Question

Locus of $z$ , if
arg$[z-(1+i\left)\right]$= $\{\begin{array}{ccc}\frac{3\pi }{4}& when& \left|z\right|\le |z-2|\\ \frac{-\pi }{4}& when& \left|z\right|>|z-4|\end{array}$

• A. straight lines passing through (2,0)
• B. straight lines passing through (2,0), (1, 1)
• C. a line segment
• D. a set of two rays

Answer 1

The correct option is D a set of two rays

Let $z=x+iy$
$z-(1+i)$
$=(x-1)+i(y-1)$
$|z|\le |z-2|$
$x^2+y^2\le (x-2{)}^2+y^2$
$(2x-2)\left(2\right)\le 0$
$x\le 1$ ...(i)
For $x\le 1$
$2(x-1{)}^2=(x-1{)}^2+(y-1{)}^2$ ...(argument is $\frac{3\pi }{4}$)
$(x-1{)}^2=(y-1{)}^2$
$(x-1)=\pm (y-1)$
$x+y=2$
$x-y=0$ Hence both are straight lines.
$|z|>|z-4|$
$x^2>(x-4{)}^2$
$(2x-4)\left(4\right)\ge 0$
$x>2.$
Hence
Even for $x>2$ we will get two straight lines.
Thus the above is a locus of two rays.
One passes through (2,0) and (1,1) while the other does not.