Problem 40 - A Graphical Approach to Precalculus with Limits

Question

Find all indicated roots and express them in rectangular form. Check your results with a calculator. The square roots of \(-1+i \sqrt{3}\).

Step-by-Step Solution

Answer

The square roots are \(\frac{\sqrt{2}}{2} + i \frac{\sqrt{6}}{2}\) and \(-\frac{\sqrt{2}}{2} - i \frac{\sqrt{6}}{2}\).

Step 1: Represent the Complex Number in Polar Form

The complex number is \(-1 + i \sqrt{3}\). First, find the modulus \(r\) using the formula \(r = \sqrt{a^2 + b^2}\). Here, \(a = -1\) and \(b = \sqrt{3}\), resulting in \(r = \sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = 2\). Next, find the argument \(\theta\) using \(\tan \theta = \frac{b}{a}\). Thus, \(\tan \theta = \frac{\sqrt{3}}{-1} = -\sqrt{3}\), which gives \(\theta = \frac{2\pi}{3}\) because the complex number is in the second quadrant.

Step 2: Apply De Moivre's Theorem to Find the Roots

To find the square roots, we apply De Moivre's theorem: \( z^{1/n} = r^{1/n} \left( \cos \left( \frac{\theta + 2k\pi}{n} \right) + i \sin \left( \frac{\theta + 2k\pi}{n} \right) \right) \), where \(n = 2\) and \(k = 0, 1\). The modulus of the roots will be \(r^{1/2} = 2^{1/2} = \sqrt{2}\).

Step 3: Calculate the First Root (k=0)

For \(k = 0\), compute:\[ \theta_0 = \frac{2\pi/3 + 2 \times 0 \times\pi}{2} = \frac{2\pi}{6} = \frac{\pi}{3} \]Now calculate the root: \[ z_0 = \sqrt{2} \left( \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} \right) = \sqrt{2} \left( \frac{1}{2} + i \frac{\sqrt{3}}{2} \right) = \frac{\sqrt{2}}{2} + i \frac{\sqrt{6}}{2} \]

Step 4: Calculate the Second Root (k=1)

For \(k = 1\), compute: \[ \theta_1 = \frac{2\pi/3 + 2 \times 1 \times \pi}{2} = \frac{2\pi}{6} + \pi = \frac{2\pi}{6} + \frac{6\pi}{6} = \frac{8\pi}{6} = \frac{4\pi}{3} \]Now calculate the root: \[ z_1 = \sqrt{2} \left( \cos \frac{4\pi}{3} + i \sin \frac{4\pi}{3} \right) = \sqrt{2} \left( -\frac{1}{2} - i \frac{\sqrt{3}}{2} \right) = -\frac{\sqrt{2}}{2} - i \frac{\sqrt{6}}{2} \]

Step 5: Check with Calculator

Input these results into a calculator to confirm: \(\frac{\sqrt{2}}{2} + i \frac{\sqrt{6}}{2}\) and \(-\frac{\sqrt{2}}{2} - i \frac{\sqrt{6}}{2}\) are indeed square roots of \(-1 + i\sqrt{3}\) when squared. Both should result in the original number \(-1 + i\sqrt{3}\).

Key Concepts

Polar Form

Complex numbers can be represented in what we call the **Polar Form**. This involves expressing the complex number in terms of a modulus (a kind of "distance" from the origin in a complex plane) and an argument (angle with the positive x-axis).

• The modulus of a complex number \( z = a + ib \) is given by \( r = \sqrt{a^2 + b^2} \). It tells us the distance from the origin to the point representing \( z \).
• The argument \( \theta \) is the angle formed with the positive x-axis. It can be found using \( \tan \theta = \frac{b}{a} \). The location in the complex plane (which quadrant) affects the precise value of \( \theta \).

For the number \(-1 + i\sqrt{3}\), we calculated the modulus as 2 and found the argument to be \( \frac{2\pi}{3} \), confirming it lies in the second quadrant. This polar representation is key for transformations and computations involving complex numbers.

De Moivre's Theorem

**De Moivre's Theorem** is a powerful tool used in finding powers and roots of complex numbers when they are in polar form. It states: \[ (r (\cos \theta + i \sin \theta))^n = r^n ( \cos(n\theta) + i \sin(n\theta)) \] This theorem is especially useful for finding roots of complex numbers. For roots, we look at the formula: \[ z^{1/n} = r^{1/n} \left( \cos \frac{\theta + 2k\pi}{n} + i \sin \frac{\theta + 2k\pi}{n} \right) \]

• Here, \( n \) is the root degree and \( k \) ranges from 0 to \( n-1 \).

In this exercise, we applied it to find the square roots of \(-1 + i\sqrt{3}\). By substituting \( n = 2 \), and testing values \( k = 0 \) and \( k = 1 \), we obtained two distinct square roots. The theorem provides the paths in the complex plane where these roots are located.

Rectangular Form

Once the complex roots are found using polar and De Moivre's forms, they can be converted back to the **Rectangular Form**. This is possibly the most familiar representation as it directly tells us the point coordinates:\( a + ib \).

• Given a polar form \( r(\cos \theta + i \sin \theta) \), you can convert it to rectangular form as follows:
• The real part \( a = r \cos \theta \)
• The imaginary part \( b = r \sin \theta \)

For example, the polar form \( \sqrt{2}( \cos \frac{\pi}{3} + i \sin \frac{\pi}{3}) \) translates to the rectangular form \( \frac{\sqrt{2}}{2} + i \frac{\sqrt{6}}{2} \).
This gives a clear, precise location of the complex number on the complex plane.

Square Roots of Complex Numbers

Finding the **Square Roots of Complex Numbers** is considerably simplified using De Moivre's Theorem and working with polar forms. Since every complex number has exactly two square roots, our goal is to find those two numbers which, when squared, return the original complex number.

• This involves finding appropriate angles \( \theta/2 \) and moduli \( r^{1/2} \).
• By applying these to De Moivre's formula with \( n = 2 \), we ensure that all roots are discovered.
• In our solution, we calculated the roots: \( \frac{\sqrt{2}}{2} + i \frac{\sqrt{6}}{2} \) and \(-\frac{\sqrt{2}}{2} - i \frac{\sqrt{6}}/2 \), which confirm the finding of square roots through recalculated checks.

Using this approach provides precise methods over basic approximation, ensuring accuracy in high-level math problems.