Roots and rotations

n-th roots, equations, the locus Arg(z − z₀) = θ and rotations in the complex plane. 7 problems with step-by-step solutions.

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

Problem 1 n-th roots

\(\sqrt[n]{r\operatorname{cis}\theta} = \sqrt[n]{r}\operatorname{cis}\dfrac{\theta + 2k\pi}{n}\), \(k = 0, 1, \dots, n - 1\).

a) Find the fourth roots of \(-16\). b) Find all solutions of \(x^6 = 64\). c) Find the cube roots of \(-8i\).

Show solution

a) \(-16 = 16\operatorname{cis}\pi\): \(2\operatorname{cis}\frac{\pi + 2k\pi}{4}\): \(\sqrt2 + \sqrt2 i\), \(-\sqrt2 + \sqrt2 i\), \(-\sqrt2 - \sqrt2 i\), \(\sqrt2 - \sqrt2 i\).

b) \(2\operatorname{cis}\frac{k\pi}{3}\): \(\pm 2\), \(\pm 1 \pm \sqrt3 i\) (all four sign combinations).

c) \(-8i = 8\operatorname{cis}\frac{3\pi}{2}\): \(2\operatorname{cis}\left(\frac{\pi}{2} + \frac{2k\pi}{3}\right)\): \(2i\), \(-\sqrt3 - i\), \(\sqrt3 - i\).

Problem 2 Square and cube roots

a) Find the square roots of \(-2 - 2\sqrt3 i\) and of \(9\operatorname{cis}\frac{2\pi}{3}\).

b) Find the cube roots of \(27i\) and show that they form an equilateral triangle.

c) Solve \(x^3 - 1 = 0\) with the root formula and by factoring.

Show solution

a) \(-2 - 2\sqrt3 i = 4\operatorname{cis}240^\circ\): \(2\operatorname{cis}120^\circ = -1 + \sqrt3 i\) and \(2\operatorname{cis}300^\circ = 1 - \sqrt3 i\). \(9\operatorname{cis}120^\circ\): \(3\operatorname{cis}60^\circ = \frac32 + \frac{3\sqrt3}{2}i\) and its opposite.

b) \(27i = 27\operatorname{cis}90^\circ\): \(3\operatorname{cis}30^\circ\), \(3\operatorname{cis}150^\circ\), \(3\operatorname{cis}270^\circ\). Same modulus, arguments differ by \(120^\circ\): an equilateral triangle on the circle of radius 3.

c) \(\operatorname{cis}\frac{2k\pi}{3}\): \(1\), \(-\frac12 \pm \frac{\sqrt3}{2}i\). Factoring: \((x - 1)(x^2 + x + 1) = 0\) gives the same.

Problem 3 Equations

a) Find the solutions of \(x^4 = 81\operatorname{cis}240^\circ\) in standard form.

b) Solve \(x^3 + i = 0\).

c) Solve \(x^6 + 64i = 0\) and give the modulus and the arguments of the roots.

Show solution

a) \(3\operatorname{cis}(60^\circ + 90^\circ k)\): \(\frac32 + \frac{3\sqrt3}{2}i\), \(-\frac{3\sqrt3}{2} + \frac32 i\), \(-\frac32 - \frac{3\sqrt3}{2}i\), \(\frac{3\sqrt3}{2} - \frac32 i\).

b) \(x^3 = -i = \operatorname{cis}270^\circ\): \(\operatorname{cis}90^\circ = i\), \(\operatorname{cis}210^\circ = -\frac{\sqrt3}{2} - \frac12 i\), \(\operatorname{cis}330^\circ = \frac{\sqrt3}{2} - \frac12 i\).

c) \(x^6 = -64i = 64\operatorname{cis}270^\circ\): \(2\operatorname{cis}(45^\circ + 60^\circ k)\), modulus 2, arguments \(45^\circ, 105^\circ, 165^\circ, 225^\circ, 285^\circ, 345^\circ\).

Problem 4 Locus of arguments

\(\operatorname{Arg}(z - z_0) = \theta\) is a ray from \(z_0\) at the angle \(\theta\) (without \(z_0\)).

a) Draw \(\operatorname{Arg}(z - 2) = \frac{\pi}{4}\) and \(\operatorname{Arg}(z + 1 - i) = \frac{2\pi}{3}\).

b) Find \(z\) with \(\operatorname{Arg}(z - 4) = \frac{3\pi}{4}\) and \(\operatorname{Arg}(z + 2) = \frac{\pi}{4}\).

c) Find \(z\) with \(\operatorname{Arg}(z - 2i) = \frac{5\pi}{4}\) and \(\operatorname{Arg}(z + 6) = 0\).

Show solution

a) The ray from \(2\) going up and right at \(45^\circ\); the ray from \(-1 + i\) going up and left at \(120^\circ\).

b) From 4: \(z = 4 + t(-1 + i)\); from \(-2\): \(z = -2 + s(1 + i)\), \(t, s > 0\). \(4 - t = -2 + s\), \(t = s\): \(t = s = 3\). \(z = 1 + 3i\).

c) From \(2i\) in the direction \(225^\circ\): \(z = -t + (2 - t)i\). On the real axis: \(t = 2\), \(z = -2\), and \(-2 > -6\). So \(z = -2\).

Problem 5 Rotations

Rotating \(z\) by \(\alpha\) counterclockwise: \(z' = z\operatorname{cis}\alpha\); clockwise: \(z'' = z\operatorname{cis}(-\alpha)\).

a) Rotate \(z = 4\operatorname{cis}100^\circ\) by \(50^\circ\) counterclockwise and clockwise.

b) Rotate \(z = 1 + \sqrt3 i\) by \(30^\circ\) counterclockwise and clockwise.

c) Rotate \(z = 2 - 3i\) by \(90^\circ\) in both directions.

Show solution

a) \(4\operatorname{cis}150^\circ = -2\sqrt3 + 2i\); \(4\operatorname{cis}50^\circ\).

b) \(z = 2\operatorname{cis}60^\circ\): \(2\operatorname{cis}90^\circ = 2i\); \(2\operatorname{cis}30^\circ = \sqrt3 + i\).

c) Counterclockwise: \(iz = 3 + 2i\). Clockwise: \(-iz = -3 - 2i\).

Problem 6 Rotations

a) Rotating \(z\) by \(30^\circ\) counterclockwise gives \(4 - i\). What do we get when \(z\) is rotated by \(120^\circ\) counterclockwise?

b) Rotate \(z = 3 + i\) by \(135^\circ\) counterclockwise.

c) Show \(z\), \(iz\), \(i^2z\), \(i^3z\) in the plane for \(z = 2 + i\).

Show solution

a) A further \(90^\circ\): multiply \(4 - i\) by \(i\): \(1 + 4i\).

b) \((3 + i)\left(-\frac{\sqrt2}{2} + \frac{\sqrt2}{2}i\right) = \frac{\sqrt2}{2}(-3 + 3i - i - 1) = \frac{\sqrt2}{2}(-4 + 2i) = -2\sqrt2 + \sqrt2 i\).

c) \(2 + i\), \(-1 + 2i\), \(-2 - i\), \(1 - 2i\): the vertices of a square centred at the origin.

Problem 7 Review test

a) Solve \(x^6 - 8i = 0\): give the modulus and the arguments of the six roots.

b) Write \(z\) with \(|z| = 6\) and \(\operatorname{Arg}(z) = \frac{2\pi}{3}\) in standard form.

c) Find \(\operatorname{Arg}(\bar z)\) for \(z = \dfrac{4}{\sqrt3 + i}\).

d) \(z = 1 - i\). Find \(z^{15}\).

e) Write \(r = \dfrac{4}{3\sin\theta - \cos\theta}\) in rectangular form.

Show solution

a) \(8i = 8\operatorname{cis}90^\circ\): modulus \(\sqrt[6]{8} = \sqrt2\), arguments \(15^\circ + 60^\circ k\): \(15^\circ, 75^\circ, 135^\circ, 195^\circ, 255^\circ, 315^\circ\).

b) \(6\left(-\frac12 + \frac{\sqrt3}{2}i\right) = -3 + 3\sqrt3 i\).

c) \(z = \frac{4}{2\operatorname{cis}30^\circ} = 2\operatorname{cis}(-30^\circ)\), \(\bar z = 2\operatorname{cis}30^\circ\): \(\operatorname{Arg}(\bar z) = \frac{\pi}{6}\).

d) \((\sqrt2\operatorname{cis}(-45^\circ))^{15} = 2^{7}\sqrt2\operatorname{cis}(-675^\circ) = 128\sqrt2\operatorname{cis}45^\circ = 128 + 128i\).

e) \(3r\sin\theta - r\cos\theta = 4\), i.e. \(3y - x = 4\): a straight line.