Question

If $Im\left(\frac{z+2i}{z+2}\right)=0$, then $z$ lies on

• A. $x^2+y^2+2x+2y=0$
• B. $x^2+y^2-2x=0$
• C. $x+y+2=0$
• D. None of these

Answer 1

The correct option is C $x+y+2=0$

Let $z=x+iy$

Then, $\frac{z+2i}{z+2}=\frac{x+iy+2i}{x+iy+2}=\frac{x+(y+2)i}{(x+2)+iy}$

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

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

Since $Im\left(\frac{z+2i}{z+2}\right)=0$

$\Rightarrow x+y+2=0$