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

Step 1: Simplify the given equation

We are given an equation involving the real part of complex fractions. We can use the property that for any complex number $w$, $\Re(w) = \Re(\overline{w})$. Also, $\overline{\left(\frac{z-1}{2z+i}\right)} = \frac{\overline{z}-1}{2\overline{z}-i}$. Therefore, the second term is the conjugate of the first term, which means both terms have the same real part.

Step 2: Express in terms of $x$ and $y$

Let $z = x + iy$. Substitute this into the simplified equation $2\Re\left(\frac{z-1}{2z+i}\right) = 2$, which simplifies to $\Re\left(\frac{z-1}{2z+i}\right) = 1$. Then, multiply the numerator and denominator by the conjugate of the denominator to isolate the real part.

Step 3: Extract the real part

After multiplying the numerator and denominator, the real part of the numerator is $(x-1)(2x) + y(y) = 2x^2 - 2x + y^2$. The denominator is $(2x)^2 + (2y+1)^2 = 4x^2 + (2y+1)^2$. Setting the real part equal to 1 gives the equation of the locus.

Step 4: Rearrange into standard circle equation

Multiply both sides by the denominator and rearrange the terms to form the general equation of a circle. This involves moving all terms to one side and simplifying.

Step 5: Find center and radius

Divide the entire equation by 2 to get $x^2 + x + y^2 + \frac{3}{2}y + \frac{1}{2} = 0$. Then, complete the square for both $x$ and $y$ terms to find the center $(a,b)$ and radius $r$. The center is $(-\frac{1}{2}, -\frac{3}{4})$ and $r^2 = \frac{5}{16}$.

Step 6: Calculate the final expression

Substitute the values of $a = -\frac{1}{2}$, $b = -\frac{3}{4}$, and $r^2 = \frac{5}{16}$ into the expression $\frac{15ab}{r^2}$ and simplify to get the final answer.

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