If the real part of ( z̅ + 2)/ ( z̅ – 1) is 4,

According to the question,

Let z = x + iy

Now,

$\frac{\bar{z}+2}{\bar{z}-1}=\frac{x-i y+2}{x-i y-1}$

$=\frac{[(x+2)-i y][(x-1)+i y]}{[(x-1)-i y][(x-1)+i y]}$

$=\frac{(x-1)(x+2)+y^{2}+i[(x+2) y-(x-1) y]}{(x-1)^{2}+y^{2}}$

According to the question, we have, real part $=4$.

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

⇒ x2 + x – 2 + y2 = 4 (x2 – 2x + 1 + y2)

⇒ 3x2 + 3y2 – 9x + 6 = 0

The equation obtained represents the equation of a circle.

Hence, locus of z is a circle.