Let the complex numbers α and 1/ᾱ lie on the circles |z - z0|² = 4 and |z - z0|² = 16 respectively,

Q: Wait, why do we reject the negative root here? Or do we even have one?

In this specific problem, we are working with the modulus squared, $|\alpha|^2$. By definition, the square of a modulus must be a non-negative real number. When we solve the resulting linear equation for $|\alpha|^2$, we get a single positive value, $1/5$. There is no quadratic equation here, so no negative roots to worry about! It's a beautiful, linear path.

Q: Why can't we just use energy conservation or some other physics principle?

That's the spirit of a physicist! However, this is a pure complex geometry problem. While you could interpret these as vectors in a 2D plane, the algebraic properties of complex numbers (like $z\bar{z} = |z|^2$) are far more efficient than coordinate geometry. Stick to the complex algebra—it's the 'energy conservation' of this problem.

Q: Is there a geometric shortcut? This algebra feels like a lot of work.

You could potentially use the concept of circle inversion, but that is often more prone to error for students. The algebraic method we used is robust, systematic, and less likely to lead to a sign error. In JEE Advanced, the most reliable path is often the one that relies on fundamental algebraic identities.

Question

Let the complex numbers $\alpha$ and $\frac{1}{\bar{\alpha}}$ lie on the circles $|z - z_0|^2 = 4$ and $|z - z_0|^2 = 16$ respectively, where $z_0 = 1 + i$. Then, the value of $100|\alpha|^2$ is.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path to JEE mastery! Today, we are going to unravel a problem that sits at the beautiful intersection of geometry and algebra.

We are looking at complex numbers $\alpha$ and $\frac{1}{\bar{\alpha}}$ dancing on two concentric circles. It might look intimidating, but I promise you, by the end of this, you will see the elegance hidden within the equations.

Imagine the complex plane with a center $z_0 = 1 + i$. Two circles are drawn around this center. The first, smaller circle has a radius squared of $4$, and the second, larger circle has a radius squared of $16$.

Our complex number $\alpha$ is resting on the smaller circle, while $\frac{1}{\bar{\alpha}}$ is on the larger one. Our mission is to find $100|\alpha|^2$. To do this, we need to translate these geometric positions into the language of algebra.

### The Algebraic Bridge

The most powerful tool in our arsenal for complex numbers is the identity $|z|^2 = z\bar{z}$. This allows us to turn a distance-based geometric statement into a product of a complex number and its conjugate.

Since $\alpha$ is on the circle $|z - z_0|^2 = 4$, we can write:

$$(\alpha - z_0)(\bar{\alpha} - \bar{z}_0) = 4$$

Expanding this, we get:

$$\alpha\bar{\alpha} - \alpha\bar{z}_0 - \bar{\alpha}z_0 + z_0\bar{z}_0 = 4$$

Recognizing that $\alpha\bar{\alpha} = |\alpha|^2$ and $z_0\bar{z}_0 = |z_0|^2$, we have:

$$|\alpha|^2 - (\alpha\bar{z}_0 + \bar{\alpha}z_0) + |z_0|^2 = 4$$

Since $z_0 = 1 + i$, we know $|z_0|^2 = 1^2 + 1^2 = 2$. Substituting this in, we get our first crucial relation:

$$|\alpha|^2 - (\alpha\bar{z}_0 + \bar{\alpha}z_0) = 2$$

Let's call this **Equation 1**. It is a simple, clean expression that captures the essence of the first circle.

### The Second Circle and the Inversion

Now, let's look at the second circle. The point $\frac{1}{\bar{\alpha}}$ lies on $|z - z_0|^2 = 16$. Using the same logic, we write:

$$\left(\frac{1}{\bar{\alpha}} - z_0ight)\left(\overline{\frac{1}{\bar{\alpha}}} - \bar{z}_0ight) = 16$$

Remember that the conjugate of $\frac{1}{\bar{\alpha}}$ is simply $\frac{1}{\alpha}$. So, the equation becomes:

$$\left(\frac{1}{\bar{\alpha}} - z_0ight)\left(\frac{1}{\alpha} - \bar{z}_0ight) = 16$$

Expanding this product, we get:

$$\frac{1}{\alpha\bar{\alpha}} - \frac{\bar{z}_0}{\bar{\alpha}} - \frac{z_0}{\alpha} + z_0\bar{z}_0 = 16$$

This simplifies to:

$$\frac{1}{|\alpha|^2} - \frac{\alpha\bar{z}_0 + \bar{\alpha}z_0}{|\alpha|^2} + 2 = 16$$

Subtracting $2$ from both sides, we get:

$$\frac{1 - (\alpha\bar{z}_0 + \bar{\alpha}z_0)}{|\alpha|^2} = 14$$

This is **Equation 2**. Notice the symmetry; the term $(\alpha\bar{z}_0 + \bar{\alpha}z_0)$ is the same as in Equation 1.

### The Grand Unification

We are almost there. From Equation 1, we know that $(\alpha\bar{z}_0 + \bar{\alpha}z_0) = |\alpha|^2 - 2$. Let's substitute this into Equation 2:

$$\frac{1 - (|\alpha|^2 - 2)}{|\alpha|^2} = 14$$

Simplifying the numerator:

$$1 - |\alpha|^2 + 2 = 14|\alpha|^2$$

$$3 - |\alpha|^2 = 14|\alpha|^2$$

$$3 = 15|\alpha|^2$$

$$|\alpha|^2 = \frac{3}{15} = \frac{1}{5}$$

The final step is just a simple calculation. We need $100|\alpha|^2$. Substituting our value:

$$100 	imes \frac{1}{5} = 20$$

And there it is! The answer is **20**. It is a beautiful result, isn't it? By trusting the algebra and staying organized, we turned a complex geometric problem into a simple, solvable equation.

Animated Solution for Mathematics - Complex Numbers: Let the complex numbers $α$ and $\frac{1}{α ˉ}$ lie on the circles $∣ z - z_{0} ∣^{2} = 4$ and $∣ z - z_{0} ∣^{2} = 16$ respectively, where $z_{0} = 1 + i$ . Then, the value of $100∣ α ∣^{2}$ is.

Enter Numerical Value:

Visualized Solution

Visualizing the Setup

Locating the Points

The Modulus Squared Property

Equation for Circle 1

Expanding Equation 1

Simplifying Equation 1

Equation for Circle 2

Expanding Equation 2

Clearing the Denominator

Eliminating the Common Term

Solving for $∣ α ∣^{2}$

Final Calculation

The Sigma Insight: Geometrical Applications of Complex Numbers

Analyzing the Setup

The Algebraic Bridge

The Second Circle and the Inversion

The Grand Unification

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