Question

If $\frac{z+1}{z+i}$ is purely imaginary, then z lies on a

• A. straight lone
• B. circle
• C. Circle with radius 1
• D. circle passing through (1, 1).

Answer 1

The correct option is B circle

$\frac{z+1}{z+i}$ is purely imaginary

$\Rightarrow Re\left(\frac{z+1}{z+i}\right)=0$

Say $z=x+iy$

then,

$Re\left(\frac{x+iy+1}{x+iy+i}\right)=0$

$\Rightarrow Re\left(\frac{(x+1)+iy}{x+i(y+1)}\right)=0$

$\Rightarrow Re\left[\frac{[(x+1)+iy][x-i(y+1)]}{x^2+{(y+1)}^2}\right]=0$

$\Rightarrow Re\left[\frac{x(x+1)+y(y+1)+i(xy-(x+1)(y+1))}{x^2+{(y+1)}^2}\right]=0$

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

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

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

$\Rightarrow x^2+y^2+2.\frac{1}{2}x+2.\frac{1}{2}y=0$

The above is a circle with centre $(\frac{-1}{2},\frac{-1}{2})$ and radius $\sqrt{\frac{1}{4}+\frac{1}{4}}=\sqrt{\frac{1}{2}}$ units

Option B.