Question

If $Re\left(\frac{z+2i}{z+4}\right)=0$ then z lies on a circle with center:

• A. (-2,-1)
• B. (-2,1)
• C. (2,-1)
• D. (2,1)

Answer 1

Answer: (A)

(-2,-1)

Let $z=x+iy$

Then

$w=\frac{z+2i}{z+4}=\frac{x+i(2+y)}{(x+4)+iy}$

Rationalizing we get:

$w=\frac{z+2i}{z+4}=\frac{x+i(2+y)}{(x+4)-iy}\times \frac{x+4-iy}{x+4+iy}=\frac{x^2+4x+y^2+2y+i(xy+2x+4y+xy+8)}{(x+4{)}^2+y^2}$

Since real part of $w$ is 0, hence

$(x+2{)}^2+(y+1{)}^2-5=0$

Hence centre of circle is $(-2,-1)$