Cartesian form: mixed practice Problem set B Answers

1. (a) \(z = 1 - 2i\)
2. (b) \(z = -1 + 2i\)

\(z = 3 - i\)

1. (a) \(x = \pm 2 \sqrt{7}i\), which is \(\pm 5.29i\) to 3 s.f.
2. (b) \(x = -2 \pm 3i\)
3. (c) \(x = \frac{3}{4} \pm \frac{\sqrt{31}}{4}i\), which is \(0.75 \pm 1.39i\) to 3 s.f.

\(x = -1 \pm \frac{1}{2}i\); points \((-1, \frac{1}{2})\) and \((-1, - \frac{1}{2})\)

\(z^* = 11 + 3i\)

1. (a) \((4 - 6a) + (a + 3)i\)
2. (b) \(9a + (2a^2 - 4)i\)

\(\dfrac{6k^2 - 5}{9k^2 + 1} - \dfrac{17k}{9k^2 + 1}\,i\)

\(a = 2\) or \(a = 8\)

\(a = -3,\ b = -2\) or \(a = 6,\ b = 1\)

\(z^*\): \((3, 2)\); \(2z\): \((6, -4)\); \(iz\): \((2, 3)\); \(w\): \((-1, 4)\); \(z + w\): \((2, 2)\); \(z - w\): \((4, -6)\); \(z\): \((3, -2)\)

\(z^2 = 0.28 - 0.96i\), \(z^3 = -0.352 - 0.936i\); all three points are 1 unit from the origin

\(z = 4 + 3i\) or \(z = -4 - 3i\)

Both sides equal \(x^2 - y^2 + 2x - (2xy + 2y)i\)

\(z = 2 - i,\ w = 1 + 2i\)