Problem 87 - Precalculus with Limits

Question

Finding the \(n\) th Roots of a Complex Number \(\operatorname{In}\) Exercises \(81-96,(\) a) use the formula on page 446 to find the indicated roots of the complex number, (b) represent each of the roots graphically, and (c) write each of the roots in standard form. Square roots of $$-25 i$$

Step-by-Step Solution

Answer

The square roots of \(-25i\) are \(5\sqrt{2} - 5 \sqrt{2}i\) and \(-5 \sqrt{2} + 5 \sqrt{2}i\) in the standard form. They're plotted on the complex plane forming a unit circle with radius \(5\sqrt{2}\) centered at the origin.

Step 1: Conversion to Polar Form

First, convert \(-25i\) to polar form. In this case, \(r = 25\) and \(\theta = -90°\) because \(-25i\) is on the negative y-axis. So the polar form of \(-25i\) is \(25 (\cos(-90°) + i \sin(-90°))\).

Step 2: Using the Root Formula

Next, use the nth root of a complex number formula, where \(n = 2\), \(r = 25\), and \(\theta = -90°\). \[ z = \sqrt[2]{25} [\cos( \frac{-90 + 2 \pi k}{2}) + i \sin (\frac{-90 + 2 \pi k}{2})]\] For \(k = 0, 1\), we have the two roots. Calculate the values precisely.

Step 3: Calculating the roots

For \(k=0\): \[z = 5[\cos (-45°) + i \sin (-45°)] = 5 \sqrt[2]{2} - 5 \sqrt[2]{2}i\]. For \(k=1\): \[ z = 5[\cos (135°) + i \sin (135°) ] = - 5 \sqrt[2]{2} + 5 \sqrt[2]{2}i\].

Step 4: Graphical representation

To plot these roots on the complex plane, identify the real part (x-coordinate) and the imaginary part (y-coordinate) for both roots. For \( z = 5 \sqrt[2]{2} - 5 \sqrt[2]{2}i\), \(x = 5 \sqrt[2]{2}\), \(y = -5 \sqrt[2]{2}\). For \(z = - 5 \sqrt[2]{2} + 5 \sqrt[2]{2}i\), \(x = -5 \sqrt[2]{2}\), \(y = 5 \sqrt[2]{2}\). They form a circle with radius \(5\sqrt[2]{2}\) centered at origin on complex plane.

Step 5: Convert Roots to Standard Form

The roots obtained above are in the standard form 'a + bi'. Specifically, they are \(5\sqrt{2} - 5 \sqrt{2}i\) and \(-5 \sqrt{2} + 5 \sqrt{2}i\).

Key Concepts

n-th roots

The concept of n-th roots in complex numbers refers to finding all possible values (roots) that, when raised to the power of \(n\), yield the original complex number. To find these n-th roots, we use a formula based on the polar representation of complex numbers. The formula for the n-th root is:

• \( z_k = \sqrt[n]{r} \left( \cos \left( \frac{\theta + 2\pi k}{n} \right) + i \sin \left( \frac{\theta + 2\pi k}{n} \right) \right) \)

Where \(k\) is an integer between 0 and \(n-1\), \(r\) is the modulus (or magnitude) of the complex number, and \(\theta\) is the argument (or angle). The challenge is to compute all n distinct roots, as done in our exercise with \(n = 2\) for \(-25i\). By systematically using this formula and changing \(k\), we can uncover all possible roots lying on a circle in the complex plane.

polar coordinates

Polar coordinates are a crucial part of understanding complex numbers. They transform the standard form \(a + bi\) into a form involving a magnitude and an angle. A complex number can be expressed in polar coordinates as:

• \(z = r(\cos \theta + i \sin \theta)\)

Where \(r\) is the magnitude (distance from origin to the point) and \(\theta\) is the angle with the positive x-axis. The conversion from rectangular form to polar involves:

• Calculating \(r = \sqrt{a^2 + b^2}\)
• Finding the angle \(\theta = \tan^{-1}\left(\frac{b}{a}\right)\)

For our exercise involving \(-25i\), the magnitude \(r\) was found to be 25, and the angle \(\theta\) was \(-90°\), placing the point on the negative y-axis.

complex plane

The complex plane is a two-dimensional plane used to visually represent complex numbers. Each complex number can be seen as a point or a vector in this plane. The horizontal axis signifies the real part, while the vertical axis represents the imaginary part.

• Real part of \(a+bi\) is \(a\) (x-axis)
• Imaginary part of \(a+bi\) is \(b\) (y-axis)

Our task shows two distinct roots: each root can be plotted as a point where its coordinates are the real and imaginary components. By plotting the calculated roots \(5 \sqrt{2} - 5 \sqrt{2}i\) and \(-5 \sqrt{2} + 5 \sqrt{2}i\), one can see the symmetric nature and the circle with radius \(5\sqrt{2}\) these points lie on. This visualization aids in better understanding the roots' distribution.

standard form

Standard form is the basic format for expressing complex numbers, displayed as \(a + bi\), where \(a\) is the real part and \(b\) is the imaginary part. This form is advantageous for straightforward operations like addition or subtraction.

• Real number part: \(a\)
• Imaginary number part: \(b\) (coupled with \(i\))

In our step-by-step solution, the calculated roots were: \(5\sqrt{2} - 5 \sqrt{2}i\) and \(-5 \sqrt{2} + 5 \sqrt{2}i\). Each is correctly arranged in standard form, illustrating their real and imaginary components. Converting from polar back to standard form can sometimes require extra steps but provides clarity for performing other arithmetic actions.