Cartesian form: mixed practice Problem set A Answers

1. (a) \(z = -3 + 2i\)
2. (b) \(z = 2 - i\)

\(z = 2 - 3i\)

1. (a) \(x = \pm 2 \sqrt{5}i\), which is \(\pm 4.47i\) to 3 s.f.
2. (b) \(x = 3 \pm i\)
3. (c) \(x = - \frac{2}{3} \pm \frac{\sqrt{11}}{3}i\), which is \(-0.667 \pm 1.11i\) to 3 s.f.

\(x = \frac{3}{2} \pm i\); points \((\frac{3}{2}, 1)\) and \((\frac{3}{2}, -1)\)

\(z^* = -4 - 11i\)

1. (a) \((3a + 2) + (2a - 2)i\)
2. (b) \(5a + (3a^2 - 2)i\)

\(\dfrac{2k^2 - 4}{4k^2 + 1} + \dfrac{9k}{4k^2 + 1}\,i\)

\(a = 1\) or \(a = 4\)

\(a = 2,\ b = -2\) or \(a = 4,\ b = -1\)

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

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

\(z = 3 - 2i\) or \(z = -3 + 2i\)

Both sides equal \(x^3 - 3xy^2 - (3x^2 y - y^3)i\)

\(z = 1 + 2i,\ w = 3 - i\)