Solving Complex Numbers Problem: Analyzing \(z^2 = i\bar{z}\) Properties

The problem asks us to consider complex numbers \(z_1, z_2, z_3\) that satisfy the equation \(z^2 = i\bar{z}\) and determine the correctness of two statements regarding their products and sums.

First, let's find the complex numbers \(z\) that satisfy the given equation \(z^2 = i\bar{z}\). We can express \(z\) in polar form, \(z = re^{i\theta}\), where \(r = |z|\) is the magnitude and \(\theta = \arg(z)\) is the argument. Since \(z\) is non-zero, \(r > 0\).

The equation becomes:

By equating the magnitudes and arguments of both sides, we get two equations:

1. Magnitude: \(r^2 = r\). Since \(r > 0\), we can divide by \(r\) to get \(r = 1\).
2. Argument: \(2\theta = \frac{\pi}{2} - \theta + 2k\pi\), where \(k\) is an integer.

Solving the argument equation:

For distinct arguments in the interval \([0, 2\pi)\), we can take \(k = 0, 1, 2\):

• For \(k=0\): \(\theta_1 = \frac{\pi}{6}\)
• For \(k=1\): \(\theta_2 = \frac{\pi}{6} + \frac{2\pi}{3} = \frac{\pi + 4\pi}{6} = \frac{5\pi}{6}\)
• For \(k=2\): \(\theta_3 = \frac{\pi}{6} + \frac{4\pi}{3} = \frac{\pi + 8\pi}{6} = \frac{9\pi}{6} = \frac{3\pi}{2}\)

The three non-zero complex numbers satisfying the equation \(z^2 = i\bar{z}\) are the roots with magnitude 1 and arguments \(\frac{\pi}{6}, \frac{5\pi}{6}, \frac{3\pi}{2}\). Let these roots be \(z_1, z_2, z_3\):

• \(z_1 = 1 \cdot e^{i\pi/6} = \cos(\frac{\pi}{6}) + i\sin(\frac{\pi}{6}) = \frac{\sqrt{3}}{2} + \frac{1}{2}i\)
• \(z_2 = 1 \cdot e^{i5\pi/6} = \cos(\frac{5\pi}{6}) + i\sin(\frac{5\pi}{6}) = -\frac{\sqrt{3}}{2} + \frac{1}{2}i\)
• \(z_3 = 1 \cdot e^{i3\pi/2} = \cos(\frac{3\pi}{2}) + i\sin(\frac{3\pi}{2}) = 0 - i = -i\)

Analyzing Statement 1: \(z_1 z_2 z_3\) is purely imaginary

Let's calculate the product \(z_1 z_2 z_3\). Using the polar form makes multiplication easier:

Adding the arguments:

So, the product is:

We can simplify \(e^{i5\pi/2}\) using the property \(e^{i(\theta + 2k\pi)} = e^{i\theta}\):

The product \(z_1 z_2 z_3 = i\). A complex number is purely imaginary if its real part is zero. Since \(i = 0 + 1i\), its real part is 0 and its imaginary part is 1. Thus, \(z_1 z_2 z_3\) is purely imaginary.

Statement 1 is correct.

Analyzing Statement 2: \(z_1z_2 + z_2z_3 + z_3z_1\) is purely real

Let's calculate the sum of the products of pairs of roots. First, calculate the individual products:

• \(z_1 z_2 = e^{i\pi/6} \cdot e^{i5\pi/6} = e^{i(\pi/6 + 5\pi/6)} = e^{i6\pi/6} = e^{i\pi} = -1\)
• \(z_2 z_3 = e^{i5\pi/6} \cdot e^{i3\pi/2} = e^{i(5\pi/6 + 3\pi/2)} = e^{i(5\pi/6 + 9\pi/6)} = e^{i14\pi/6} = e^{i7\pi/3}\). Since \(7\pi/3 = 2\pi + \pi/3\), \(e^{i7\pi/3} = e^{i\pi/3} = \cos(\pi/3) + i\sin(\pi/3) = \frac{1}{2} + \frac{\sqrt{3}}{2}i\)
• \(z_3 z_1 = e^{i3\pi/2} \cdot e^{i\pi/6} = e^{i(3\pi/2 + \pi/6)} = e^{i(9\pi/6 + \pi/6)} = e^{i10\pi/6} = e^{i5\pi/3}\). Since \(5\pi/3 = 2\pi - \pi/3\), \(e^{i5\pi/3} = e^{-i\pi/3} = \cos(-\pi/3) + i\sin(-\pi/3) = \cos(\pi/3) - i\sin(\pi/3) = \frac{1}{2} - \frac{\sqrt{3}}{2}i\)

Now, let's sum these products:

The result is 0. A complex number is purely real if its imaginary part is zero. Since \(0 = 0 + 0i\), its imaginary part is 0. Thus, \(z_1z_2 + z_2z_3 + z_3z_1\) is purely real.

Statement 2 is correct.

Conclusion

Both Statement 1 (\(z_1 z_2 z_3\) is purely imaginary) and Statement 2 (\(z_1z_2 + z_2z_3 + z_3z_1\) is purely real) are correct based on our analysis of the complex numbers satisfying \(z^2 = i\bar{z}\).

Therefore, the correct option is the one stating that both 1 and 2 are correct.

Revision Table: Key Concepts for Complex Numbers

Additional Information: Roots of Complex Equations and Properties

The equation \(z^2 = i\bar{z}\) is an example of an equation involving both \(z\) and \(\bar{z}\). Solving such equations often involves converting to polar form, which simplifies expressions involving magnitudes and arguments.

The three roots we found, \(e^{i\pi/6}, e^{i5\pi/6}, e^{i3\pi/2}\), are distinct and have magnitude 1. Geometrically, they lie on the unit circle in the complex plane.

• The product \(z_1 z_2 z_3\) resulted in \(i\). This means the product of these three complex numbers lies on the positive imaginary axis.
• The sum of pairwise products \(z_1z_2 + z_2z_3 + z_3z_1\) resulted in \(0\). This is a purely real number.

These properties are specific to the complex numbers satisfying the given equation. For arbitrary complex numbers, these statements would not generally hold true.