Solving complex number equations often involves manipulating the properties of complex numbers and utilizing various techniques to simplify and isolate the unknown variable. One such equation that presents an interesting challenge is $z^3 = \overline{z}$, where $z$ represents a complex number and $\overline{z}$ denotes its complex conjugate. This equation requires us to understand the relationship between a complex number and its conjugate, as well as the properties of exponents and complex multiplication. We will explore a step-by-step solution to this equation and illustrate the key concepts involved.
Understanding Complex Numbers and Their Conjugates
Before diving into the solution, let's define some key concepts:
- Complex Number: A complex number is a number that can be expressed in the form $z = a + bi$, where $a$ and $b$ are real numbers, and $i$ is the imaginary unit ($i^2 = -1$).
- Complex Conjugate: The complex conjugate of a complex number $z = a + bi$ is denoted as $\overline{z}$ and is defined as $\overline{z} = a - bi$. In other words, the conjugate is obtained by changing the sign of the imaginary part.
The key property we will use is that the product of a complex number and its conjugate is always a real number:
$z \cdot \overline{z} = (a + bi)(a - bi) = a^2 - (bi)^2 = a^2 + b^2$
Solving the Equation $z^3 = \overline{z}$
Let's break down the solution process:
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Express z in polar form: Any complex number can be expressed in polar form as $z = r(\cos \theta + i \sin \theta)$, where $r$ is the magnitude of the complex number and $\theta$ is its angle (argument) with respect to the positive real axis.
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Find the complex conjugate: The conjugate of $z$ in polar form is simply $\overline{z} = r(\cos \theta - i \sin \theta)$.
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Substitute into the equation: Substituting the polar form of $z$ and $\overline{z}$ into the equation $z^3 = \overline{z}$, we get:
$ [r(\cos \theta + i \sin \theta)]^3 = r(\cos \theta - i \sin \theta)$
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Simplify using De Moivre's Theorem: De Moivre's Theorem states that for any complex number in polar form and any integer $n$,
$[r(\cos \theta + i \sin \theta)]^n = r^n(\cos n\theta + i \sin n\theta)$
Applying this theorem to our equation, we get:
$r^3(\cos 3\theta + i \sin 3\theta) = r(\cos \theta - i \sin \theta)$
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Equate real and imaginary parts: For two complex numbers to be equal, their real and imaginary parts must be equal. Therefore, we can equate the real and imaginary parts of both sides of the equation:
- Real parts: $r^3 \cos 3\theta = r \cos \theta$
- Imaginary parts: $r^3 \sin 3\theta = -r \sin \theta$
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Solve for r and θ:
- Case 1: r = 0 If $r = 0$, then $z = 0$, which is a solution to the equation.
- Case 2: r ≠ 0 Divide both sides of the equations by $r$ (since $r \neq 0$):
- $r^2 \cos 3\theta = \cos \theta$
- $r^2 \sin 3\theta = -\sin \theta$
- Solve for θ: Divide the second equation by the first equation:
- $\tan 3\theta = -\tan \theta$
- Using trigonometric identities, we can rewrite this as $\tan 3\theta + \tan \theta = 0$
- This simplifies to $\frac{\sin 4\theta}{\cos 3\theta \cos \theta} = 0$
- Therefore, $\sin 4\theta = 0$, which means $4\theta = k\pi$, where $k$ is an integer.
- This gives us $\theta = \frac{k\pi}{4}$
- Solve for r: Substitute the values of $\theta$ back into either of the equations for $r$ and solve. This will yield different values for $r$ depending on the value of $k$.
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General solution: Combining both cases, the general solution to the equation $z^3 = \overline{z}$ is:
- Case 1: $z = 0$
- Case 2: $z = r(\cos \frac{k\pi}{4} + i \sin \frac{k\pi}{4})$, where $k$ is an integer and $r$ is a solution to the equation $r^2 \cos \frac{3k\pi}{4} = \cos \frac{k\pi}{4}$.
Conclusion
Solving complex number equations like $z^3 = \overline{z}$ involves a combination of understanding the properties of complex numbers, utilizing various techniques like polar form and De Moivre's Theorem, and carefully manipulating trigonometric identities to simplify the equation and obtain a general solution. This process demonstrates the intricate relationship between complex numbers and their conjugates, and highlights the power of algebraic and trigonometric tools in solving complex problems. The solution reveals that the equation has infinitely many solutions, each represented by a complex number with a specific magnitude and angle, along with the trivial solution of $z = 0$.
