Problem 31 - Advanced Engineering Mathematics

Question

$$\begin{array}{l}(-1+\sqrt{3} i)^{1 / 2}=2^{1 / 2}\left[\cos \left(\frac{\pi}{3}+k \pi\right)+i \sin \left(\frac{\pi}{3}+k \pi\right)\right], \quad k=0,1 \\\w_{0}=2^{1 / 2}\left[\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}\right]=\frac{\sqrt{2}}{2}+\frac{\sqrt{6}}{2} i \\\w_{2}=2^{1 / 2}\left[\cos \frac{4 \pi}{3}+i \sin \frac{4 \pi}{3}\right]=-\frac{\sqrt{2}}{2}-\frac{\sqrt{6}}{2} i\end{array}$$

Step-by-Step Solution

Answer

The square roots are \(w_0 = \frac{\sqrt{2}}{2} + \frac{\sqrt{6}}{2}i\) and \(w_1 = -\frac{\sqrt{2}}{2} - \frac{\sqrt{6}}{2}i\).

Step 1: Understand the Problem

The problem involves finding the square root of a complex number \(-1 + \sqrt{3}i\) in polar form. We need to express this complex number as \(2^{1/2}\left[\cos\left(\frac{\pi}{3} + k\pi\right) + i \sin\left(\frac{\pi}{3} + k\pi\right)\right]\) for two values of \(k\), specifically \(k=0\) and \(k=1\).

Step 2: Convert the Complex Number to Polar Form

First, convert the complex number \(-1 + \sqrt{3}i\) to its polar form. Calculate the modulus \(r = \sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{4} = 2\). Next, find the argument \(\theta = \tan^{-1}\left(\frac{\sqrt{3}}{-1}\right)\), which gives \(\theta = \frac{2\pi}{3}\) since the complex number is in the second quadrant.

Step 3: Apply De Moivre's Theorem

To find the square root, apply De Moivre's theorem: \((r \text{cis} \theta)^{1/2} = r^{1/2} \text{cis}(\theta/2)\). Our value is \(2^{1/2}\text{cis}\left(\frac{\pi}{3} + k\pi\right)\), where \(k=0,1\). This gives two possible results due to the periodic nature of cosine and sine functions.

Step 4: Calculate \(w_0\) and \(w_1\)

For \(k=0\):\[ w_0 = 2^{1/2} \left[ \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} \right] = \frac{\sqrt{2}}{2} + \frac{\sqrt{6}}{2} i \]For \(k=1\):\[ w_1 = 2^{1/2} \left[ \cos \frac{4\pi}{3} + i \sin \frac{4\pi}{3} \right] = -\frac{\sqrt{2}}{2} - \frac{\sqrt{6}}{2} i \]The calculation of these values involves using standard cosine and sine values for these angles.

Key Concepts

Polar Form

When we talk about the polar form of complex numbers, we are looking at a way to express these numbers using angles and distances. In contrast to the rectangular form, which uses the real and imaginary parts, polar form makes use of the magnitude (also known as modulus) and the angle (argument). The formula for polar form is:

• \[ z = r \times (\cos \theta + i \sin \theta) \] where:
  • \( z \) is the complex number
  • \( r \) is the modulus
  • \( \theta \) is the argument

This form is particularly useful for multiplying and dividing complex numbers, finding powers, and roots, as it simplifies these operations greatly. In our exercise, we are supposed to express \(-1 + \sqrt{3}i\) in polar form. To do this, we first find the modulus and argument. This conversion helps us to easily apply De Moivre's Theorem later.

De Moivre's Theorem

De Moivre's Theorem is a magical shortcut that helps us quickly find powers and roots of complex numbers when they are in polar form. According to De Moivre, for a complex number in polar form, the nth power is given by:

• \[( r(\cos \theta + i \sin \theta) )^n = r^n (\cos (n\theta) + i \sin (n\theta)) \]

This theorem significantly simplifies computations that involve powers and roots, as in our exercise, where we need to find the square root of a complex number.For the square root, specifically, you decide what number (let's use 1/2, since we are finding the square root) you need to use with the power, and that's where De Moivre's Theorem comes in. So, it helps us find different roots by adjusting the angle appropriately, giving us several solutions due to the periodicity of sine and cosine!

Complex Modulus

The complex modulus, often denoted as \( r \), is the distance of the complex number from the origin in the complex plane. It gives you an idea of how "large" the complex number is. To find the modulus of a complex number \( a + bi \), we use the formula:

• \[ r = \sqrt{a^2 + b^2} \]

In our exercise, for the number \(-1 + \sqrt{3}i\), we calculate:

• \[ r = \sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2 \]

The modulus is key when converting a complex number into its polar form and when using De Moivre's theorem. It tells us how "far" the complex number is from the origin, thus setting the "length" of our number in polar coordinates.

Complex Argument

The complex argument is the angle, \( \theta \), that a complex number makes with the positive real axis in the complex plane, usually expressed in radians. It's a measure of the "direction" of the number. The argument is calculated with the formula:

• \[ \theta = \tan^{-1}(\frac{b}{a}) \]

However, it also depends on the quadrant where the complex number lies, adjusting the angle accordingly to ensure it falls within the correct range.In our problem, for \(-1 + \sqrt{3}i\), the argument can be found as:

• \[ \theta = \tan^{-1}(\frac{\sqrt{3}}{-1}) \]Since the number is in the second quadrant, we adjust to obtain:\[ \theta = \frac{2\pi}{3} \]

This angle helps with the conversion to polar form and assists in using De Moivre's theorem to find powers or roots of the complex number.