Question

If lm $\left(\frac{z+2i}{z+2}\right)=0$ then z lies on the curve

• A.
• B.
• 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, $z+2i=x+iy+2i=x+(y+2)iz+2=x+iy+2(x+2)+iy=\frac{[x+(y+2\left)i\right]\left[\right(x+2\left)iy\right]}{(x+2{)}^2+y^2}=\frac{(x^2+y^2+2x+2y)+i(2x+2y+4)}{(x+2{)}^2+y^2}sincelm\left(\frac{z+2i}{z+2}\right)=0\Rightarrow x+y+2=0$ which represents a straight line