Question

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

• A. straight line
• B. circle
• C. circle with radius $=$ 1/$\sqrt{2}$
• D. circle passing through the origin

Answer 1

Answer: (B), (C), (D)

The correct options are
B circle
C circle with radius $=$ 1/$\sqrt{2}$
D circle passing through the origin

$\frac{z+1}{z+i}$
Let $z=x+iy$
Hence
$\frac{(x+1)+iy}{(x+i(y+1)}$
$=\frac{\left[\right(x+1)+iy][x-i(y+1\left)\right]}{x^2+(y+1{)}^2}$
$=\frac{\left[x\right(x+1)+y(y+1\left)\right]+i[xy-(x+1\left)\right(y+1\left)\right]}{x^2+(y+1{)}^2}$
It is given that $\frac{z+1}{z+i}$ is purely imaginary.
Hence
$\frac{x(x+1)+y(y+1)}{x^2+(y+1{)}^2}=0$
Therefore
$x(x+1)+y(y+1)=0$
$x^2+x+y^2+y=0$
$(x+\frac{1}{2}{)}^2+(y+\frac{1}{2}{)}^2-\frac{1}{4}-\frac{1}{4}=0$
$(x+\frac{1}{2}{)}^2+(y+\frac{1}{2}{)}^2={\frac{1}{\sqrt{2}}}^2$
Now $z=x+iy$
Hence $z$ lies on a circle centered at $(\frac{-1}{2},\frac{-1}{2})$ and has radius of $\frac{1}{\sqrt{2}}$
Now let $x=0,y=0$ thus $z=origin$
Therefore
$\frac{1}{4}+\frac{1}{4}$
$=\frac{1}{2}$
$=RHS$
Hence the circle passes through origin.