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 and center (a, b), then 15ab/r^2 is equal to :

The correct answer is Option 3. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q4 – JEE-MAIN 2025

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Home Maths Complex & QuadraticQuestion #95

If the locus of

z \in C

, such that

Re (\frac{z - 1}{2 z + i}) + Re (\frac{\bar{z} - 1}{2 \bar{z} - i}) = 2

, is a circle of radius

r

and center

(a, b)

, then

\frac{15 a b}{r^{2}}

is equal to :

Clue 1 of 3: Properties of Complex Numbers

Recognize the relationship between $Re (w)$ and $Re (\bar{w})$ to simplify the initial equation.

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Step 1: Simplify the given equation using complex number properties✦ Active

The given equation is $Re (\frac{z - 1}{2 z + i}) + Re (\frac{\bar{z} - 1}{2 \bar{z} - i}) = 2$ . Let $w = \frac{z - 1}{2 z + i}$ . Then $\bar{w} = \frac{\bar{z} - 1}{2 \bar{z} - i}$ . The equation becomes $Re (w) + Re (\bar{w}) = 2$ . Since $Re (\bar{w}) = Re (w)$ , this simplifies to $2 Re (w) = 2$ , or $Re (w) = 1$ . Thus, we need to solve $Re (\frac{z - 1}{2 z + i}) = 1$ .

Step 2: Convert to Cartesian coordinates and find the locus○ Expand

Let $z = x + i y$ . Substitute into the simplified equation:

Re (\frac{(x - 1) + i y}{2 x + i (2 y + 1)}) = 1

Multiply the numerator and denominator by the conjugate of the denominator to find the real part:

Re (\frac{((x - 1) + i y) (2 x - i (2 y + 1))}{(2 x)^{2} + (2 y + 1)^{2}}) = 1

The real part of the numerator is $2 x (x - 1) + y (2 y + 1) = 2 x^{2} - 2 x + 2 y^{2} + y$ . So, the equation becomes:

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

This implies $2 x^{2} + 2 y^{2} - 2 x + y = 4 x^{2} + 4 y^{2} + 4 y + 1$ . Rearranging the terms gives the equation of a circle:

2 x^{2} + 2 y^{2} + 2 x + 3 y + 1 = 0

Dividing by 2, we get the standard form: $x^{2} + y^{2} + x + \frac{3}{2} y + \frac{1}{2} = 0$ .

Step 3: Determine center, radius, and the final expression○ Expand

Comparing $x^{2} + y^{2} + x + \frac{3}{2} y + \frac{1}{2} = 0$ with the general circle equation $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ , we find:

2 g = 1 \Rightarrow g = \frac{1}{2}

2 f = \frac{3}{2} \Rightarrow f = \frac{3}{4}

c = \frac{1}{2}

The center of the circle is $(a, b) = (- g, - f) = (- \frac{1}{2}, - \frac{3}{4})$ . The radius $r$ is given by $r = \sqrt{g^{2} + f^{2} - c}$ :

r = \sqrt{{(\frac{1}{2})}^{2} + {(\frac{3}{4})}^{2} - \frac{1}{2}} = \sqrt{\frac{1}{4} + \frac{9}{16} - \frac{1}{2}} = \sqrt{\frac{4 + 9 - 8}{16}} = \sqrt{\frac{5}{16}}

So, $r^{2} = \frac{5}{16}$ . Now, we calculate the expression $\frac{15 a b}{r^{2}}$ :

\frac{15 a b}{r^{2}} = \frac{15 \times (- \frac{1}{2}) \times (- \frac{3}{4})}{\frac{5}{16}} = \frac{15 \times \frac{3}{8}}{\frac{5}{16}} = \frac{45}{8} \times \frac{16}{5} = 9 \times 2 = 18

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Maths Question 4 – JEE-MAIN 2025

Clue 1 of 3: Properties of Complex Numbers

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