Problem 30 - Precalculus : Building Concepts and Connections

Question

Use De Moivre's Theorem to find each expression. $$(2-2 i)^{4}$$

Step-by-Step Solution

Answer

The expression \((2 - 2i)^{4}\) simplifies to \(-32\) when solved using De Moivre's Theorem.

Step 1: Convert the complex number to polar form

The first step is to convert the complex number into polar form. The polar form of a complex number is \(r(\cos \Theta + i \sin \Theta)\), r is the modulus and \(\Theta\) is the angle made with the positive x-axis. Modulus r = \(\sqrt{(real\ part)^{2} + (imaginary\ part)^{2}}\) and \(\Theta = \arctan(\frac{imaginary\ part}{real\ part})\). So for our complex number \(2 - 2i\), the modulus r = \( \sqrt{2^{2} + ((-2)^{2})} = \sqrt{8} = 2\sqrt{2}\) and \(\Theta = \arctan(\frac{-2}{2}) = - 45^{\circ}\). That means the polar form of the complex number \(2 - 2i\) is \(2\sqrt{2}(\cos(-45^{\circ}) - i\sin(45^{\circ}))\).

Step 2: Apply the De Moivre's Theorem

According to the De Moivre's Theorem, to raise a complex number to a power, we just need to raise the modulus to the power and multiply the argument by the power. So, \((2\sqrt{2} (\cos(-45)-i\sin(-45))^ {4}\) = \((2\sqrt{2})^{4} (\cos(-4*45^{\circ}) - i\sin(-4*45^{\circ}))\). That simplifies to \(32(\cos(-180^{\circ}) - i\sin(-180^{\circ}))\).

Step 3: Convert back to Cartesian form

The last step is to convert back the solution into Cartesian form (a + ib). So, \(32(\cos(-180^{\circ}) - i\sin(-180^{\circ}))\) converts to \(32(-1) = -32\).

Key Concepts

Complex Numbers

Complex numbers are expressions that include both a real and an imaginary part. For instance, in the case of the complex number \(2 - 2i\):

• The real part is 2.
• The imaginary part is \-2i (where \(i\) is the imaginary unit defined as the square root of \-1).

Complex numbers are typically written in the form \(a + bi\), where \(a\) is the real part, and \(b\) is the coefficient of the imaginary part \(i\). These numbers are fundamental in advanced mathematics due to their applications in fields like engineering and physics.
They allow for a broader range of solutions and analyses than just real numbers alone. When dealing with complex numbers, we often need to understand them in both Cartesian form and other representations, such as the polar form.

Polar Form

Polar form is another way to represent complex numbers, highlighting their magnitude and angle. This approach is valuable when performing operations like multiplication and exponentiation.
In polar form, a complex number is expressed as \(r(\cos \Theta + i \sin \Theta)\), where:

• \(r\) is the modulus or magnitude, calculated as \(\sqrt{(\text{real part})^2 + (\text{imaginary part})^2}\).
• \(\Theta\) is the argument or the angle with the positive x-axis, found using \( \Theta = \arctan(\text{im_part}/\text{re_part})\).

For example, in converting \(2 - 2i\) to polar form, we find \(r = 2\sqrt{2}\), and \(\Theta = -45^\circ\). Thus, the polar form is \(2\sqrt{2} (\cos(-45^\circ) + i \sin(-45^\circ))\).
This representation simplifies the process of raising complex numbers to any power, thanks to De Moivre's Theorem.

Cartesian Form

After performing complex number operations in polar form, it’s often necessary to convert the result back to the more common Cartesian form—this allows for easier interpretation and application. In Cartesian form, a complex number appears as \(a + ib\), where \(a\) is the real component, and \(b\) represents the imaginary part.
For instance, when raising \(2 - 2i\) to the fourth power using De Moivre's Theorem, we eventually convert the polar result back to understand it in its Cartesian form. In this exercise:

• After applying the theorem, \(32(\cos(-180^\circ) + i \sin(-180^\circ))\) becomes \(32(-1) + i(0)\).
• This simplifies to \(-32\), dropping the imaginary component entirely in the final result.

This conversion back to Cartesian form presents a clear end-point for the calculation, encapsulating the operation's physical and real-world meaning.