A point satisfies $$ \arg\left(\frac {z+i}{z-i}\right ) =\frac \pi 4 $$ We are asked to find the perimeter of the locus of the point.
I did actually got an answer but it isn't correct says my book.
Here's my attempt:
I put $z=x+iy$
$$\frac {x+i(y+1)}{x+i(y-1)} = \frac {x+i(y+1)}{x+i(y-1)} \cdot \frac {x-i(y-1)}{x-i(y-1)} \\ $$ On solving and taking the argument I get $$ x^2+y^2-2x-1=0$$ Obviously that represents a circle with radius $\sqrt 2$. So I conclude the perimeter must be $2\times \pi \sqrt 2$.
The answer is $\frac {3\pi} {\sqrt 2}$. Please do include what is the mistake in my method.
I heard the teacher saying that an easy and quick answer can be given by simple rotation. Please add an alternative answer also.
Edit: Here's a link with the same answer as my answer above.
Simplifying we get
$$ \frac{z+i}{z-i} = \frac{x^2+y^2-1+2xi}{x^2+(y-1)^2} $$
Therefore
$$ \arg\left( \frac{z+i}{z-1} \right) = \arctan\left(\frac{2x}{x^2+y^2-1} \right) = \frac{\pi}{4} $$
Or
$$ \frac{2x}{x^2+y^2-1} = 1 $$
An argument of $\frac{\pi}{4}$ implies that both the real and imaginary parts of the complex number $w=\frac{z+1}{z-1}$ have to be positive, therefore $2x > 0$ and $x^2+y^2-1>0$
This implies your locus is the set of points such that $$ \{x,y: (x-1)^2 + y^2 = 2; x > 0, x^2+y^2 > 1 \} $$
Do you see why it's not the whole circle?
EDIT: Here's a graph of the locus
Note the center at $(1,0)$. You only include part of the circle on the right of the $y$-axis (where $x>0$). This turns out to be $3/4$ of the entire circle.
I'll leave it to you to write a formal proof.