Problem 63 - Precalculus with Limits

Question

In Exercises 59-64, (a) write the trigonometric forms of the complex numbers, (b) perform the indicated operation using the trigonometric forms, and (c) perform the indicated operation using the standard forms, and check your result with that of part (b). \(\dfrac{3\ +\ 4i}{1\ -\ \sqrt{3}i}\)

Step-by-Step Solution

Answer

Trigonometric form of \(3+4i\) is \(5e^{i\arctan(\frac{4}{3})}\), that of \(1-\sqrt{3}i\) is \(2e^{i\arctan(-\sqrt{3})}\). The result of the operation in trigonometric form is \(\frac{5}{2}e^{i[\arctan(\frac{4}{3}) - \arctan(-\sqrt{3})]}\). The result of the operation in standard form can be obtained by solving \(\frac{(3 + 4i)(1 + \sqrt{3}i)}{1^2 + (-\sqrt{3})^2}\).

Step 1: Write the Trigonometric Forms

First convert the complex numbers from rectangular form to polar form using the formula \(r=\sqrt{a^2+b^2}\) and \(\theta = \arctan\frac{b}{a}\). With \(z_1 = 3+4i\), we find \(r_1 = \sqrt{3^2 + 4^2} = 5\) and \(\theta_1 = \arctan(\frac{4}{3})\). With \(z_2 = 1-\sqrt{3}i\), we find \(r_2= \sqrt{1^2 + (-\sqrt{3})^2} = 2\) and \(\theta_2 = \arctan(-\sqrt{3})\). Now, convert to trigonometric form using Euler's formula \(re^{i\theta}\). So, \(z_1 = 5e^{i\arctan(\frac{4}{3})}\) and \(z_2 = 2e^{i\arctan(-\sqrt{3})}\).

Step 2: Perform Operation in Trigonometric Form

Now, we need to perform the operation (division, in our case) in the trigonometric form. For the division of two complex numbers in polar form, the formula is \(\frac{z_1}{z_2} = \frac{r_1}{r_2}e^{i(\theta_1 - \theta_2)}\)Thus, \(\frac{z_1}{z_2} = \frac{5}{2}e^{i[\arctan(\frac{4}{3}) - \arctan(-\sqrt{3})]}\).

Step 3: Perform Operation in Standard Form and check

Next, perform the operation in the standard form. The formula for dividing complex numbers is \(\frac{a + bi}{c + di} = \frac{(a + bi)(c - di)}{c^2 + d^2}\). Hence, \(\frac{3 + 4i}{1 - \sqrt{3}i} = \frac{(3 + 4i)(1 + \sqrt{3}i)}{1^2 + (-\sqrt{3})^2}\)Performing the operation, we obtain the result. Now, convert this result back to the polar form to compare it with the result from Step 2. If they are the same, then the operation has been correctly performed.

Key Concepts

Polar Form

The polar form of a complex number provides a way to express the number using a magnitude and angle, rather than just real and imaginary parts. This can often simplify mathematical operations such as multiplication and division.
To convert from rectangular to polar form:

• Calculate the magnitude, \( r \), using \( r = \sqrt{a^2 + b^2} \). Here, \( a \) and \( b \) are the real and imaginary parts, respectively.
• Determine the angle, \( \theta \), with \( \theta = \arctan\left(\frac{b}{a}\right) \).

In the given exercise:

• The complex number \( z_1 = 3 + 4i \) is transformed to \( 5e^{i\arctan\left(\frac{4}{3}\right)} \).
• Similarly, \( z_2 = 1 - \sqrt{3}i \) becomes \( 2e^{i\arctan(-\sqrt{3})} \).

This transformation not only gives us a new form but also prepares us for efficient arithmetic operations on these numbers.

Rectangular Form

Rectangular form, also known as Cartesian form, expresses complex numbers as \( a + bi \), where \( a \) is the real part and \( b \) is the imaginary part. This is a straightforward and intuitive way to think about complex numbers.
Using rectangular form can be helpful for computations requiring addition and subtraction:

• The simple components \( a \) and \( b \) make it easy to visualize numbers on the complex plane, where \( a \) is the horizontal axis, and \( b \) is the vertical axis.

In this exercise:

• \( z_1 = 3 + 4i \) and \( z_2 = 1 - \sqrt{3}i \).

These forms initially allow us to employ algebraic methods before transforming the numbers for operations like division using their polar forms.

Complex Number Division

Dividing complex numbers in rectangular form can be cumbersome, but polar form streamlines the process significantly. In polar form, you divide by adjusting the magnitudes and subtracting the angles:

• The formula for division is \( \frac{z_1}{z_2} = \frac{r_1}{r_2}e^{i(\theta_1 - \theta_2)} \).
• Here, \( r_1 \) and \( r_2 \) are the magnitudes, and \( \theta_1 \) and \( \theta_2 \) are the angles of \( z_1 \) and \( z_2 \) respectively.

This conversion helps in directly obtaining a result without complex algebraic manipulation.
Considering our example again, we find:

• \( \frac{3 + 4i}{1 - \sqrt{3}i} \) translates to \( \frac{5}{2}e^{i(\arctan(\frac{4}{3}) - \arctan(-\sqrt{3}))} \).

By visualizing division as manipulating distances and angles, the problem becomes simpler and more intuitive.

Euler's Formula

Euler's formula bridges complex analysis and trigonometry by providing an elegant expression for complex exponentiation. It states \( e^{i\theta} = \cos\theta + i\sin\theta \). This facilitates the transformation between exponential and trigonometric forms.

• By applying Euler's formula, complex numbers in polar form are written as \( re^{i\theta} \).
• This representation makes operations like multiplication and division much simpler, as angles and magnitudes deal separately.

In context, Euler's formula helps us rewrite the complex numbers from the problem:

• \( z_1 = 5e^{i\arctan(\frac{4}{3})} \)
• \( z_2 = 2e^{i\arctan(-\sqrt{3})} \)

Thus, Euler’s formula doesn't only convert forms but also makes computations more efficient and visually intuitive in many scenarios, demonstrating the unity of geometry and algebra through complex numbers.