Problem 54 - Algebra and Trigonometry

Question

Write each expression in rectangular form \(x+\) yi and in exponential form \(r e^{i \theta} .\) $$ (\sqrt{3}-i)^{6} $$

Step-by-Step Solution

Answer

Rectangular form: \(-64\). Exponential form: \(64 e^{-i \pi}\).

Step 1: Convert to Polar Form

First, convert \(\sqrt{3} - i\) to polar form. Identify \(r\) which is the modulus: \[r = \sqrt{(\sqrt{3})^2 + (-1)^2} = \sqrt{3 + 1} = 2.\] Next, find the argument \(\theta\): \[ \theta = \tan^{-1} \left(\frac{-1}{\sqrt{3}} \right) = -\frac{\pi}{6}.\] Therefore, the polar form is \[ 2 e^{-i \pi / 6 }. \]

Step 2: Apply De Moivre's Theorem

Use De Moivre's Theorem to raise the number to the 6th power: \[ (2 e^{-i \pi / 6 })^6 = 2^6 e^{-i 6 ( \pi / 6 )} = 64 e^{-i \pi}. \]

Step 3: Simplify the Exponential Form

Simplify the expression \(64 e^{-i \pi}\) using the exponential definition: \[ 64 e^{-i \pi} = 64 (\cos(-\pi) + i\sin(-\pi)) = 64 (-1 + i \cdot 0) = -64. \]

Step 4: Write in Rectangular Form

Since \(-64\) is real, the rectangular form is \(-64+i \cdot 0 = -64.\)

Key Concepts

Polar Form

Polar form represents complex numbers in terms of their magnitude and angle. It’s incredibly useful for multiplication and division.

Here’s how you convert a complex number to polar form:

• Identify the magnitude or modulus, which is the distance from the origin. For a complex number \( a + bi \), it's given by \( r = \sqrt{a^2 + b^2} \).
• Determine the argument, which is the angle \( \theta \), using \( \theta = \tan^{-1}\left(\frac{b}{a}\right) \).

In the exercise, \( \sqrt{3} - i \) is first converted to polar form. The modulus \( r \) is found as \sqrt{2}\ and the argument \( \theta \) as \( -\frac{\pi}{6} \). Therefore, its polar form is \ 2e^{-i\pi/6 } \.

De Moivre's Theorem

De Moivre's Theorem is crucial for raising complex numbers to a power. It states:

For a complex number \( r \text{cis} \theta \), \( (r \text{cis} \theta)^n = r^n \text{cis}(n \theta) \).

In the exercise, the theorem is used to raise the polar form of \( \sqrt{3} - i \) to the 6th power. Applying this, we get:

• \$r^6 = 2^6 = 64\$.
• And, the angle becomes \ -6 \frac{\pi}{6} = -\pi\.

This results in \( 64e^{-i \pi} \), which is the simplified exponential form.

Rectangular Form

Rectangular form writes complex numbers as \( a + bi \) where \( a \) is the real part and \( b \) is the imaginary part. It's the most straightforward way to represent complex numbers but less convenient for some operations.

In the exercise, we need to convert the exponential form \( 64e^{-i\pi} \) back to rectangular form:

• Use Euler's formula: \ e^{i\theta} = \cos{\theta} + i\sin{\theta}, \ which gives \ 64e^{-i\pi} = 64 (\cos{ -\pi} + i\sin{ -\pi}).\
• Since \ \cos{ -\pi} = -1 \ and \ \sin{ -\pi} = 0 \, the result simplifies to \ 64 (-1 + 0i) = -64. \.

Thus, in rectangular form, it is \ -64 \.

Exponential Form

Exponential form of a complex number utilizes Euler's formula: \( re^{i\theta} \), which makes many operations simpler. It's especially handy for multiplication, division, and exponentiation.

The exercise simplifies \( (2e^{-i\pi/6})^6 \) to \( 64e^{-i\pi} \) using De Moivre's Theorem.

• Exponential form captures both the magnitude \(r\) and the direction (angle) \(\theta\) succinctly.
• It ties together polar and rectangular forms through Euler's relationship: \ e^{i\theta} = \cos{\theta} + i\sin{\theta}.

This allows us to convert back to rectangular form easily, as shown in the solution. Therefore, \(64e^{-i\pi}\rightarrow -64 \).