Question

Solve the equation z^4=-2\sqrt{3}-2i Sketch the solution in the complex plane.

Answer 1

Answer on Question #61150 - Math - Algebra

Question

Solve the equation

$$z ^ {4} = - 2 \sqrt {3} - 2 i$$

Sketch the solution in the complex plane.

Solution

The absolute value (the modulus) of the complex number $-2\sqrt{3} - 2i$ is

$$\rho (z ^ {4}) = \sqrt {\left(- 2 \sqrt {3}\right) ^ {2} + (- 2) ^ {2}} = 4$$

The argument of the complex number $-2\sqrt{3} - 2i$ is

$$\theta (z ^ {4}) = - \pi + \tan^ {- 1} \left(\frac {- 2}{- 2 \sqrt {3}}\right) = - \pi + \frac {\pi}{6} = - \frac {5 \pi}{6} \mathrm {o r} 2 \pi - \frac {5 \pi}{6} = \frac {7 \pi}{6}.$$

The absolute value (the modulus) of the complex number $z = \sqrt[4]{-2\sqrt{3} - 2i}$ is

$$\rho (z) = \sqrt [ 4 ]{4} = \sqrt {2} \approx 1. 4 1$$

The argument of the complex number $z = \sqrt[4]{-2\sqrt{3} - 2i}$ is

$$\theta (z) = \frac {\frac {7 \pi}{6} + 2 \pi k}{4}, k = 0, 1, 2, 3.$$

Solutions to the equation (1) are given by

$$z = \rho (z) \cdot (\cos \theta (z) + i \sin \theta (z)),$$

where $\rho (z)$ and $\theta (z)$ are defined by means of formulae (2) and (3).

If $k = 0$ , then $\theta(z) = \theta_1 = \frac{\frac{7\pi}{6}}{4} = \frac{7\pi}{24} \approx 52.5{}^\circ$ ,

$\cos \frac{7\pi}{24} = \frac{1}{2}\sqrt{2 - \sqrt{2 - \sqrt{3}}} \approx 0.61$ (see http://mathworld.wolfram.com/TrigonometryAnglesPi24.html),

$\sin \frac{7\pi}{24} = \frac{1}{2}\sqrt{2 + \sqrt{2 - \sqrt{3}}} \approx 0.79$ (see http://mathworld.wolfram.com/TrigonometryAnglesPi24.html),

and a solution will be

$$z = z _ {1} = \rho (z) \cdot (\cos \theta_ {1} + i \sin \theta_ {1}) = \sqrt {2} \left(\cos \frac {7 \pi}{2 4} + i \sin \frac {7 \pi}{2 4}\right) \approx 0. 8 6 + 1. 1 2 i.$$

If $k = 1$ , then $\theta(z) = \theta_2 = \frac{\frac{7\pi}{6} + 2\pi}{4} = \frac{19\pi}{24} \approx 142.5{}^\circ$ ,

$$\cos \frac {1 9 \pi}{2 4} = \cos \left(\pi - \frac {5 \pi}{2 4}\right) = - \cos \frac {5 \pi}{2 4} = - \frac {1}{2} \sqrt {2 + \sqrt {2 - \sqrt {3}}} \approx - 0. 7 9$$

(see http://mathworld.wolfram.com/TrigonometryAnglesPi24.html),

$$\sin \frac {19\pi}{24} = \sin \left(\pi - \frac {5\pi}{24}\right) = \sin \frac {5\pi}{24} = \frac {1}{2} \sqrt {2 - \sqrt {2 - \sqrt {3}}} \approx 0.61$$

(see http://mathworld.wolfram.com/TrigonometryAnglesPi24.html),

and a solution will be

$$z = z _ {2} = \rho (z) \cdot (\cos \theta_ {2} + i \sin \theta_ {2}) = \sqrt {2} \left(\cos \frac {19\pi}{24} + i \sin \frac {19\pi}{24}\right) \approx - 1.12 + 0.86i.$$

If $k = 2$, then $\theta (z) = \theta_{3} = \frac{\frac{7\pi}{6} + 4\pi}{4} = \frac{31\pi}{24}\approx 232.5{}^{\circ}$,

$$\cos \frac {31\pi}{24} = \cos \left(\pi + \frac {7\pi}{24}\right) = - \cos \frac {7\pi}{24} = - \frac {1}{2} \sqrt {2 - \sqrt {2 - \sqrt {3}}} \approx - 0.61$$

(see http://mathworld.wolfram.com/TrigonometryAnglesPi24.html),

$$\sin \frac {31\pi}{24} = \sin \left(\pi + \frac {7\pi}{24}\right) = - \sin \frac {7\pi}{24} = - \frac {1}{2} \sqrt {2 + \sqrt {2 - \sqrt {3}}} \approx - 0.79$$

(see http://mathworld.wolfram.com/TrigonometryAnglesPi24.html),

and a solution will be

$$z = z _ {3} = \rho (z) \cdot (\cos \theta_ {3} + i \sin \theta_ {3}) = \sqrt {2} \left(\cos \frac {31\pi}{24} + i \sin \frac {31\pi}{24}\right) \approx - 0.86 - 1.12i$$

If $k = 3$, then $\theta (z) = \theta_{4} = \frac{\frac{7\pi}{6} + 6\pi}{4} = \frac{43\pi}{24}\approx 322.5{}^{\circ}$,

$$\cos \frac {43\pi}{24} = \cos \left(\frac {48\pi - 5\pi}{24}\right) = \cos \left(2\pi - \frac {5\pi}{24}\right) = \cos \left(- \frac {5\pi}{24}\right) = \cos \frac {5\pi}{24} = \frac {1}{2} \sqrt {2 + \sqrt {2 - \sqrt {3}}} \approx 0.79$$

(see http://mathworld.wolfram.com/TrigonometryAnglesPi24.html),

$$\sin \frac {43\pi}{24} = \sin \left(\frac {48\pi - 5\pi}{24}\right) = \sin \left(2\pi - \frac {5\pi}{24}\right) = \sin \left(- \frac {5\pi}{24}\right) = - \sin \frac {5\pi}{24} = - \frac {1}{2} \sqrt {2 - \sqrt {2 - \sqrt {3}}} \approx - 0.61$$

(see http://mathworld.wolfram.com/TrigonometryAnglesPi24.html),

and a solution will be

$$z = z _ {4} = \rho (z) \cdot (\cos \theta_ {4} + i \sin \theta_ {4}) = \sqrt {2} \left(\cos \frac {43\pi}{24} + i \sin \frac {43\pi}{24}\right) \approx 1.12 - 0.86i$$

Table 1. Arguments of the solutions

Figure 1. The solutions to the equation $z^4 = -2\sqrt{3} - 2i$ in the complex plane

Now show how to deduce auxiliary formulae, for example,

$\cos \frac{5\pi}{24} = \frac{1}{2}\sqrt{2 + \sqrt{2 - \sqrt{3}}}, \sin \frac{5\pi}{24} = \frac{1}{2}\sqrt{2 - \sqrt{2 - \sqrt{3}}}$ .

It is known that $\cos 30{}^{\circ} = \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}$ , $\sin 30{}^{\circ} = \sin \frac{\pi}{6} = \frac{1}{2}$ .

Using half-angle formulae

$\sin \frac{2\pi}{24} = \sin 15{}^{\circ} = \sin \frac{30{}^{\circ}}{2} = \sqrt{\frac{1 - \cos 30{}^{\circ}}{2}} = \sqrt{\frac{1 - \frac{\sqrt{3}}{2}}{2}} = \frac{1}{2}\sqrt{2 - \sqrt{3}},$

$\cos \frac{2\pi}{24} = \cos 15{}^{\circ} = \cos \frac{30{}^{\circ}}{2} = \sqrt{\frac{1 + \cos 30{}^{\circ}}{2}} = \sqrt{\frac{1 + \frac{\sqrt{3}}{2}}{2}} = \frac{1}{2}\sqrt{2 + \sqrt{3}}.$

Using reduction formulae

$\cos 75{}^{\circ} = \sin \left(90{}^{\circ} - 75{}^{\circ}\right) = \sin 15{}^{\circ} = \frac{1}{2}\sqrt{2 - \sqrt{3}}$ ,

$\sin 75{}^{\circ} = \sin \left(90{}^{\circ} - 15{}^{\circ}\right) = \cos 15{}^{\circ} = \frac{1}{2}\sqrt{2 + \sqrt{3}}$

Using half-angle formulae finally obtain

$$\sin 37.5{}^{\circ} = \sin \frac{5\pi}{24} = \sin \left(\frac{5\pi/12}{2}\right) = \sin \left(\frac{75{}^{\circ}}{2}\right) = \sqrt{\frac{1 - \cos 75{}^{\circ}}{2}} = \sqrt{\frac{1 - \frac{1}{2}\sqrt{2 - \sqrt{3}}}{2}} = \frac{1}{2} \sqrt{2 - \sqrt{2 - \sqrt{3}}},$$ $$\cos 37.5{}^{\circ} = \cos \frac{5\pi}{24} = \cos \left(\frac{5\pi/12}{2}\right) = \cos \left(\frac{75{}^{\circ}}{2}\right) = \sqrt{\frac{1 + \cos 75{}^{\circ}}{2}} = \sqrt{\frac{1 + \frac{1}{2}\sqrt{2 - \sqrt{3}}}{2}} = \frac{1}{2} \sqrt{2 + \sqrt{2 - \sqrt{3}}}.$$

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