Problem 35 - Algebra and Trigonometry

Question

\(25-48=\) Write the complex number in polar form with argument \(\theta\) between 0 and 2\(\pi .\) $$ 4 \sqrt{3}-4 i $$

Step-by-Step Solution

Answer

Polar form is \( 8\left(\cos\frac{11\pi}{6} + i\sin\frac{11\pi}{6}\right).\)

Step 1: Determine the Modulus

To write a complex number in polar form, first find the modulus. The modulus of a complex number \( a + bi \) is given by \( \sqrt{a^2 + b^2} \). For the complex number \( 4\sqrt{3} - 4i \), calculate the modulus: \[ \sqrt{(4\sqrt{3})^2 + (-4)^2} = \sqrt{48 + 16} = \sqrt{64} = 8. \] So, the modulus is 8.

Step 2: Determine the Argument

Next, find the argument \( \theta \). The argument \( \theta \) is given by \( \theta = \tan^{-1}\left(\frac{b}{a}\right) \), where \( a \) and \( b \) are the real and imaginary parts, respectively. For \( 4\sqrt{3} - 4i \), we calculate: \[ \theta = \tan^{-1}\left(\frac{-4}{4\sqrt{3}}\right) = \tan^{-1}\left(\frac{-1}{\sqrt{3}}\right). \] The angle with tangent \( \frac{-1}{\sqrt{3}} \) is \( -\frac{\pi}{6} \). Since we need \( 0 \leq \theta < 2\pi \), add \( 2\pi \) to get \( \theta = \frac{11\pi}{6} \).

Step 3: Write in Polar Form

Combine the modulus and argument to express the complex number in polar form: \( r(\cos\theta + i\sin\theta) \). Thus, the polar form of \( 4\sqrt{3} - 4i \) is: \[ 8\left(\cos\frac{11\pi}{6} + i\sin\frac{11\pi}{6}\right). \] Alternatively, you can write this using Euler's formula as \( 8e^{i\frac{11\pi}{6}} \).

Key Concepts

Modulus of a Complex Number

The modulus of a complex number is like its "distance" from the origin in a coordinate plane. Think of it as finding out how far away the number is from the center, just like measuring the length of a line from the center to the point. To find the modulus of a complex number given in the form \( a + bi \), use the formula: \[ \sqrt{a^2 + b^2} \] Here, \( a \) represents the real part and \( b \) represents the imaginary part of the number. This formula is really a direct application of the Pythagorean theorem.
For instance, with the complex number \( 4\sqrt{3} - 4i \), locate the real part \( 4\sqrt{3} \) and the imaginary part \( -4 \). Next, plug them into the formula: \[ \sqrt{(4\sqrt{3})^2 + (-4)^2} = \sqrt{48 + 16} = \sqrt{64} = 8. \] Therefore, the modulus is 8. This tells us that the point \( 4\sqrt{3} - 4 \) is 8 units away from the origin in the complex plane.

Argument of a Complex Number

The argument of a complex number is like a direction. It tells us which way to go from the origin to reach our complex number on the plane. The argument \( \theta \) is the angle made with the positive real axis. This angle can vary between \( 0 \) and \( 2\pi \) for full coverage of the circle. To find the argument for a number \( a + bi \), use: \[ \theta = \tan^{-1}\left(\frac{b}{a}\right) \] This formula calculates the angle from the positive x-axis to the line connecting the origin and the point \( a + bi \).
For \( 4\sqrt{3} - 4i \), insert \( a = 4\sqrt{3} \) and \( b = -4 \) to gain the argument:\[ \theta = \tan^{-1}\left(\frac{-4}{4\sqrt{3}}\right) = \tan^{-1}\left(\frac{-1}{\sqrt{3}}\right) \] Here, \( \tan^{-1}\left(\frac{-1}{\sqrt{3}}\right) = -\frac{\pi}{6} \). Adjust the angle to ensure it's within the range \(0\) to \(2\pi\), so add \(2\pi \) to get \( \theta = \frac{11\pi}{6} \). This angle points us directly to \( 4\sqrt{3} - 4i \) on the complex plane.

Trigonometric Form of Complex Numbers

In mathematics, expressing a complex number in trigonometric, or polar, form simplifies many operations, like multiplication and division. We combine both the modulus and argument to rewrite a complex number as: \[ r(\cos \theta + i\sin \theta) \] This expression uses the magnitude \( r \), which is our modulus, and the angle \( \theta \), which is our argument.
For the number \( 4\sqrt{3} - 4i \), we've found \( r = 8 \) and \( \theta = \frac{11\pi}{6} \). Thus, its polar form becomes: \[ 8\left(\cos \frac{11\pi}{6} + i\sin \frac{11\pi}{6}\right) \] It may also be expressed using Euler's formula, which states \( e^{i\theta} = \cos \theta + i\sin \theta \). Hence, another way to write it is: \[ 8e^{i\frac{11\pi}{6}} \] This trigonometric form not only is a neat representation but also facilitates complex calculations, especially when dealing with powers and roots of complex numbers.