Solving the Quadratic Equation (z^2 4z (4 - i) 0): A Detailed Guide

Mathematics is an integral part of engineering and science, where solving equations is a fundamental skill. This article presents a comprehensive approach to solving the quadratic equation (z^2 4z (4 - i) 0), highlighting the techniques and steps involved. We will explore the application of the quadratic formula, substitution, and complex number theory.

Introduction to the Equation

The given equation, (z^2 4z (4 - i) 0), is a quadratic equation in the variable (z). The standard form of a quadratic equation is (az^2 bz c 0). Here, (a 1), (b 4), and (c 4 - i), where (i) is the imaginary unit. Plugging these values into the quadratic formula, (z frac{-b pm sqrt{b^2 - 4ac}}{2a}), we can solve for (z).

Using the Quadratic Formula

First, let's apply the quadratic formula step by step:

Identify the coefficients: (a 1), (b 4), and (c 4 - i). Calculate the discriminant: (b^2 - 4ac). Solve for (z) using the quadratic formula: (z frac{-4 pm sqrt{4^2 - 4 cdot 1 cdot (4 - i)}}{2 cdot 1})

Calculating the discriminant:

(4^2 - 4 cdot 1 cdot (4 - i) 16 - 4(4 - i) 16 - 16 4i 4i)

Therefore, the discriminant is (4i).

Applying the quadratic formula:

(z frac{-4 pm sqrt{4i}}{2})

Since the discriminant is (4i), we need to handle the square root of a complex number. This requires us to express the square root of a complex number in terms of its magnitude and argument.

Substitution and Simplification

Instead of expanding out the right side and solving a messy quadratic, we introduce a substitution: (z k - 2). This substitution simplifies the equation by transforming it into a more manageable form. Let's see why this substitution is useful:

Creating a difference of squares on the right-hand side. Matching the constant term on both sides of the equation.

Now, let's proceed with the substitution:

Given (z k - 2), substitute and solve for (k)

(k - 2)**2 4(k - 2) (4 - i) 0)

Expanding the equation:

(k^2 - 4k 4 4k - 8 4 - i 0)

(k^2 0 - i 0)

(k^2 i)

Therefore, (k sqrt{i}).

Resolving (sqrt{i})

To resolve (sqrt{i}), we can use DeMoivre's theorem or any other technique. Here, we use DeMoivre's theorem:

(sqrt{i} sqrt{e^{ipi/2}} e^{ipi/4} cos(pi/4) isin(pi/4) frac{sqrt{2}}{2} ifrac{sqrt{2}}{2})

Thus, the two solutions for (k) are:

(k frac{sqrt{2}}{2} ifrac{sqrt{2}}{2})

(k frac{sqrt{2}}{2} - ifrac{sqrt{2}}{2})

Back Substitution

Now, substitute back for (z):

If (k frac{sqrt{2}}{2} ifrac{sqrt{2}}{2}), then:

(z left(frac{sqrt{2}}{2} ifrac{sqrt{2}}{2}right) - 2)

(z -frac{3sqrt{2}}{2} ifrac{sqrt{2}}{2})

And, if (k frac{sqrt{2}}{2} - ifrac{sqrt{2}}{2}), then:

(z left(frac{sqrt{2}}{2} - ifrac{sqrt{2}}{2}right) - 2)

(z -frac{3sqrt{2}}{2} - ifrac{sqrt{2}}{2})

Therefore, the solutions to the equation (z^2 4z (4 - i) 0) are:

(z -2 frac{sqrt{2}}{2} ifrac{sqrt{2}}{2})

(z -2 - frac{sqrt{2}}{2} - ifrac{sqrt{2}}{2})

Conclusion

We explored the process of solving a quadratic equation involving complex numbers. Using the quadratic formula and employing a substitution to simplify the equation, we found the solutions. Understanding these methods is crucial in various fields such as electrical engineering, physics, and more.

For more insights on solving equations and complex number manipulations, refer to the resources linked below.