Modulus and the complex plane

Modulus, distance, equations with the modulus, lines and circles in the complex plane. 7 problems with step-by-step solutions.

Solve each problem in your notebook first, then open the solution.

Problem 1 Modulus

a) Find \(|z|\) for \(z = 6 - 8i\), \(-5 + 12i\), \(-9i\), \(\frac12 + \frac{\sqrt3}{2}i\), \(\frac{1 + i}{\sqrt2}\).

b) Find \(|z|\) for \(z = \dfrac{(3 + 4i)(5 - 12i)}{8 + 6i}\).

c) Find \(|\bar z|\) for \(z = \dfrac{(2 - i)^2}{1 - i}\).

Show solution

a) \(10\); \(13\); \(9\); \(1\); \(1\).

b) \(\frac{5\cdot 13}{10} = \frac{13}{2}\).

c) \(\frac{5}{\sqrt2} = \frac{5\sqrt2}{2}\).

Problem 2 Modulus of powers

a) Find \(|z^{20}|\) for \(z = 1 - \sqrt3 i\).

b) Find \(|z|\) for \(z = 1 + \cos\alpha + i\sin\alpha\), \(\alpha \in (0, \pi)\).

c) Find \(|z|\) for \(z = \dfrac{(1 + 2i)^3(3 - i)}{(2 - i)^2}\).

Show solution

a) \(|z| = 2\), \(|z^{20}| = 2^{20}\).

b) \(|z|^2 = (1 + \cos\alpha)^2 + \sin^2\alpha = 2 + 2\cos\alpha = 4\cos^2\frac\alpha2\). For \(\alpha \in (0, \pi)\), \(\cos\frac\alpha2 > 0\): \(|z| = 2\cos\frac\alpha2\).

c) \(\frac{(\sqrt5)^3\sqrt{10}}{5} = \sqrt5\cdot\sqrt{10} = 5\sqrt2\).

Problem 3 Equations with modulus

a) Find \(z\) with \(|z| = z + 2 - 4i\).

b) Find \(z\) with \(|z| + z = 8 + 4i\).

c) Find \(|z|\) if \(z\bar z + |z| = 12\).

d) \(z = a + 2i\) and \(|\bar z + iz| = 2\sqrt2\). Find real \(a\).

Show solution

a) \(z = a + bi\): \(\sqrt{a^2 + b^2} = a + 2 + (b - 4)i\). So \(b = 4\) and \(\sqrt{a^2 + 16} = a + 2\): \(a^2 + 16 = a^2 + 4a + 4\), \(a = 3\). \(z = 3 + 4i\).

b) \(b = 4\), \(\sqrt{a^2 + 16} + a = 8\): \(a^2 + 16 = 64 - 16a + a^2\), \(a = 3\). \(z = 3 + 4i\).

c) \(|z|^2 + |z| - 12 = 0\), \(|z| = 3\).

d) \(\bar z + iz = a - 2i + ai - 2 = (a - 2) + (a - 2)i\). Modulus \(\sqrt2|a - 2| = 2\sqrt2\), so \(a = 4\) or \(a = 0\).

Problem 4 Distance

a) Find the distance between \(z_1 = 1 + 7i\) and \(z_2 = 4 + 3i\).

b) Show that \(A(1 + 2i)\), \(B(4 + 6i)\), \(C(-3 + 5i)\) form an isosceles triangle.

c) \(z = 2 + 2i\) is rotated \(90^\circ\) about the origin to \(w\). Find \(|z - w|\).

d) Find the distance from the origin to \(z\) if \((3 - 4i)z = 10 + 5i\).

Show solution

a) \(|z_1 - z_2| = |-3 + 4i| = 5\).

b) \(|AB| = |3 + 4i| = 5\), \(|AC| = |-4 + 3i| = 5\), \(|BC| = |-7 - i| = \sqrt{50}\). Two equal sides.

c) \(w = iz = -2 + 2i\), \(|z - w| = |4| = 4\).

d) \(|z| = \frac{|10 + 5i|}{|3 - 4i|} = \frac{5\sqrt5}{5} = \sqrt5\).

Problem 5 Locus: lines

a) Find the equation of the locus \(|z - 2 - i| = |z + 1 - 3i|\).

b) Find \(z\) with \(|z - 4| = |z|\) and \(|z - 2i| = |z - 2 - 4i|\).

c) Describe the region \(|z - 3| < |z + 1|\).

Show solution

a) \((x - 2)^2 + (y - 1)^2 = (x + 1)^2 + (y - 3)^2 \Rightarrow -4x - 2y + 5 = 2x - 6y + 10 \Rightarrow 6x - 4y + 5 = 0\). It is the perpendicular bisector of the segment.

b) The first gives \(x = 2\). The second: \(x^2 + (y - 2)^2 = (x - 2)^2 + (y - 4)^2\), with \(x = 2\): \(4 + (y - 2)^2 = (y - 4)^2\), \(y = 2\). \(z = 2 + 2i\).

c) Points closer to 3 than to \(-1\): the half-plane \(x > 1\).

Problem 6 Locus: circles

a) Describe and draw: \(|z| = 3\), \(|z - 2i| = 1\), \(|z + 1 - i| = 2\).

b) Draw the regions: \(|z| \ge 2\), \(|z - 1 + i| < 3\), \(2 \le |z + 2| \le 4\).

c) Find the minimum and maximum of \(|z - (8 + 6i)|\) for \(|z| \le 4\).

d) Find the maximum and minimum of \(|z - 5 + 12i|\) for \(|z| \le 3\).

Show solution

a) Circles: centre 0, radius 3; centre \(2i\), radius 1; centre \(-1 + i\), radius 2.

b) Outside (and on) the circle of radius 2; inside the circle with centre \(1 - i\), radius 3; the ring around \(-2\) between radii 2 and 4.

c) \(|8 + 6i| = 10\). Minimum \(10 - 4 = 6\), maximum \(10 + 4 = 14\).

d) \(|5 - 12i| = 13\): maximum \(16\), minimum \(10\).

Problem 7 Mixed problems

a) Find \(|z|\) if \(z = i + \dfrac{i}{1 + \frac{2i}{3 - 4i}}\).

b) Find \(\operatorname{Im}(z)\) if \((z + 3i)(i - 2) = (z + 1)(2i - 1)\).

c) Solve \(z^2 - 4zi - 5 = 0\).

d) Find \(z\) with \(|z - 3i| = |z + 2|\) and \(|z - 1 - i| = |\bar z + 3|\).

Show solution

a) \(\frac{2i}{3 - 4i} = \frac{2i(3 + 4i)}{25} = \frac{-8 + 6i}{25}\). \(1 + \) that \(= \frac{17 + 6i}{25}\). \(\frac{i\cdot 25}{17 + 6i} = \frac{25i(17 - 6i)}{325} = \frac{150 + 425i}{325} = \frac{6}{13} + \frac{17}{13}i\). \(z = \frac{6}{13} + \frac{30}{13}i\), \(|z| = \frac{\sqrt{936}}{13} = \frac{6\sqrt{26}}{13}\).

b) \(zi - 2z - 3 - 6i = 2zi - z + 2i - 1 \Rightarrow -zi - z = 2 + 8i \Rightarrow z = -\frac{2 + 8i}{1 + i} = -\frac{(2 + 8i)(1 - i)}{2} = -5 - 3i\). Im \(= -3\).

c) \((z - 2i)^2 = 5 - 4 = 1\): \(z = 2i \pm 1\).

d) First: \(x^2 + (y - 3)^2 = (x + 2)^2 + y^2 \Rightarrow 4x + 6y = 5\). Second: \(|\bar z + 3| = |z + 3|\): \((x - 1)^2 + (y - 1)^2 = (x + 3)^2 + y^2 \Rightarrow 8x + 2y = -7\). So \(x = -\frac{13}{10}\), \(y = \frac{17}{10}\): \(z = -\frac{13}{10} + \frac{17}{10}i\).