To find the number of complex numbers \( z \), that satisfy \( |z| = 1 \) and
\(\left| \frac{z}{\overline{z}} + \frac{\overline{z}}{z} \right| = 1,\)
we need to approach the problem using the properties of complex numbers and their magnitudes.
Given that \( |z| = 1 \), this implies \( z \) lies on the unit circle in the complex plane. So, if \( z = a + ib \), then \( a^2 + b^2 = 1 \).
The expression \( \frac{z}{\overline{z}} + \frac{\overline{z}}{z} \) simplifies as follows:
Let \( z = e^{i\theta} \), thus \( \overline{z} = e^{-i\theta} \).
Then,
\(\frac{z}{\overline{z}} = e^{2i\theta}\) and \(\frac{\overline{z}}{z} = e^{-2i\theta}\).
Therefore:
\(\frac{z}{\overline{z}} + \frac{\overline{z}}{z} = e^{2i\theta} + e^{-2i\theta} = 2\cos(2\theta).\)
We need the absolute value of the above expression:
\(|2\cos(2\theta)| = 1.\)
This simply happens when:
\(\cos(2\theta) = \pm \frac{1}{2}.\)
The solutions to \(\cos(2\theta) = \pm \frac{1}{2}\) are:
\[ 2\theta = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{5\pi}{3}, \frac{4\pi}{3}, \frac{7\pi}{3}, \frac{8\pi}{3}, \cdots \]
Hence, \( \theta = \frac{\pi}{6}, \frac{\pi}{3}, \frac{5\pi}{6}, \frac{2\pi}{3}, \frac{7\pi}{6}, \frac{4\pi}{3}, \frac{11\pi}{6}, \frac{5\pi}{3} \).
These angles represent 8 distinct solutions for \( z \), all having \( |z| = 1 \).
Therefore, the number of complex numbers \( z \) satisfying both conditions is 8.
We are given that \( |z| = 1 \). This means that the complex number \( z \) lies on the unit circle in the complex plane.
Any complex number on the unit circle can be written as: \( z = e^{i\theta} \), where \( \theta \) is the argument (angle) of \( z \).
The equation provided is:
\[ \left| \frac{z}{\overline{z}} + \frac{\overline{z}}{z} \right| = 1 \]
We know:
Now substitute into the equation:
\[ \frac{z}{\overline{z}} = \frac{e^{i\theta}}{e^{-i\theta}} = e^{2i\theta}, \quad \frac{\overline{z}}{z} = \frac{e^{-i\theta}}{e^{i\theta}} = e^{-2i\theta} \]
Now the equation becomes:
\[ \left| e^{2i\theta} + e^{-2i\theta} \right| = 1 \]
We use the identity: \( e^{ix} + e^{-ix} = 2\cos x \)
Applying this, we get:
\[ e^{2i\theta} + e^{-2i\theta} = 2\cos(2\theta) \Rightarrow |2\cos(2\theta)| = 1 \]
Divide both sides by 2:
\[ |\cos(2\theta)| = \frac{1}{2} \]
The cosine of an angle is \( \pm \frac{1}{2} \) at the following values within \( 0 \leq 2\theta < 2\pi \):
So we have 8 distinct values of \( 2\theta \) that satisfy the condition.
Divide each \( 2\theta \) by 2 to get the angle \( \theta \):
There are 8 distinct complex numbers \( z \) that satisfy the given condition.

What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,