Question

Solve for all possible values of the real numbers \(x\) and \(y\) in the following equations. $$(x+i y)^{3}=-1$$

Short answer

The solutions are \(x, y\) = (-1, 0), (\frac{1}{2}, -\frac{\sqrt{3}}{2}), (\frac{1}{2}, \frac{\sqrt{3}}{2}).

Step-by-step solution

Identify the form of the complex number

The given equation is \((x + i y)^3 = -1\). Express \(-1\) in its polar form, \(-1 = e^{i\pi}\).

Apply roots of unity

To solve for \((x + iy)^3 = e^{i\pi}\), we apply the roots of unity. The solutions are given by \(x + iy = e^{i(\pi + \frac{2k\pi}{3})}, k = 0, 1, 2\).

Calculate the roots

Calculate each root corresponding to different values of \k.\ For \k = 0\, \(x + iy = e^{i\pi} = -1\). For \k = 1\, \(x + iy = e^{i\frac{5\pi}{3}}\). For \k = 2\, \(x + iy = e^{i\frac{7\pi}{3}}\).

Simplify the solutions

Express each exponential term in its rectangular form: \(-1\) corresponds to \(-1 + 0i\); \e^{i\frac{5\pi}{3}} = \cos(\frac{5\pi}{3}) + i\sin(\frac{5\pi}{3}) = \frac{1}{2} - i\frac{\sqrt{3}}{2}\; \e^{i\frac{7\pi}{3}} = \cos(\frac{7\pi}{3}) + i\sin(\frac{7\pi}{3}) = \frac{1}{2} + i\frac{\sqrt{3}}{2}\.

Extract real and imaginary parts

From the simplified forms, the solutions for \(x, y\) are: \-1 + 0i\ gives \x = -1\, \y = 0\; \frac{1}{2} - i\frac{\sqrt{3}}{2}\ gives \x = \frac{1}{2}\, \y = -\frac{\sqrt{3}}{2}\; \frac{1}{2} + i\frac{\sqrt{3}}{2}\ gives \x = \frac{1}{2}\, \y = \frac{\sqrt{3}}{2}\.

Key concepts

complex number polar form

In solving problems involving complex numbers, transforming them into their polar form can make calculations much simpler. A complex number in the form \(z = x + iy\) can be represented in polar form as \(z = re^{i\theta}\). Here, \(r\) is the magnitude (or modulus) and \(\theta\) is the argument (or angle).\r
Translating into polar form involves:\r

\r
• Calculating the modulus: \(r = \sqrt{x^2 + y^2}\)
\r
• Finding the argument: \(\theta = \text{atan2}(y, x)\)
\r

\rExpressing \(-1\) in polar form, we get \(-1 = e^{i\pi}\), because the modulus is 1 and the argument is \(\pi\) radians.

roots of unity

The concept of roots of unity is essential when solving equations involving powers of complex numbers. The \(n\)-th roots of unity are the solutions to the equation \(z^n = 1\). These solutions can be found using the formula: \r
\[z_k = e^{i\frac{2k\pi}{n}} \; \text{for}\; k = 0, 1, 2, \, ...,\, n-1 \]\r
In our problem, we need to find the cube roots of \(-1\). So we solve \r\((x+i y)^3 = e^{i\pi}\). The cube roots are given by: \r\[x + iy = e^{i\left(\pi + \frac{2k\pi}{3}\right)} \;\text{for}\; k = 0,1,2 \]\r Calculating for each value of \(k\) we get: \r

\r
• For \(k = 0\): \(x + iy = e^{i\left(\pi + \frac{2 \cdot 0\pi}{3}\right)} = e^{i\pi} = -1\)
\r
• For \(k = 1\): \(x + iy = e^{i\left(\pi + \frac{2\pi}{3}\right)} = e^{i\frac{5\pi}{3}}\)
\r
• For \(k = 2\): \(x + iy = e^{i\left(\pi + \frac{4\pi}{3}\right)} = e^{i\frac{7\pi}{3}}\)
\r

Euler's formula

Euler's formula, \(e^{i\theta} = \cos(\theta) + i\sin(\theta)\), provides a crucial bridge between exponential and trigonometric representations of complex numbers. This theorem is pivotal for transforming complex numbers between polar and rectangular forms.\r
Applying Euler's formula to our cube roots problem, we need to express each root \(e^{i\theta}\) in rectangular form:\r

\r
• For \(e^{i\pi}\): Using Euler's formula, \(e^{i\pi} = \cos(\pi) + i\sin(\pi) = -1 + 0i\)
\r
• For \(e^{i\frac{5\pi}{3}}\): \(e^{i\frac{5\pi}{3}} = \cos\left(\frac{5\pi}{3}\right) + i\sin\left(\frac{5\pi}{3}\right) = \frac{1}{2} - i\frac{\sqrt{3}}{2}\)
\r
• For \(e^{i\frac{7\pi}{3}}\): \(e^{i\frac{7\pi}{3}} = \cos\left(\frac{7\pi}{3}\right) + i\sin\left(\frac{7\pi}{3}\right) = \frac{1}{2} + i\frac{\sqrt{3}}{2}\)
\r

\rFinally, extracting the real and imaginary parts from these simplified forms gives us the solutions for \(x\) and \(y\):\r

\r
• \(-1 + 0i\): \(x = -1\), \(y = 0\)
\r
• \(\frac{1}{2} - i\frac{\sqrt{3}}{2}\): \(x = \frac{1}{2}\), \(y = -\frac{\sqrt{3}}{2}\)
\r
• \(\frac{1}{2} + i\frac{\sqrt{3}}{2}\): \(x = \frac{1}{2}\), \(y = \frac{\sqrt{3}}{2}\)
\r