Problem 78 - Precalculus with Limits

Question

Finding a Power of a Complex Number In Exercises \(65-80\) , use DeMoivre's Theorem to find the indicated power of the complex number. Write the result in standard form. $$(\sqrt{5}-4 i)^{3}$$

Step-by-Step Solution

Answer

The 3rd power of the complex number \( (\sqrt{5}-4i) \) is equal to \((\sqrt{21})^3 \, [\cos(3\arctan(-4/\sqrt{5})) + i\sin(3\arctan(-4/\sqrt{5}))]\).

Step 1: Find the Polar Form

We first convert the complex number \(\sqrt{5}-4i\) into its polar form. This is given by \( r\, cis\, θ \), where \( r = \sqrt{(\sqrt{5})^2 + (-4)^2} = \sqrt{21} \), and \( θ = \arctan(\frac{-4}{\sqrt{5}}) \). This gives us \( r\, cis\, θ = \sqrt{21}\,cis(\arctan(-4/\sqrt{5}))\).

Step 2: Apply De Moivre's Theorem

Now, we can apply De Moivre's theorem to find the third power of the complex number, it states that \((r\, cis\, θ)^n = r^n\, cis(nθ)\). Therefore, \( (\sqrt{21}\,cis(\arctan(-4/\sqrt{5})))^3 = (\sqrt{21})^3\,cis(3\arctan(-4/\sqrt{5}))\).

Step 3: Convert Back to Standard Form

We have used De Moivre's theorem to evaluate \( (\sqrt{5}-4i)^3 \) in polar form. The last step is to convert it back to standard form. Since \( cisθ = cos θ + i sin θ \), we get \((\sqrt{21})^3[cis(3\arctan(-4/\sqrt{5}))) = (\sqrt{21})^3 \, [\cos(3\arctan(-4/\sqrt{5})) + i\sin(3\arctan(-4/\sqrt{5}))]\). This is our answer in standard form.

Key Concepts

Complex Numbers

Complex numbers are numbers that have both a real part and an imaginary part. They are expressed in the form \( a + bi \), where \( a \) is the real part, and \( b \) is the imaginary part. The imaginary part is accompanied by the imaginary unit \( i \), where \( i \) is defined as \( \sqrt{-1} \).
This means that \( i^2 = -1 \). When dealing with complex numbers, it is important to remember that they can be added, subtracted, and multiplied just like numerical values. However, division involves multiplying by the conjugate to eliminate the imaginary part from the denominator.

• Example: \( 3 + 2i \) is a complex number where \( 3 \) is the real part, and \( 2i \) is the imaginary part.
• To add complex numbers, add the real parts together and the imaginary parts together. For instance, \((3 + 2i) + (1 + 4i) = 4 + 6i\).
• To multiply complex numbers: use the distributive property (FOIL method) and remember to replace \( i^2 \) with \(-1\). E.g., \((3 + 2i)(1 + 4i) = 3 + 12i + 2i + 8 (-1) = -5 + 14i\).

Polar Form

The polar form of a complex number is a way to express it using its magnitude and angle. This form is especially useful when performing operations such as multiplication or finding powers of complex numbers. The polar form of a complex number \( a + bi \) is expressed as \( r \text{cis}(\theta) \), where:

• \( r \) is the magnitude of the complex number, calculated as \( \sqrt{a^2 + b^2} \).
• \( \theta \) is the angle (argument) and can be found using \( \theta = \arctan\left( \frac{b}{a} \right) \).
• "cis" stands for \( \cos(\theta) + i \sin(\theta) \).

Using De Moivre's Theorem, which relates powers of complex numbers in polar form: \((r\,\text{cis}\,\theta)^n = r^n\,\text{cis}(n\theta)\), we can simplify the calculations effectively. The conversion to polar form and using De Moivre's Theorem makes finding powers and roots much simpler.

Standard Form

The standard form of a complex number is \( a + bi \), where both \( a \) and \( b \) are real numbers. After computing in polar form, it's crucial to convert results back to standard form for clarity, especially in most textbook exercises.

The conversion process includes handling trigonometric forms in polar notation, where \( \text{cis}(\theta) \) translates back to \( \cos(\theta) + i \sin(\theta) \). This means calculations done in polar form must be broken back into separate real and imaginary parts to get the result in standard form.

• For example, if you have \( r^n \text{cis}(n\theta) \), its standard form becomes \( r^n [ \cos(n\theta) + i \sin(n\theta) ] = x + yi \).
• The real part \( x \) is calculated as \( r^n \cos(n\theta) \), and the imaginary part \( y \) is \( r^n \sin(n\theta) \).

This step is crucial in mathematics because standard form is more intuitive and ensures all expressions are easily comparable, enhancing comprehension and coherence in problem-solving.