Demystifying the Equation: w 5/4 Equal to -2 – A Journey into Complex Solutions

The equation w 5/4 equal to -2 presents an interesting challenge in mathematics. At first glance, it might seem puzzling because raising any real number to the 5/4 power typically yields a nonnegative result. In this post, we explain why there is no real solution for this equation and illustrate how to find its complex solutions by expressing the negative number in polar form.

Understanding the Equation: w 5/4 Equal to -2

When we write the equation in exponent notation, it reads as:

w54=−2.w^{\frac{5}{4}} = -2.w45​=−2.

For any positive real number $w$, w54w^{\frac{5}{4}}w45​ is positive. If we were negative, the operation becomes ambiguous because the 5/4 power involves taking a fourth root (an even index) of a number, which is not defined for negative values in the set of real numbers. Thus, no real number $w$ satisfies the equation, and we must seek solutions in the complex plane.

Solving for w Using Complex Numbers

To solve the equation w54=−2w^{\frac{5}{4}} = -2w45​=−2 in the complex domain, we first express $-2$ in polar form. Recall that any complex number can be written as:

reiθ,r e^{i\theta},reiθ,

where $r$ is the magnitude and $\theta$ is the argument (angle).

For $-2$:

• The magnitude is $r=2$.
• The principal argument is $\theta =\pi$ (since $-2$ lies on the negative real axis).

Thus, we can write:

−2=2 ei(π+2πk),k∈Z.-2 = 2\,e^{i(\pi + 2\pi k)}, \quad k \in \mathbb{Z}.−2=2ei(π+2πk),k∈Z.

The general solution for $w$ is found by raising both sides of the equation to the power 45\frac{4}{5}54​:

w=(2 ei(π+2πk))45=245 ei45(π+2πk).w = \left(2\,e^{i(\pi + 2\pi k)}\right)^{\frac{4}{5}} = 2^{\frac{4}{5}}\, e^{i\frac{4}{5}(\pi + 2\pi k)}.w=(2ei(π+2πk))54​=254​ei54​(π+2πk).

Since the exponent $5/4$ originally suggests that there will be 5 distinct solutions (because the exponent’s denominator is 5 when written in lowest terms), we typically consider $k=0,1,2,3,4$ to obtain all unique complex roots.

Interpretation of the Complex Solutions

The solutions for $w$ are given by:

wk=245 ei(4π5+8πk5),k=0,1,2,3,4.w_k = 2^{\frac{4}{5}}\, e^{i\left(\frac{4\pi}{5} + \frac{8\pi k}{5}\right)}, \quad k = 0, 1, 2, 3, 4.wk​=254​ei(54π​+58πk​),k=0,1,2,3,4.

These five distinct complex numbers represent all the solutions to the equation w54=−2w^{\frac{5}{4}} = -2w45​=−2. Each solution is evenly spaced around a circle in the complex plane with a radius 2452^{\frac{4}{5}}254​.

Frequently Asked Questions (FAQs)

1. Why is there no real solution for w 5/4 equal to -2?
   Raising a real number to the 5/4 power always results in a nonnegative number if the number is positive, and if the base is negative, the operation becomes undefined (due to the even root in the exponent). Hence, no real $w$ can satisfy the equation w54=−2w^{\frac{5}{4}} = -2w45​=−2.

2. How do we solve w54=−2w^{\frac{5}{4}} = -2w45​=−2 using complex numbers?
   We express $-2$ in polar form as 2 ei(π+2πk)2\,e^{i(\pi + 2\pi k)}2ei(π+2πk) and then raise both sides to the 45\frac{4}{5}54​ power. This yields w=245 ei45(π+2πk)w = 2^{\frac{4}{5}}\, e^{i\frac{4}{5}(\pi + 2\pi k)}w=254​ei54​(π+2πk) for $k=0,1,2,3,4$, which are the five complex solutions.

3. What does the notation 2452^{\frac{4}{5}}254​ represent?
   It represents the 5th root of 242^424 (or the 5th root of 16). This is the magnitude of each complex solution and is a positive real number.

4. How many solutions does the equation have, and why?
   The equation w54=−2w^{\frac{5}{4}} = -2w45​=−2 has 5 distinct complex solutions because the fractional exponent 54\frac{5}{4}45​ implies that the equation can be rewritten in a form that produces 5 roots in the complex plane (reflecting the periodicity of the complex exponential function).

5. In what areas of mathematics or engineering might such equations be useful?
   Equations with fractional exponents and complex solutions frequently appear in fields like signal processing, control systems, and anywhere complex dynamics are studied. They also serve as important examples in courses on complex analysis and advanced algebra.

Conclusion

The equation w 5/4 equal to -2 may seem perplexing at first due to its fractional exponent and negative right-hand side. However, by moving into the realm of complex numbers, we can solve it using polar representation and exponentiation rules. The five complex solutions derived not only illustrate the beauty of complex analysis but also demonstrate how extending the number system allows us to tackle problems that have no real solutions.