Let z 1, z 2 and z 3 be non-zero complex numbers satisfying z 2 = i z̅ , where i = √-1
Consider the following statements: 1. z 1 z 2 z 3 is purely imaginary. 2. z 1z 2 + z 2z 3 + z 3z 1 is purely real.
Both 1 and 2
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:
\[ (re^{i\theta})^2 = i (re^{-i\theta}) \] \[ r^2 e^{i2\theta} = e^{i\pi/2} r e^{-i\theta} \] \[ r^2 e^{i2\theta} = r e^{i(\pi/2 - \theta)} \]By equating the magnitudes and arguments of both sides, we get two equations:
- Magnitude: \(r^2 = r\). Since \(r > 0\), we can divide by \(r\) to get \(r = 1\).
- Argument: \(2\theta = \frac{\pi}{2} - \theta + 2k\pi\), where \(k\) is an integer.
Solving the argument equation:
\[ 3\theta = \frac{\pi}{2} + 2k\pi \] \[ \theta = \frac{\pi}{6} + \frac{2k\pi}{3} \]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:
\[ z_1 z_2 z_3 = e^{i\pi/6} \cdot e^{i5\pi/6} \cdot e^{i3\pi/2} \] \[ z_1 z_2 z_3 = e^{i(\pi/6 + 5\pi/6 + 3\pi/2)} \]Adding the arguments:
\[ \frac{\pi}{6} + \frac{5\pi}{6} + \frac{3\pi}{2} = \frac{6\pi}{6} + \frac{3\pi}{2} = \pi + \frac{3\pi}{2} = \frac{2\pi + 3\pi}{2} = \frac{5\pi}{2} \]So, the product is:
\[ z_1 z_2 z_3 = e^{i5\pi/2} \]We can simplify \(e^{i5\pi/2}\) using the property \(e^{i(\theta + 2k\pi)} = e^{i\theta}\):
\[ e^{i5\pi/2} = e^{i(2\pi + \pi/2)} = e^{i2\pi} \cdot e^{i\pi/2} = 1 \cdot i = i \]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:
\[ z_1z_2 + z_2z_3 + z_3z_1 = (-1) + \left(\frac{1}{2} + \frac{\sqrt{3}}{2}i\right) + \left(\frac{1}{2} - \frac{\sqrt{3}}{2}i\right) \] \[ z_1z_2 + z_2z_3 + z_3z_1 = -1 + \frac{1}{2} + \frac{1}{2} + \frac{\sqrt{3}}{2}i - \frac{\sqrt{3}}{2}i \] \[ z_1z_2 + z_2z_3 + z_3z_1 = -1 + 1 + 0i \] \[ z_1z_2 + z_2z_3 + z_3z_1 = 0 \]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.
| Statement | Result | Purely Real/Imaginary? | Correctness |
|---|---|---|---|
| \(z_1 z_2 z_3\) | \(i\) | Purely Imaginary | Correct |
| \(z_1z_2 + z_2z_3 + z_3z_1\) | \(0\) | Purely Real | Correct |
Revision Table: Key Concepts for Complex Numbers
| Concept | Description | Formula/Property |
|---|---|---|
| Complex Number | A number of the form \(x + iy\), where \(x\) and \(y\) are real numbers, and \(i^2 = -1\). | \(z = x + iy\) |
| Purely Real Number | A complex number where the imaginary part is zero. | \(z = x + 0i = x\) |
| Purely Imaginary Number | A complex number where the real part is zero. | \(z = 0 + iy = iy\) |
| Polar Form | Representing a complex number by its magnitude \(r\) and argument \(\theta\). | \(z = r(\cos\theta + i\sin\theta) = re^{i\theta}\) |
| Complex Conjugate | For \(z = x + iy\), the conjugate is \(\bar{z} = x - iy\). In polar form, if \(z = re^{i\theta}\), then \(\bar{z} = re^{-i\theta}\). | \(\bar{z} = x - iy\) or \(re^{-i\theta}\) |
| Product in Polar Form | Multiply magnitudes and add arguments. | \(z_1 z_2 = r_1 r_2 e^{i(\theta_1 + \theta_2)}\) |
| De Moivre's Theorem | For an integer \(n\), \( (e^{i\theta})^n = e^{in\theta}\). | \((\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)\) |
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.
Important Questions from Properties of Complex Numbers
If z z̅ = |z + z̅ |, where z = x + iy, i = \(\sqrt{-1}\), then the locus of z is a pair of:
What is the value of \(\sqrt{12+5 i}+\sqrt{12-5 i}\) where \(i=\sqrt{-1}\) ?
If z is a complex number such that \(\frac{z-1}{z+1}\) is purely imaginary, then what is |z| equal to ?
What is the real part of (sin x + icos x) 3
What is z 1+ z 2+ z 3equal to?
