Question

Consider a complex number $w=\frac{z-i}{2z+1}$, where $z=x+iy$ and $x,yϵR$. If the complex number w is purely imaginary then the locus of z is,

• A. a straight line
• B. a circle with centre $(\frac{-1}{4},\frac{-1}{2})$ and radius $\frac{\sqrt{3}}{4}$
• C. a circle with centre $(\frac{1}{4},\frac{1}{2})$ and passing through origin.
• D. neither a circle nor a straight line.

Answer 1

The correct option is B a circle with centre $(\frac{-1}{4},\frac{-1}{2})$ and radius $\frac{\sqrt{3}}{4}$

$w=\frac{z-1}{2z+1}$

$=\frac{(x+iy)-i}{2(x+iy)+1}=\frac{x+i(y-1)}{(2x+1)+i2y}$

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

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

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

it is purely imaginary s. Re (z) = 0

$\frac{x(2x+1)2y(y+1)}{(2x+1{)}^{2}4y^2}=0$

$2x^2+x+2y^2+2y=0$

$2x^2+2y^2+x+y=0$

$x^2+y^2+\frac{x}{2}+y=0$

$C:(\frac{-1}{4},\frac{-1}{2})$

$r=\sqrt{\frac{1}{16}+\frac{1}{8}-0}=\sqrt{\frac{1+2}{16}}=\frac{\sqrt{3}}{4}$