If Re((z−1)/(2z+i)) = 1, Where z = x + iy, Then the Point (x, y) Lies On

JEE Main 2020 — 7 January, Morning Shift. Previous Year Question.

Problem

If $\operatorname{Re}\left(\dfrac{z-1}{2z+i}\right)=1$, where $z=x+iy$, then the point $(x,y)$ lies on:

Key insight. Rationalising the fraction by multiplying through by the denominator’s conjugate turns the whole expression into standard $a+ib$ form — from there, “the real part equals $1$” is just one algebraic equation in $x$ and $y$, which simplifies directly into a recognisable conic.

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Approach

Substitute $z=x+iy$ into the fraction, then multiply numerator and denominator by the conjugate of the denominator to separate the expression into real and imaginary parts. Set the real part equal to $1$, clear the denominator, and simplify into the general circle equation $x^2+y^2+2gx+2fy+c=0$. Read off the centre and radius, and compute the diameter to match against the given options.

Solution

Step 1 — Substitute z = x + iy

$$\frac{z-1}{2z+i} = \frac{(x-1)+iy}{2x+i(2y+1)}$$

Step 2 — Rationalise by multiplying by the conjugate of the denominator

$$\frac{\left[(x-1)+iy\right]\left[2x-i(2y+1)\right]}{4x^2+(2y+1)^2}$$

Step 3 — Expand the numerator and extract the real part

$$\left[(x-1)+iy\right]\left[2x-i(2y+1)\right] = 2x(x-1)+y(2y+1) + i\left[2xy-(x-1)(2y+1)\right]$$

So the real part of the whole fraction is:

$$\operatorname{Re}\left(\frac{z-1}{2z+i}\right) = \frac{2x(x-1)+y(2y+1)}{4x^2+(2y+1)^2} = \frac{2x^2-2x+2y^2+y}{4x^2+(2y+1)^2}$$

Step 4 — Set this equal to 1 and cross-multiply

$$2x^2-2x+2y^2+y = 4x^2+(2y+1)^2 = 4x^2+4y^2+4y+1$$

Step 5 — Rearrange into the general circle form

$$2x^2-2x+2y^2+y-4x^2-4y^2-4y-1 = 0$$

$$-2x^2-2x-2y^2-3y-1 = 0$$

Dividing by $-2$:

$$x^2+y^2+x+\frac32y+\frac12 = 0$$

Step 6 — Identify the centre and radius

Comparing with $x^2+y^2+2gx+2fy+c=0$: $2g=1\implies g=\dfrac12$; $2f=\dfrac32\implies f=\dfrac34$; $c=\dfrac12$.

Centre $=(-g,-f) = \left(-\dfrac12,-\dfrac34\right)$

$$\text{Radius} = \sqrt{g^2+f^2-c} = \sqrt{\frac14+\frac{9}{16}-\frac12} = \sqrt{\frac{4}{16}+\frac{9}{16}-\frac{8}{16}} = \sqrt{\frac{5}{16}} = \frac{\sqrt5}{4}$$

Step 7 — Compute the diameter

$$\text{Diameter} = 2\times\frac{\sqrt5}{4} = \frac{\sqrt5}{2}$$

Answer

Common mistakes

• Trying to extract the real part without rationalising the denominator first. A fraction with a complex denominator doesn’t have an obvious “real part” until it’s rewritten in the form $a+ib$ — multiplying by the conjugate is the essential first step, not an optional shortcut.
• Mixing up the centre formula. For $x^2+y^2+2gx+2fy+c=0$, the centre is $(-g,-f)$ — using $(g,f)$ directly (without the negative signs) gives a centre matching none of the given options, including the tempting-but-wrong option that names $\left(-\dfrac12,-\dfrac32\right)$ instead of the correct $\left(-\dfrac12,-\dfrac34\right)$.

Practise next

• If $\operatorname{Re}\left(\dfrac{z+1}{z-i}\right)=2$, where $z=x+iy$, find the locus of $(x,y)$, using the same rationalise-and-simplify approach.