Question

Locate the complex numbers $z=x+iy$ such that
$|z-i|=1,\arg \frac{z}{z+i}=\frac{\pi }{2}.$

Answer 1

Here are two conditions to satisfied.
$|z-i|=1\Rightarrow x^2+{(y-1)}^2=1$ ......$\left(1\right)$
It represents a circle with centre at $(0,1)$ and radius $1$. It clearly touches $x-$axis as $r=k=1$. Hence all points on the boundary of this circle are the solutions of this.
Again $\arg \frac{z}{z+i}=\frac{\pi }{2}$
$\therefore \frac{z}{z+i}=r(\cos \frac{\pi }{2}+i\sin \frac{\pi }{2})$
$0+ir$
$\therefore R.P.=0$, $I.P.=r=+ive$ ......$\left(2\right)$
Now $\frac{z}{z+i}=\frac{x+iy}{x+i(y+1)}.\frac{x-i(y+1)}{x-i(y+1)}$
$=\frac{x^2+y^2+y-ix}{x^2+{(y+1)}^2}$
Condition $\left(2\right)\Rightarrow x^2+y^2+y=0$
It represents a circle with centre at $(0,-\frac{1}{2})$ and radius $\frac{1}{2}$, i.e. $x^2+{(y+\frac{1}{2})}^2={\left(\frac{1}{2}\right)}^2$
Also I.P. $=-x>0$ or $x<0$
Thus we have two circles $x^2+(y-1{)}^2=1^2$
and $x^2+{(y+\frac{1}{2})}^2={\left(\frac{1}{2}\right)}^2$ with $x<0$
Because of the condition $x<0$, only left hand portion of the second circle will be valid except the points $P$ and $Q$ as for both $x=0$. Also $P(0,0)$ does not satisfy 1st circle.