Problem 42 - Contemporary Precalculus

Question

In Exercises \(37-52,\) express the number in polar form. $$-4-4 \sqrt{3} i$$

Step-by-Step Solution

Answer

Question: Express the complex number \(-4 - 4\sqrt{3}i\) in polar form. Answer: The polar form of the given complex number is \(8(\cos{\frac{4\pi}{3}} + i\sin{\frac{4\pi}{3}})\).

Step 1: Identify the real and imaginary parts

We know that a complex number is written in the form \(a + bi\). So, in this case, we have \(a = -4\) and \(b = -4\sqrt{3}\).

Step 2: Calculate the magnitude (r)

The magnitude of a complex number \(z = a + bi\) can be calculated as \(r = \sqrt{a^2 + b^2}\). Therefore, for our given number, we have: $$r = \sqrt{(-4)^2 + (-4\sqrt{3})^2}$$ $$r = \sqrt{16 + 48}$$ $$r = \sqrt{64}$$ $$r = 8$$

Step 3: Calculate the argument (theta)

The argument (theta) of a complex number \(z = a + bi\) can be found using the arctangent function as \(\theta = \arctan(\frac{b}{a})\). In our case, we have: $$\theta = \arctan\Big(\frac{-4\sqrt{3}}{-4}\Big)$$ $$\theta = \arctan(\sqrt{3})$$ As we know, \(\arctan(\sqrt{3})\) corresponds to \(\frac{\pi}{3}\), however, the complex number lies in the third quadrant, so we need to add \(\pi\): $$\theta = \frac{\pi}{3} + \pi$$ $$\theta = \frac{4\pi}{3}$$

Step 4: Express the complex number in polar form

Now that we have the magnitude and the argument, we can represent the complex number in polar form as \(z = r(\cos{\theta} + i\sin{\theta})\). Our answer comes out to be: $$z = 8(\cos{\frac{4\pi}{3}} + i\sin{\frac{4\pi}{3}})$$

Key Concepts

Polar Form of Complex Numbers

Understanding complex numbers can seem challenging, but representing them in polar form often simplifies matters, particularly when dealing with multiplication, division, or finding powers and roots. To convert a complex number to polar form, we focus on its magnitude and argument.

A complex number is typically written as \( z = a + bi \), where \( a \) is the real part and \( b \) is the imaginary part. In polar form, however, it is represented as \( z = r(\cos{\theta} + i\sin{\theta}) \), where \( r \) is the magnitude and \( \theta \) is the argument of the complex number.

When given a complex number like \( -4 - 4\sqrt{3}i \), we identify the real and imaginary parts, and then proceed to find its magnitude and argument (more on these in subsequent sections). Combining the calculated magnitude and argument, you can express the number in polar form, which, for the provided example, the polar form is \( 8(\cos{\frac{4\pi}{3}} + i\sin{\frac{4\pi}{3}}) \). This representation can offer significant insight, especially when working with the complex plane, also known as the Argand plane.

Magnitude of Complex Numbers

The magnitude (or modulus) of a complex number is a measure of its distance from the origin in the complex plane. To compute the magnitude of a complex number \( z = a + bi \), we use the formula \( r = \sqrt{a^2 + b^2} \), which is derived from the Pythagorean theorem.

For instance, if we take the complex number \( -4 - 4\sqrt{3}i \), its magnitude is calculated as follows:

• Real part, \( a = -4 \)
• Imaginary part, \( b = -4\sqrt{3} \)
• Magnitude, \( r = \sqrt{(-4)^2 + (-4\sqrt{3})^2} = \sqrt{64} = 8 \)

This result, \( r = 8 \), tells us that the complex number is 8 units away from the origin on the complex plane. The magnitude is always a non-negative number and is particularly useful when we compare the 'sizes' of different complex numbers or perform operations with them.

Argument of Complex Numbers

The argument of a complex number is the angle it makes with the positive real axis, usually denoted as \( \theta \). It can be thought of as the direction from the origin to the number in the complex plane. The argument is typically found using the arctangent function as \( \theta = \arctan(\frac{b}{a}) \), where \( a \) and \( b \) are the real and imaginary parts of the complex number, respectively.

Consider our example \( -4 - 4\sqrt{3}i \). The argument is calculated by:

• \( \theta = \arctan(\frac{-4\sqrt{3}}{-4}) \)
• \( \theta = \arctan(\sqrt{3}) \), which is \( \frac{\pi}{3} \) radians
• However, since the complex number is in the third quadrant, we add \( \pi \) to the result for the correct direction, ending up with \( \theta = \frac{4\pi}{3} \) radians.

This adjustment is crucial because the arctangent function only gives results between \( -\frac{\pi}{2} \) and \( \frac{\pi}{2} \), sometimes called a principal value. Yet, a complex number may reside in any of the four quadrants, so we may have to add \( \pi \) or use other methods to find the correct angle. This angle allows us to accurately represent the direction of our complex number on the complex plane.