Question

If $z$ lies on' the curve $|z-2i|=Imgz,$ then

• A. the curve is a circle
• B. minimum of $\left|z\right|=1$
• C. the maximum argument of $z$ is $\frac{\pi }{2}$
• D. the maximum argument of $z$ is $\frac{3\pi }{4}$

Answer 1

Answer: (B), (D)

The correct options are
B minimum of $\left|z\right|=1$
D the maximum argument of $z$ is $\frac{3\pi }{4}$

$|z-2i|=Imgz$
Substituting $z=x+iy$, we have
$x^2+{(y-2)}^2=y^2$
$\Rightarrow x^2=4(y-1)$ . It represents a parabola with vertex $V(0,1)$ and latus rectum $4$.
$\therefore$ $z$ lies on the parabola .
Minimum of $\left|z\right|$ corresponds to $OV$ which is $1$,
when $x=\pm y$
we get ${(y-2)}^2$ = 0 .It has equal roots
$\therefore$ $y=x$ and $y=-x$ are the equations of the tangents from the origin $O$ to the parabola .
$\therefore$ maximum argument corresponds to $OA(i.e.,y=x)$ which is $\frac{3\pi }{4}$