \pm i\)

3. (c) \(3x^2 + 4x + 5 = 0\)

   • \(b^2 - 4ac = -44 < 0\), so the roots are complex.
   • From the calculator (complex mode): \(x = -0.667 \pm 1.11i\)

Solve \(4x^2 - 12x + 13 = 0\), and show the solutions on an Argand diagram.

• \(b^2 - 4ac = -64 < 0\)
• From the calculator (complex mode): \(x = \frac{3}{2} \pm i\)
• Plot \((\frac{3}{2}, 1)\) and \((\frac{3}{2}, -1)\), a conjugate pair.

Given that \(z = 2 + 5i - \dfrac{12}{1 + i}\), find \(z^*\).

• \(\dfrac{12}{1 + i} = \dfrac{12(1 - i)}{(1 + i)(1 - i)} = \dfrac{12 - 12i}{2} = 6 - 6i\)
• \(z = 2 + 5i - (6 - 6i) = -4 + 11i\)
• \(z^* = -4 - 11i\)

\(a\) is a real number, \(z = 3a - 2i\) and \(w = 1 + ai\). Write each of these in terms of \(a\), in the form \(x + iy\).

1. (a) \(z + 2w\)

   • \(z + 2w = 3a - 2i + 2 + 2ai\)
   • Real parts: \(3a + 2\). Imaginary parts: \(-2 + 2a\).
   • \(z + 2w = (3a + 2) + (2a - 2)i\)

2. (b) \(zw\)

   • \(zw = (3a - 2i)(1 + ai) = 3a + 3a^2 i - 2i - 2ai^2\)
   • \(i^2 = -1\), so \(-2ai^2 = +2a\).
   • \(zw = 5a + (3a^2 - 2)i\)

\(k\) is a real number. Write \(\dfrac{k + 4i}{2k - i}\) in the form \(x + iy\).

• Multiply top and bottom by \(2k + i\).
• Top: \((k + 4i)(2k + i) = 2k^2 + ki + 8ki + 4i^2 = (2k^2 - 4) + 9ki\)
• Bottom: \((2k - i)(2k + i) = 4k^2 + 1\)
• \(\dfrac{2k^2 - 4}{4k^2 + 1} + \dfrac{9k}{4k^2 + 1}\,i\)

\(a\) is a real number and \(z = \dfrac{a + 3i}{a - 2i}\). Find the values of \(a\) for which \(\text{Im}(z) = 1\).

• \(z = \dfrac{(a + 3i)(a + 2i)}{(a - 2i)(a + 2i)} = \dfrac{a^2 - 6 + 5ai}{a^2 + 4}\)
• \(\text{Im}(z) = \dfrac{5a}{a^2 + 4} = 1\), so \(a^2 - 5a + 4 = 0\)
• \(a = 1\) or \(a = 4\)

\(a\) and \(b\) are real numbers, \(z = 2 + ai\) and \(w = b + i\). Given that \(zw = -6 - 2i\), find the possible values of \(a\) and \(b\).

• \(zw = (2 + ai)(b + i) = 2b - a + (2 + ab)i = -6 - 2i\)
• Real: \(2b - a = -6\), so \(a = 2b + 6\). Imaginary: \(2 + ab = -2\), so \(ab = -4\).
• \(b(2b + 6) = -4\), so \(b^2 + 3b + 2 = 0\) and \(b = -2\) or \(b = -1\).
• \(b = -2: a = 2\). \(b = -1: a = 4\).

\(z = -2 + 3i\) and \(w = 4 + i\). Show \(z\), \(z^*\), \(2z\), \(iz\), \(w\), \(z + w\) and \(z - w\) on one Argand diagram, and write down the coordinates of each point.

• \(z^* = -2 - 3i,\ 2z = -4 + 6i,\ iz = i(-2 + 3i) = -3 - 2i\)
• \(z + w = 2 + 4i,\ z - w = -6 + 2i\)
• Plot: \(z^*\) at \((-2, -3)\), \(2z\) at \((-4, 6)\), \(iz\) at \((-3, -2)\), \(w\) at \((4, 1)\), \(z + w\) at \((2, 4)\), \(z - w\) at \((-6, 2)\).

Let \(z = 0.6 + 0.8i\). Use technology to find \(z^2\) and \(z^3\), and show \(z\), \(z^2\) and \(z^3\) on one Argand diagram. What do you notice about the distance of the three points from the origin?

• From the calculator (complex mode): \(z^2 = -0.28 + 0.96i\) and \(z^3 = -0.936 + 0.352i\).
• Plot \((0.6, 0.8)\), \((-0.28, 0.96)\) and \((-0.936, 0.352)\).
• \(|z| = \sqrt{0.6^2 + 0.8^2} = 1\), and the same holds for \(z^2\) and \(z^3\): the points lie on a circle of radius 1.

Find the two complex numbers \(z\) such that \(z^2 + zz^* = 18 - 12i\).

• \(z^2 + zz^* = (x^2 - y^2 + 2ixy) + (x^2 + y^2) = 2x^2 + 2ixy\)
• Real: \(2x^2 = 18\), so \(x = \pm 3\). Imaginary: \(2xy = -12\), so \(xy = -6\).
• \(x = 3: y = -2\). \(x = -3: y = 2\).
• \(z = 3 - 2i\) or \(z = -3 + 2i\)

By writing \(z = x + iy\), show that \((z^*)^3 = (z^3)^*\).

• \(z^3 = (x + iy)^3 = x^3 + 3x^2(iy) + 3x(iy)^2 + (iy)^3 = x^3 - 3xy^2 + (3x^2 y - y^3)i\)
• \((z^3)^* = x^3 - 3xy^2 - (3x^2 y - y^3)i\)
• \((z^*)^3 = (x - iy)^3 = x^3 - 3x^2(iy) + 3x(iy)^2 - (iy)^3 = x^3 - 3xy^2 - (3x^2 y - y^3)i\)
• The two expressions are the same, so \((z^*)^3 = (z^3)^*\).

Solve the simultaneous equations \(2z + iw = 3 + 7i\) and \(z - 2iw = -1 - 4i\) for the complex numbers \(z\) and \(w\).

• Multiply the first equation by 2: \(4z + 2iw = 6 + 14i\). Add the second: \(5z = 5 + 10i\).
• \(z = 1 + 2i\)
• Put \(z\) in the first equation: \(iw = 3 + 7i - 2(1 + 2i) = 1 + 3i\).
• \(w = \dfrac{1 + 3i}{i} = \dfrac{(1 + 3i)(-i)}{1} = 3 - i\)