Problem 55 - A Graphical Approach to Precalculus with Limits

Question

Find all \(n\) complex solutions of each equation of the form \(x^{n}=k\) $$x^{2}=-1$$

Step-by-Step Solution

Answer

The solutions are \(i\) and \(-i\).

Step 1: Understand the Problem

We need to find all complex numbers \(x\) that satisfy the equation \(x^2 = -1\). This means finding all numbers whose square is equal to \(-1\).

Step 2: Use Complex Number Representation

Represent a complex number \(x\) in polar form as \(x = re^{i\theta}\), where \(r\) is the magnitude and \(\theta\) is the argument of the complex number. Here, we know that the magnitude squared should equal 1 so \(r=1\). Therefore, the complex number can be represented as \(e^{i\theta}\).

Step 3: Calculate Possible Angles

Since \(x^2 = -1\), we have \((e^{i\theta})^2 = e^{i2\theta} = -1\). In polar form, \(-1\) can be represented as \(e^{i(\pi + 2k\pi)}\), where \(k\) is any integer because each full rotation around the circle (\(2\pi\)) gets us back to \(-1\).

Step 4: Solve for \(\theta\)

From the equation \(e^{i2\theta} = e^{i(\pi+2k\pi)}\), equate the exponents: \(2\theta = \pi + 2k\pi\). We solve for \(\theta\) as \(\theta = \frac{\pi}{2} + k\pi\).

Step 5: Find Distinct Solutions

For \(n=2\), there are two unique solutions for \(\theta\) within one full rotation (\(0\) to \(2\pi\)). Setting \(k = 0\), we have \(\theta = \frac{\pi}{2}\). Setting \(k = 1\), we find \(\theta = \frac{3\pi}{2}\). These correspond to the complex numbers \(e^{i\frac{\pi}{2}} = i\) and \(e^{i\frac{3\pi}{2}} = -i\).

Step 6: Verify the Solutions

Check each solution: \((i)^2 = -1\) and \((-i)^2 = -1\). Both numbers satisfy the original equation, so these are the correct solutions.

Key Concepts

Complex Solutions

Complex solutions involve finding solutions to equations in the complex number plane. When dealing with complex numbers, we step beyond the real numbers, which only cover the simple number line to include numbers like \(i\), the imaginary unit. In the context of the exercise, we need to find solutions to the equation \(x^2 = -1\), which have no real solutions.

• Since \(x^2 = -1\), we're looking for numbers that when squared equals \(-1\).
• No real number squared gives a negative result, thus solutions must be complex.

This is one reason why complex numbers are valuable; they provide solutions where real numbers cannot. This leads us to use polar form to identify these complex solutions.

Polar Form

The polar form of a complex number offers a way to express complex numbers using a combination of a magnitude and an angle. In our exercise, we represent the complex number in polar form so it can be more easily manipulated mathematically.

• A complex number \(x = a + bi\) can be represented as \(x = re^{i\theta}\) in polar form.
• Here, \(r\) is the magnitude of the complex number, and \(\theta\) is its argument (angle).

For the equation \(x^2 = -1\), by using the polar form \(re^{i\theta}\), it becomes easier to solve for \(x\) by equating magnitudes and arguments to known values, which allows us to systematically find all possible solutions.

Magnitude and Argument

Understanding magnitude and argument is crucial when working with complex numbers in polar form. Magnitude, denoted as \(r\), is the distance from the origin to the complex number on the complex plane, while argument, denoted as \(\theta\), is the angle measured from the positive real axis to the line representing the complex number.

• For \(x^2 = -1\), we find that magnitude \(r = 1\) as \(-1\) lies on the unit circle.
• The argument changes since it represents the direction of the complex number.

By solving \(2\theta = \pi + 2k\pi\), where \(k\) is an integer, we find distinct angles yielding the complex numbers \(i\) and \(-i\), both having a magnitude of 1.

Euler's Formula

Euler's formula provides a powerful connection between complex exponentials and trigonometry. It states that any complex number can be expressed as \(e^{i\theta} = \cos(\theta) + i\sin(\theta)\). This form is quite handy in solving equations like \(x^2 = -1\).

• Euler's formula relates exponential functions to trigonometric functions, bridging gaps between different mathematical concepts.
• By expressing \(-1\) as \(e^{i\pi}\), we convert trigonometric tasks into easier algebraic manipulations.

Euler's formula simplifies complex exponentials to help us easily solve for potential solutions in the form of \(e^{i\theta}\), leading directly to the solutions \(i\) and \(-i\).