Problem 37 - Precalculus : Building Concepts and Connections

Question

Find the square roots of each complex number. Round all numbers to three decimal places. $$1+\sqrt{3} i$$

Step-by-Step Solution

Answer

The square roots of the complex number \(1+ \sqrt{3}i \) are approximately \(1.366 + 0.586i \) and \( -0.366 - 1.366i \) .

Step 1: Convert to Polar Form

First, determine the modulus \( r \) and argument \( \theta \) for the complex number. The modulus \( r \) is calculated by \( r = \sqrt{a^2 + b^2} \), where \( a \) and \( b \) are the real and imaginary parts of the complex number, respectively. So, \( r = \sqrt{1^2 + (\sqrt{3})^2} = 2 \). Then, \( \theta = tan^{-1}(b/a) = tan^{-1}(\sqrt{3}/1) = \pi/3 \). Therefore, the complex number in polar form becomes \( 2(cos(\pi/3) + i \cdot sin(\pi/3)) \).

Step 2: Calculate the square roots

The square roots of the complex number are calculated by halving the angle \( \theta \) and taking the square root of the modulus \( r \). So the arguments of the square roots will be \( \pi/6 \) and \( \pi/6 + \pi = 7\pi/6 \). The modulus of the square roots is \( \sqrt{2} \).

Step 3: Convert back to Rectangular Form

We then convert our solution from polar to rectangular form using the formula \( r(cos \theta + i \cdot sin \theta) = r \cdot cos \theta + r \cdot i \cdot sin \theta \). Doing so gives us the square roots \( \sqrt{2}(cos(\pi/6) + i \cdot sin(\pi/6)) \approx 1.366 + 0.586i \) and \( \sqrt{2}(cos(7\pi/6) + i \cdot sin(7\pi/6)) \approx -0.366 - 1.366i \).

Key Concepts

Polar Form Conversion

Understanding how to convert complex numbers to polar form is essential for simplifying many problems involving complex numbers, especially when finding roots or powers. To convert a complex number from its rectangular or standard form, which is expressed as \( a + bi \) where \( a \) and \( b \) are real numbers, to its polar form, the following steps are undertaken.

To begin, we need to calculate two important properties: the modulus and the argument. The modulus (or absolute value) of a complex number is found by the formula \( r = \sqrt{a^2 + b^2} \), representing the distance from the origin (0,0) to the point \( (a, b) \) on the complex plane. The argument is the angle \( \theta \) made with the positive x-axis, and it can be calculated using \( \tan^{-1}(b / a) \), albeit considering the quadrant in which the complex number is located is crucial for getting the correct angle.

Once both modulus and argument are known, the complex number in polar form is expressed as \( r(\cos(\theta) + i\cdot\sin(\theta)) \), often abbreviated to \( r\cis(\theta) \) where \( \cis \) stands for \( \cos + i\cdot\sin \) as seen in the problem's step-by-step solution. This form is particularly helpful for multiplication, division, and finding powers and roots of complex numbers.

Calculating Modulus and Argument

The modulus and argument are two fundamental properties of complex numbers that give us an alternative way to express and manipulate them. As part of the exercise, let's explore these concepts in depth.

The modulus of a complex number is simply its 'size', akin to the magnitude of a vector. It's determined by calculating the square root of the sum of the squares of the real and imaginary parts. To memorize this, you might think of it as the hypotenuse of a right triangle whose legs are the real and imaginary components. So, for the complex number \( 1+\sqrt{3}i \), the modulus is \( \sqrt{1^2 + (\sqrt{3})^2} = 2 \).

In contrast, the argument, denoted as \( \theta \), is the angle formed with the positive real axis. To calculate the angle, we use the inverse tangent function, considering the signs of \( a \) and \( b \) for correct placement in the complex plane. For \( 1+\sqrt{3}i \), we get \( \theta = \tan^{-1}(\sqrt{3}/1) = \pi/3 \). The argument is crucial in locating the position of the complex number on the polar coordinate system. Together, the modulus and argument allow us to represent complex numbers on a plane effectively and perform operations like finding roots, as demonstrated in our exercise solution.

Rectangular Form Conversion

Once we've operated on complex numbers in polar form—such as finding their roots or powers—converting the result back to rectangular form often makes interpretation easier, especially when dealing with real-world applications. The conversion from polar to rectangular form requires us to re-engage with trigonometry. The formula encapsulates a return trip from the modulus and argument back to the familiar \( a + bi \) format.

Using the formula \( r(\cos(\theta) + i\cdot\sin(\theta)) = r\cdot\cos(\theta) + r\cdot i\cdot\sin(\theta) \), we express the polar form in terms of its real and imaginary components. The real part \( a \) is calculated by \( r\cdot\cos(\theta) \) and the imaginary part \( b \) by \( r\cdot\sin(\theta) \). With this, the calculations yield a complex number in rectangular form that is easy to read and directly applicable—for example, in describing electrical currents or waves.

In our exercise, the two square roots of the complex number \(1+\sqrt{3}i\) were given in polar form. By applying the conversion formula, the results are returned to rectangular form, producing \( \sqrt{2}(\cos(\pi/6) + i\cdot\sin(\pi/6)) \) which approximates to \( 1.366 + 0.586i \) and \( \sqrt{2}(\cos(7\pi/6) + i\cdot\sin(7\pi/6)) \) which approximates to \( -0.366 - 1.366i \), showing the square roots in the standard form most are familiar with in the real and imaginary number context.