keep z= x+iy and square both the sides to get,
(√2x+2y+√2-1)(√2x-2y+√2+1) = 0
Hence the the locus of z is pair of straight lines i.e.
all solutions to,
√2x+2y+√2-1 = 0 , √2x-2y+√2+1=0
\hspace{-16}$Find Complex no. $\mathbf{z}$ Which satisfy the equation \\\\ $\mathbf{\left|\frac{z-\bar{z}-i}{z+\bar{z}+2}\right|=\frac{\sqrt{2}}{2}}$
Is the locus of ' z ' coming out to be straight line? If it is correct, solution follows.
keep z= x+iy and square both the sides to get,
(√2x+2y+√2-1)(√2x-2y+√2+1) = 0
Hence the the locus of z is pair of straight lines i.e.
all solutions to,
√2x+2y+√2-1 = 0 , √2x-2y+√2+1=0