Cartesian form: mixed practice Problem set B Worked solutions

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

1. (a) \(3z^* - z = 2 + 8i\)

   • Put \(z = x + iy\): \(3x - 3iy - x - iy = 2 + 8i\), that is \(2x - 4iy = 2 + 8i\).
   • Real: \(2x = 2\). Imaginary: \(-4y = 8\).
   • \(x = 1,\ y = -2\), so \(z = 1 - 2i\)

2. (b) \(iz + 3z^* = -5 - 7i\)

   • \(i(x + iy) + 3x - 3iy = 3x - y + (x - 3y)i = -5 - 7i\)
   • Real: \(3x - y = -5\). Imaginary: \(x - 3y = -7\).
   • \(x = -1,\ y = 2\), so \(z = -1 + 2i\)

Find the complex number \(z\) such that \(4z - 3 = 2iz + 7 - 10i\).

• Collect the terms in \(z\): \(4z - 2iz = 3 + 7 - 10i\), so \((4 - 2i)z = 10 - 10i\).
• \(z = \dfrac{(10 - 10i)(4 + 2i)}{(4 - 2i)(4 + 2i)} = \dfrac{40 + 20i - 40i - 20i^2}{20} = \dfrac{60 - 20i}{20}\)
• \(z = 3 - i\)

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

1. (a) \(x^2 + 28 = 0\)

   • \(b^2 - 4ac = -112 < 0\), so the roots are complex.
   • From the calculator (complex mode): \(x = \pm 5.29i\)

2. (b) \(x^2 + 4x + 13 = 0\)

   • \(b^2 - 4ac = -36 < 0\), so the roots are complex.
   • From the calculator (complex mode): \(x = -2 \pm 3i\)

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

   • \(b^2 - 4ac = -31 < 0\), so the roots are complex.
   • From the calculator (complex mode): \(x = 0.75 \pm 1.39i\)

Solve \(4x^2 + 8x + 5 = 0\), and show the solutions on an Argand diagram.

• \(b^2 - 4ac = -16 < 0\)
• From the calculator (complex mode): \(x = -1 \pm \frac{1}{2}i\)
• Plot \((-1, \frac{1}{2})\) and \((-1, - \frac{1}{2})\), a conjugate pair.

Given that \(z = 3 - 7i + \dfrac{20}{2 - i}\), find \(z^*\).

• \(\dfrac{20}{2 - i} = \dfrac{20(2 + i)}{(2 - i)(2 + i)} = \dfrac{40 + 20i}{5} = 8 + 4i\)
• \(z = 3 - 7i + (8 + 4i) = 11 - 3i\)
• \(z^* = 11 + 3i\)

\(a\) is a real number, \(z = 4 + ai\) and \(w = 2a - i\). Write each of these in terms of \(a\), in the form \(x + iy\).

1. (a) \(z - 3w\)

   • \(z - 3w = 4 + ai - 6a + 3i\)
   • Real parts: \(4 - 6a\). Imaginary parts: \(a + 3\).
   • \(z - 3w = (4 - 6a) + (a + 3)i\)

2. (b) \(zw\)

   • \(zw = (4 + ai)(2a - i) = 8a - 4i + 2a^2 i - ai^2\)
   • \(i^2 = -1\), so \(-ai^2 = +a\).
   • \(zw = 9a + (2a^2 - 4)i\)

\(k\) is a real number. Write \(\dfrac{2k - 5i}{3k + i}\) in the form \(x + iy\).

• Multiply top and bottom by \(3k - i\).
• Top: \((2k - 5i)(3k - i) = 6k^2 - 2ki - 15ki + 5i^2 = (6k^2 - 5) - 17ki\)
• Bottom: \((3k + i)(3k - i) = 9k^2 + 1\)
• \(\dfrac{6k^2 - 5}{9k^2 + 1} - \dfrac{17k}{9k^2 + 1}\,i\)

\(a\) is a real number and \(z = \dfrac{a + 6i}{a - 4i}\). Find the values of \(a\) for which \(\text{Im}(z) = 1\).

• \(z = \dfrac{(a + 6i)(a + 4i)}{(a - 4i)(a + 4i)} = \dfrac{a^2 - 24 + 10ai}{a^2 + 16}\)
• \(\text{Im}(z) = \dfrac{10a}{a^2 + 16} = 1\), so \(a^2 - 10a + 16 = 0\)
• \(a = 2\) or \(a = 8\)

\(a\) and \(b\) are real numbers, \(z = 3 + ai\) and \(w = b + i\). Given that \(zw = -3 + 9i\), find the possible values of \(a\) and \(b\).

• \(zw = (3 + ai)(b + i) = 3b - a + (3 + ab)i = -3 + 9i\)
• Real: \(3b - a = -3\), so \(a = 3b + 3\). Imaginary: \(3 + ab = 9\), so \(ab = 6\).
• \(b(3b + 3) = 6\), so \(b^2 + b - 2 = 0\) and \(b = -2\) or \(b = 1\).
• \(b = -2: a = -3\). \(b = 1: a = 6\).

\(z = 3 - 2i\) and \(w = -1 + 4i\). Show \(z\), \(z^*\), \(2z\), \(iz\), \(w\), \(z + w\) and \(z - w\) on one Argand diagram, and write down the coordinates of each point.

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

Let \(z = 0.8 - 0.6i\). 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 \(z^3 = -0.352 - 0.936i\).
• Plot \((0.8, -0.6)\), \((0.28, -0.96)\) and \((-0.352, -0.936)\).
• \(|z| = \sqrt{0.8^2 + (-0.6)^2} = 1\), 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 \(z^2 + zz^* = 32 + 24i\).

• \(z^2 + zz^* = (x^2 - y^2 + 2ixy) + (x^2 + y^2) = 2x^2 + 2ixy\)
• Real: \(2x^2 = 32\), so \(x = \pm 4\). Imaginary: \(2xy = 24\), so \(xy = 12\).
• \(x = 4: y = 3\). \(x = -4: y = -3\).
• \(z = 4 + 3i\) or \(z = -4 - 3i\)

By writing \(z = x + iy\), show that \((z^2 + 2z)^* = (z^*)^2 + 2z^*\).

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

Solve the simultaneous equations \(3z - iw = 8 - 4i\) and \(z + 2iw = -2 + i\) for the complex numbers \(z\) and \(w\).

• Multiply the first equation by 2: \(6z - 2iw = 16 - 8i\). Add the second: \(7z = 14 - 7i\).
• \(z = 2 - i\)
• Put \(z\) in the second equation: \(2iw = -2 + i - (2 - i) = -4 + 2i\), so \(iw = -2 + i\).
• \(w = \dfrac{-2 + i}{i} = (-2 + i)(-i) = 1 + 2i\)