Sketching a set of complex numbers and deducing the value of $|z +1 - i|$ for such numbers

Question

The point $P$ represents the complex number $z$.

a) Given that $\arg(\frac{z-2i}{z+2}) = \frac{\pi}{2}$ , sketch the locus of $P$.

Ok so I've sketched this and this is what it looks like :

b) Deduce the value of $|z + 1 - i|$

From my understanding $|z + 1 - i|$ represents the distance between any point on the locus and the point $(-1,i)$ (the dot in the diagram). I wasn't sure how to work this out because I thought the answer would vary as it's any point on the locus to that point.

The answer says it's $\sqrt{2}$.

I don't understand this. Can someone explain to me what the meaning of this is?

Answer 1

You can't get the second part because the first part is incorrect. The locus of $P$ is the circle with the points $2i$, $-2$ lying on its diameter. This is a consequence of the fact that a triangle inscribed into a circle and having one of its sides as a diameter is a right triangle.

Once you have the correct circle in (1), the conclusion in (2) flows naturally.

Answer 2

Some ideas:

Well, we know that $\;\arg z=\frac\pi2\iff \text{Re}\,z=0\;,\;\;\text{Im}\,z>0\;$ (principal values of arctangent, of course) , so putting$\;z=x+iy\;,\;x,y\in\Bbb R\;$:

$$\frac{z-2i}{z+2}\frac{\overline z+2}{\overline z+2}=\frac{|z|^2+2z-2i\overline z-4i}{|z+2|^2}=\frac{x^2+y^2+2x+2yi-2y-2xi-4i}{(x+2)^2+y^2}=$$

$$=\frac{(x+1)^2+(y-1)^2-2-2(x-y+2)i}{(x+2)^2+y^2}\implies$$

$$\begin{cases}(x+1)^2+(y-1)^2=2\\{}\\x-y+2<0\end{cases}\;,\;\;x\neq-2\;\;or\;\;y\neq 0$$

From the above I see a circle centered at $\;(-1,1)\sim-1+i\;$ and with radius $\;\sqrt2\;$ , and s.t. the point is above the line $\;y=x+2\;$, whose intersection points on the two axis are antipode points of the circle (i.e., straight line segment joining these two points is a diameter of the circle)...