# Find the perimeter of the locus of the point satisfying this condition

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.

• But $$arc(z)=\arctan\left(\frac{b}{a}\right)$$ if $z=a+bi$ Apr 5, 2018 at 12:27
• $a,b$ are real numbers, and $i^2=-1$ Apr 5, 2018 at 12:29
• @Dr.SonnhardGraubner I took tan(pi/4) = 1 while solving Apr 5, 2018 at 12:29
• You've shown that every point that satisfies the condition lies on the circle, but what you're saying is that every point on the circle satisfies the condition, a stronger statement. Apr 5, 2018 at 12:39
• @saulspatz But why does this method eliminate the other points? Apr 5, 2018 at 12:40

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.

• I follow upto "lies in the first quadrant". Apr 5, 2018 at 12:45
• @SmarthBansal If $x+iy$ is in the first quadrant, $x\ge 0, y \ge 0$ That's the definition of "first quadrant," isn't it? Apr 5, 2018 at 12:51
• I see, so it's actually the portion of the circle in the first quadrant? Apr 5, 2018 at 12:55
• @Dylan I took the liberty of inserting the -1 you accidentally omitted in the very last inequality. Apr 5, 2018 at 13:04
• @Dylan If the portion of the circle in the first quadrant is considered then it's perimeter comes out to be half of the actual answer. Apr 5, 2018 at 13:07