Question

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

• 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

Answer: (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.