Problem 72 - 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. $$\left[3\left(\cos 60^{\circ}+i \sin 60^{\circ}\right)\right]^{4}$$

Step-by-Step Solution

Answer

The power of the given complex number is -81/2 - 41i\sqrt{3}/2.

Step 1: Convert the complex number to polar form

The complex number is already in polar form, as \(3(\cos 60^{\circ}+i \sin 60^{\circ})\). Here, the modulus is 3 and the angle is 60 degrees. So it's not necessary to convert.

Step 2: Apply DeMoivre's theorem

DeMoivre's theorem says if a complex number is in polar form z=r(cos θ +i sin θ), then for any integer n, [r(cos θ +i sin θ)]^n =[r^n(cos nθ +i sin nθ)]. Applying DeMoivre's theorem to our complex number, we get \[[3(\cos 60^{\circ}+i \sin 60^{\circ})]^4 =[3^4(\cos 4*60^{\circ} + i \sin 4*60^{\circ})]=[81(\cos 240^{\circ} +i \sin 240^{\circ})] \].

Step 3: Convert the result back to standard form

The resultant complex number is in polar form and needs to be converted to standard (Cartesian) form. We remember that \[\cos 240^{\circ} = -1/2 \]and\[ \sin 240^{\circ} = -\sqrt{3}/2.\]Thus the conversion formula gives\[81[(-1/2) + i(-\sqrt{3}/2)] = -81/2 - 81i\sqrt{3}/2.\]

Key Concepts

Complex Numbers

Complex numbers extend the idea of one-dimensional numbers like real numbers to two dimensions using an imaginary part. This is because complex numbers comprise both real and imaginary parts. A typical complex number is often denoted as \( a + bi \), where:

• \( a \) represents the real part
• \( bi \) represents the imaginary part, where \( i \) is the square root of -1

The imaginary unit \( i \) is not a real number and, by mathematical convention, \( i^2 \) is defined as \(-1\). Thus, complex numbers allow for the square root of negative numbers, expanding the capabilities of number systems. In our example, converting a complex number to polar form involves expressing it in terms of a modulus and angle, as seen in DeMoivre's Theorem.

Polar Form

Polar form provides a different way to represent complex numbers, emphasizing their magnitude and direction. Specifically, a complex number described in standard form \( a + bi \) can also be expressed in polar form as \( r(\cos \theta + i \sin \theta) \).

• \( r \) is the modulus (or magnitude) and is calculated as \( \sqrt{a^2 + b^2} \).
• \( \theta \) is the argument (or angle) of the complex number, typically measured in radians or degrees.

Polar form is particularly useful in simplifying the process of multiplying complex numbers. In the example exercise, the complex number is already in polar form, with a defined modulus and angle. DeMoivre's Theorem utilizes this form to efficiently compute powers of complex numbers.

Standard Form

Standard form (also known as Cartesian form) presents complex numbers in the familiar \( a + bi \) notation, where it consists of:

• \( a \) (the real component)
• \( bi \) (the imaginary component)

When converting from polar form back to standard form, mathematical functions \( \cos \theta \) and \( \sin \theta \) are pivotal. For instance, using the angle \( 240^\circ \) from the solved exercise, you find the real part by multiplying the modulus with \( \cos 240^\circ \) and the imaginary part with \( \sin 240^\circ \). Thus, the polar result \( 81(\cos 240^\circ + i \sin 240^\circ) \) translates into the standard form \( -\frac{81}{2} - \frac{81\sqrt{3}}{2}i \), providing a clear representation of the complex number in its rectangular coordinates.