Problem 68 - Algebra and Trigonometry

Question

\(65-76=\) Find the indicated power using DeMoivre's Theorem. $$ (1-i)^{8} $$

Step-by-Step Solution

Answer

The result of \((1-i)^8\) is 16.

Step 1: Convert to Polar Form

The first step in using DeMoivre's Theorem is to express the complex number \(1 - i\) in polar form. A complex number \(a + bi\) can be converted to polar form \(r(\cos \theta + i\sin \theta)\) where \(r = \sqrt{a^2 + b^2}\) and \(\theta = \tan^{-1}(b/a)\).For \(1 - i\), \(a = 1\) and \(b = -1\). Calculate the modulus: \[r = \sqrt{1^2 + (-1)^2} = \sqrt{2}\]Calculate the argument:\[\theta = \tan^{-1}(-1/1) = \tan^{-1}(-1) = -\frac{\pi}{4}\]Thus, the polar form is: \[\sqrt{2}(\cos(-\frac{\pi}{4}) + i\sin(-\frac{\pi}{4}))\].

Step 2: Apply DeMoivre's Theorem

DeMoivre's Theorem states that for a complex number in polar form \(r(\cos \theta + i\sin \theta)\), its \(n\)-th power is:\[r^n \left(\cos(n\theta) + i\sin(n\theta)\right)\]Apply the theorem for \((1-i)^8\):We have \(r = \sqrt{2}\), \(\theta = -\frac{\pi}{4}\), and \(n = 8\).First, determine the modulus:\[r^n = (\sqrt{2})^8 = 2^4 = 16\]Determine the argument:\[n\theta = 8 \times -\frac{\pi}{4} = -2\pi\]

Step 3: Calculate Result in Polar Form

Substitute back into the polar form with the new modulus and argument:\[(1-i)^8 = 16 \left(\cos(-2\pi) + i\sin(-2\pi)\right)\]

Step 4: Simplify Using Trigonometric Identities

Since \(\cos(-2\pi) = 1\) and \(\sin(-2\pi) = 0\), the expression simplifies to:\[16(1 + i \times 0) = 16\]

Step 5: Write the Final Answer

After simplifying, we find the result:\[\boxed{16}\]

Key Concepts

Complex Numbers

Complex numbers are an extension of the real numbers and include the imaginary unit, denoted by \(i\), where \(i^2 = -1\). A complex number is generally expressed in the form \(a + bi\), where \(a\) is the real part and \(b\) is the imaginary part. In the exercise, you are dealing with the complex number \(1-i\).
This number can be thought of as a point on the complex plane, with coordinates \((1, -1)\). The real part \(a = 1\) places it on the horizontal axis, while the imaginary part \(b = -1\) places it on the vertical axis.
By understanding complex numbers in this way, you can perform operations like addition, subtraction, multiplication, and division, as well as more advanced processes like conversion to polar form.

Polar Form

The polar form of a complex number provides a different way to visualize and manipulate these numbers. Instead of representing them with real and imaginary parts, we use a modulus (or magnitude) and an argument (or angle).
The modulus \(r\) is the distance from the origin to the point \((a, b)\) in the complex plane, found using the formula \(r = \sqrt{a^2 + b^2}\). For \(1 - i\), we find \(r = \sqrt{2}\).
The argument \(\theta\) is the angle made with the positive real axis, calculated with \(\theta = \tan^{-1}\left(\frac{b}{a}\right)\). Here, \(\theta = -\frac{\pi}{4}\).
In polar form, the complex number is expressed as \(r(\cos \theta + i\sin \theta)\). This form simplifies raising the numbers to powers, as shown through DeMoivre's Theorem.

Modulus and Argument Conversion

Converting a complex number from its rectangular form \(a + bi\) to polar form \(r(\cos \theta + i\sin \theta)\) involves finding the modulus and argument, which transforms multiplication and powers of complex numbers into simpler arithmetic.

**Modulus Calculation**
- The modulus \(r\) equals the square root of the sum of the squares of the real and imaginary parts: \(r = \sqrt{a^2 + b^2}\).
- For \(1 - i\), we compute \(r = \sqrt{1^2 + (-1)^2} = \sqrt{2}\).

**Argument Calculation**
- The argument \(\theta\) is the arctangent of the imaginary part over the real part: \(\theta = \tan^{-1}\left(\frac{b}{a}\right)\).
- Here, \(\theta = \tan^{-1}\left(\frac{-1}{1}\right) = -\frac{\pi}{4}\).
Once converted, DeMoivre's Theorem helps handle complex powers with ease: raising the modulus to the power and multiplying the argument by the power.