If the locus of z ∈ C, such that Re((z-1)/(2z+i)) + Re((z̄-1)/(2z̄-i)) = 2 is a circle of radius r a

Q: Wait, why can we just say $2\text{Re}(\frac{z-1}{2z+i}) = 2$? Is that always true?

Yes! It is a beautiful property of complex numbers. For any complex number $w$, $\text{Re}(w) = \text{Re}(\bar{w})$. Since the second term in your equation is the conjugate of the first, adding them is just adding the real part to itself. It is a gift from the symmetry of the complex plane—always look for these shortcuts before diving into heavy algebra!

Q: Why do we have to multiply by the conjugate? Can't we just equate the real parts directly?

That is a classic trap! You cannot equate real parts directly when the denominator is complex. The real part of a fraction $\frac{A}{B}$ is not simply $\frac{\text{Re}(A)}{\text{Re}(B)}$. You must make the denominator a real number first to isolate the real part of the entire expression. Think of it as clearing the 'fog' of the imaginary unit from the denominator.

Q: I got the equation $2x^2 + 2y^2 + 2x + 3y + 1 = 0$. How do I find the center and radius from this?

You are almost there! Divide the entire equation by 2 to get the standard form $x^2 + y^2 + x + \frac{3}{2}y + \frac{1}{2} = 0$. Now compare this to the general circle equation $x^2 + y^2 + 2gx + 2fy + c = 0$. The center is $(-g, -f)$ and the radius is $\sqrt{g^2 + f^2 - c}$. Just be careful with the signs when you extract $g$ and $f$!

Question

If the locus of $z \in \mathbb{C}$, such that $	ext{Re}(\frac{z-1}{2z+i}) + 	ext{Re}(\frac{\bar{z}-1}{2\bar{z}-i}) = 2$ is a circle of radius $r$ and center $(a, b)$ then $\frac{15ab}{r^2}$ is equal to :

Options:

(A) 24

(B) 12

(C) 18

(D) 16

Sigma Percentile · Accepted Answer

### The Geometry of Complex Numbers: A Journey into Loci

Welcome, fellow traveler of the JEE Advanced path. Today, we are not just solving an equation; we are uncovering a hidden geometry.

Often, when we see complex numbers, we panic. We see $z$, we see $	ext{Re}$, we see conjugates, and we feel the urge to start calculating immediately. But hold on.

In mathematics, as in life, observation is the precursor to mastery. Before you touch your pen to paper, look at the structure.

### Phase 1: The Symmetry Insight

Look at the given equation: $	ext{Re}(\frac{z-1}{2z+i}) + 	ext{Re}(\frac{\bar{z}-1}{2\bar{z}-i}) = 2$. Do you see the elegance?

The second term is the complex conjugate of the first. There is a profound symmetry here. We know that for any complex number $w$, the real part of $w$ is identical to the real part of its conjugate, $\bar{w}$.

Therefore, $	ext{Re}(w) + 	ext{Re}(\bar{w}) = 2	ext{Re}(w)$. By recognizing this, we instantly collapse the complexity.

We are left with $2	ext{Re}(\frac{z-1}{2z+i}) = 2$, which simplifies to the clean, manageable statement: $	ext{Re}(\frac{z-1}{2z+i}) = 1$. This is the moment where the problem stops being a monster and starts being a puzzle.

### Phase 2: The Cartesian Leap

Now, we must translate this into the language of the Cartesian plane. We substitute $z = x + iy$.

Our numerator becomes $(x-1) + iy$, and our denominator becomes $2x + i(2y+1)$. We are now staring at a fraction with an imaginary unit in the denominator.

This is a "no-go" zone in complex algebra. To fix this, we perform the ritual of rationalization. We multiply the numerator and the denominator by the conjugate of the denominator: $2x - i(2y+1)$.

This is the key that unlocks the door. By doing this, the denominator transforms into a purely real number: $(2x)^2 + (2y+1)^2$, which is $4x^2 + (2y+1)^2$.

### Phase 3: The Algebraic Grind

I know this part feels tedious, but stay with me. We only care about the real part of the numerator.

When we multiply the complex numbers, the real part is formed by the product of the real components and the product of the imaginary components (remembering that $i^2 = -1$).

Expanding this, we get the numerator's real part as $(x-1)(2x) + y(2y+1)$, which simplifies to $2x^2 - 2x + 2y^2 + y$. Now, we set this equal to our denominator (since the whole expression equals 1).

We have:

$$\frac{2x^2 - 2x + 2y^2 + y}{4x^2 + (2y+1)^2} = 1$$

Cross-multiplying gives us $2x^2 - 2x + 2y^2 + y = 4x^2 + 4y^2 + 4y + 1$.

### Phase 4: The Circle Emerges

Now, we gather our terms. Moving everything to one side, we arrive at $2x^2 + 2y^2 + 2x + 3y + 1 = 0$.

Divide by 2, and we see the familiar face of a circle: $x^2 + y^2 + x + \frac{3}{2}y + \frac{1}{2} = 0$.

Comparing this to the general form $x^2 + y^2 + 2gx + 2fy + c = 0$, we identify $g = \frac{1}{2}$, $f = \frac{3}{4}$, and $c = \frac{1}{2}$.

The center $(a, b)$ is $(-g, -f)$, which gives us $(-\frac{1}{2}, -\frac{3}{4})$. The radius squared, $r^2$, is calculated via $g^2 + f^2 - c$, resulting in $\frac{1}{4} + \frac{9}{16} - \frac{1}{2} = \frac{5}{16}$.

### The Final Victory

We have arrived at the finish line. We need to evaluate $\frac{15ab}{r^2}$.

Substituting our values:

$$\frac{15(-\frac{1}{2})(-\frac{3}{4})}{\frac{5}{16}} = \frac{\frac{45}{8}}{\frac{5}{16}} = \frac{45}{8} 	imes \frac{16}{5} = 18$$

And there it is. **18**. It wasn't just about the calculation; it was about seeing the symmetry, handling the algebra with care, and trusting the process. You have conquered this locus.

Animated Solution for Mathematics - Complex Numbers: If the locus of $z \in C$ , such that $Re (\frac{z - 1}{2 z + i}) + Re (\frac{z ˉ - 1}{2 z ˉ - i}) = 2$ is a circle of radius $r$ and center $(a, b)$ then $\frac{15 ab}{r ^{2}}$ is equal to :

Select Answer:

Visualized Solution

The Symmetry Property

Cartesian Substitution

Rationalizing the Fraction

Finding the Real Part

Formulating the Algebraic Equation

Expanding and Rearranging

Standard Circle Equation

Identifying the Center $(a, b)$

Calculating the Radius Squared $r^{2}$

Final Evaluation of $\frac{15 ab}{r ^{2}}$

The Sigma Insight: Geometrical Applications of Complex Numbers

The Geometry of Complex Numbers

Phase 1

Phase 2

Phase 3

Phase 4

The Final Victory

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