Animated Solution for Mathematics - Complex Numbers: The number of complex numbers z, satisfying $|z|=1$ and $|\frac{z}{\overline{z}}+\frac{\overline{z}}{z}|=1$ is:

A Journey into the Argand Plane

Welcome, fellow traveler of the mathematical realms. Today, we are not just solving an equation; we are embarking on a geometric exploration. We are looking for the number of complex numbers $z$ that satisfy two specific conditions.

At first glance, this might look like a dry algebraic exercise, but I want you to see it as a dance between the unit circle and the hyperbola. Let us break this down, step by step, and uncover the hidden symmetry.

The Foundation — The Unit Circle

We begin with the condition $|z|=1$. In the Argand plane, this is the most beautiful and fundamental shape: the unit circle centered at the origin. Every complex number $z$ that satisfies this condition lies exactly on the circumference of this circle.

Recall the golden rule of complex numbers: $|z{|}^2=z\bar{z}$. Since $|z|=1$, it follows immediately that $z\bar{z}=1$.

This is our key. Whenever you see $z\bar{z}$ in an expression, you can replace it with $1$. This is the secret weapon that will simplify our intimidating second condition.

The Algebraic Dance

Now, let us look at the second condition: $|\frac{z}{\bar{z}}+\frac{\bar{z}}{z}|=1$. It looks messy, doesn't it? Fractions within a modulus can be daunting.

But remember, in JEE Advanced, complexity is often just a mask for simplicity. Let us combine these fractions by finding a common denominator:

Now, apply our golden rule. Since $z\bar{z}=1$, the denominator vanishes! We are left with the much cleaner expression: $|z^2+{\bar{z}}^2|=1$.

Take a moment to appreciate this. We have stripped away the complexity. We are no longer dealing with fractions; we are dealing with the sum of squares of a complex number and its conjugate.

The Cartesian Shift

To truly understand the geometry, we must move from the abstract to the concrete. Let us represent $z$ in its Cartesian form: $z=x+iy$. Consequently, its conjugate is $\bar{z}=x-iy$.

Let us calculate the squares:

When we add these two expressions, the imaginary parts, $2ixy$ and $-2ixy$, cancel out perfectly. This is the elegance of symmetry! We are left with:

Substituting this back into our modulus equation, we get $|2(x^2-y^2)|=1$, which simplifies to:

The Intersection of Curves

We have arrived at the heart of the problem. We are looking for the intersection of two sets of points. First, the unit circle: $x^2+y^2=1$. Second, the two rectangular hyperbolas defined by $|x^2-y^2|=\frac{1}{2}$.

This gives us two distinct cases to solve.

We have the system: 1) $x^2+y^2=1$ 2) $x^2-y^2=\frac{1}{2}$

Adding these equations, we get $2x^2=\frac{3}{2}$, so $x^2=\frac{3}{4}$. This gives $x=\pm \frac{\sqrt{3}}{2}$.

Subtracting them, we get $2y^2=\frac{1}{2}$, so $y^2=\frac{1}{4}$. This gives $y=\pm \frac{1}{2}$. This yields four points: $(\frac{\sqrt{3}}{2},\frac{1}{2}),(\frac{\sqrt{3}}{2},-\frac{1}{2}),(-\frac{\sqrt{3}}{2},\frac{1}{2}),(-\frac{\sqrt{3}}{2},-\frac{1}{2})$.

We have the system: 1) $x^2+y^2=1$ 2) $x^2-y^2=-\frac{1}{2}$

Adding these, we get $2x^2=\frac{1}{2}$, so $x^2=\frac{1}{4}$. This gives $x=\pm \frac{1}{2}$.

Subtracting them, we get $2y^2=\frac{3}{2}$, so $y^2=\frac{3}{4}$. This gives $y=\pm \frac{\sqrt{3}}{2}$. This yields another four points: $(\frac{1}{2},\frac{\sqrt{3}}{2}),(\frac{1}{2},-\frac{\sqrt{3}}{2}),(-\frac{1}{2},\frac{\sqrt{3}}{2}),(-\frac{1}{2},-\frac{\sqrt{3}}{2})$.

The Final Count

By carefully analyzing the intersection of our circle and our hyperbolas, we have identified $4+4=8$ distinct points in the Argand plane. Each of these points corresponds to a valid complex number $z$ that satisfies both original conditions.

The total number of such complex numbers is 8.