Finding the roots of a polynomial equation, especially one like $x^4 + 1 = 0$, can be a fascinating challenge. This equation, at first glance, might seem simple, but it hides a surprising depth in its solution. The roots of this equation are complex numbers, and understanding how to extract them requires delving into the realm of complex algebra. Let's explore how to find these roots and unravel the intricacies behind them.
The Fundamental Theorem of Algebra and Complex Numbers
The Fundamental Theorem of Algebra guarantees that a polynomial equation of degree n has exactly n roots, counting multiplicity. Since our equation, $x^4 + 1 = 0$, is a fourth-degree polynomial, we know it must have four roots. However, these roots might not be real numbers – they could be complex numbers.
Complex numbers are expressed in the form $a + bi$, where a and b are real numbers, and i is the imaginary unit, defined as $i = \sqrt{-1}$. Complex numbers are crucial for solving polynomial equations that don't have real-valued solutions.
Finding the Roots of $x^4 + 1 = 0$
To find the roots of $x^4 + 1 = 0$, we can employ several techniques, but we'll focus on two common methods:
1. Using De Moivre's Theorem
De Moivre's Theorem provides a powerful tool for finding roots of complex numbers. It states that for any complex number in polar form, $z = r(\cos \theta + i \sin \theta)$, and any integer n, the following holds:
$z^n = r^n (\cos n \theta + i \sin n \theta)$
To apply De Moivre's Theorem to find the roots of $x^4 + 1 = 0$, we first need to rewrite the equation in polar form.
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Find the modulus: The modulus of a complex number z is denoted by |z| and is calculated as the square root of the sum of the squares of its real and imaginary parts. Since our equation has a real coefficient and no imaginary term, the modulus of the root is:
$|x| = \sqrt{x^2 + 0^2} = \sqrt{x^2} = |x|$
As our equation is $x^4 + 1 = 0$, we can rewrite it as $x^4 = -1$. Since $-1$ is a real number, we can find its modulus:
$|-1| = \sqrt{(-1)^2 + 0^2} = \sqrt{1} = 1$
Therefore, the modulus of the root x is 1.
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Find the argument: The argument of a complex number is the angle it makes with the positive real axis in the complex plane. To find the argument of $-1$, we can consider its position on the complex plane. It lies on the negative real axis, making an angle of 180 degrees (or $\pi$ radians) with the positive real axis.
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Express the equation in polar form: Now we can express $-1$ in polar form:
$-1 = 1 (\cos \pi + i \sin \pi)$
Our equation now becomes:
$x^4 = 1 (\cos \pi + i \sin \pi)$
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Apply De Moivre's Theorem: We are looking for the fourth roots of -1, so we need to find the values of x that satisfy this equation. Using De Moivre's Theorem:
$x^4 = 1 (\cos \pi + i \sin \pi)$
$x = \sqrt[4]{1} [\cos ( \frac{\pi + 2k\pi}{4} ) + i \sin( \frac{\pi + 2k\pi}{4}) ]$
where k is an integer, and we've used the fact that the fourth root of 1 is 1.
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Find the four roots: By plugging in different values for k, we can find the four distinct roots:
- k = 0: $x_1 = \cos (\frac{\pi}{4}) + i \sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i$
- k = 1: $x_2 = \cos (\frac{3\pi}{4}) + i \sin(\frac{3\pi}{4}) = -\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i$
- k = 2: $x_3 = \cos (\frac{5\pi}{4}) + i \sin(\frac{5\pi}{4}) = -\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i$
- k = 3: $x_4 = \cos (\frac{7\pi}{4}) + i \sin(\frac{7\pi}{4}) = \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i$
2. Using Factorization
We can also find the roots of $x^4 + 1 = 0$ by factoring the polynomial.
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Recognize a pattern: Notice that we can rewrite $x^4 + 1$ as a difference of squares:
$x^4 + 1 = (x^2 + 1)^2 - (x^2)^2$
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Apply the difference of squares formula: The difference of squares factorization states:
$a^2 - b^2 = (a + b)(a - b)$
Applying this to our equation:
$(x^2 + 1)^2 - (x^2)^2 = (x^2 + 1 + x^2)(x^2 + 1 - x^2) = (2x^2 + 1)(1) = 2x^2 + 1$
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Solve for the roots: Now we have a simpler quadratic equation:
$2x^2 + 1 = 0$
Solving for x:
$x^2 = -\frac{1}{2}$
$x = \pm \sqrt{-\frac{1}{2}} = \pm \frac{i}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2} i$
We get two solutions:
- $x = \frac{\sqrt{2}}{2} i$
- $x = -\frac{\sqrt{2}}{2} i$
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Finding the remaining roots: Since we know $x^4 + 1$ has four roots, we need to find the remaining two. We can do this by considering that the roots of a polynomial equation always come in conjugate pairs. If a + bi is a root, then a - bi is also a root.
Since we already found two roots with imaginary components, their conjugates must also be roots:
- $x = \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} i$
- $x = -\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} i$
Conclusion
Finding the roots of $x^4 + 1 = 0$ reveals the elegance and power of complex numbers. While the equation appears simple, its solutions demonstrate the necessity of exploring the complex number system to fully understand the behavior of polynomials. By utilizing techniques like De Moivre's Theorem and factorization, we can successfully extract the roots of this equation and gain a deeper appreciation for the interconnectedness of algebra and complex analysis.
