Which way is the semi-circle with complex equation $\arg{\frac{z – z_1}{z – z_2}} = \frac{\pi}{2}$

algebra-precalculuscirclescomplex numbersgeometry

Good day

I was learning about the equation for a semi-circle in the Argand Plane and learnt it to be: $$\arg{\frac{z – z_1}{z – z_2}} = \frac{\pi}{2}$$

But which way is the semi-circle?
Given two points, It could be like this:

enter image description here

or this:

enter image description here

How can I find out what the direction of a semi – circle (tilted or horizontal or in the the $3$rd quadrant etc., in other words, any general semi – circle) is from just the equation?

Thanks

EDIT: As the answer pointed out, we can think of this as $\arg{z – z_1} – \arg{z – z_2}$. So, we can just check the case where $\arg{z – z_1} > \arg{z – z_2}$ and we'll have the answer. I tried doing that.

Consider the first figure (on left side), call the upper point $z_1$ and the lower: $z_2$. Consider any point on the semi-circle $z$. Now $z – z_1$ will have a negative real part and a negative imaginary part. So $z – z_1$ will lie in the third quadrant and thus have negative argument. Also, $z – z_2$ will have a negative real part and a positive imaginary part, so it will lie in the second quadrant and have a positive argument. So, $\arg{z – z_1} < \arg{z – z_2}$. This case is ruled out.

Consider the second case, now $z – z_1$ will have positive real part and a negative imaginary part, so it will lie in the fourth quadrant. $z – z_2$ will have a positive real part and a positive imaginary part so it will lie in the first quadrant. So, again, $\arg{z – z_1} < \arg{z – z_2}$. But according to the answers, the circle will go in the anti – clockwise direction? What am I doing wrong?

Thanks again

Best Answer

Argument of a ratio is viewed as a difference between the two arguments.

The first angle from $z_1$ is obtuse and the second angle from $z_2$ is acute. We go anti-clockwise from $z_1$ to $z_2$ to achieve this.

For example, referring to the first picture with $z_1 = 3i$ and $z_2 = i$. Let $z=-1+2i$. Then we have $z-z_1=-1-i$, $\arg(z-z_1) = \frac{5\pi}{4}$ and $\arg(z-z_2)=\arg(-1+i)=\frac{3\pi}{4}$, the difference would be $\frac{\pi}2$.

Also note that $$-\frac{3\pi}4-\frac{3\pi}4=-\frac{6\pi}{4} \equiv 2\pi - \frac{6\pi}{4}\equiv \frac{\pi}{2} \pmod{2\pi}$$

If we look at the second picture and if we let $z=1+2i$, then $z-z_1=1-i$, $\arg(z-z_1) = -\frac{\pi}4$ and $z-z_2=1+i$ and $\arg(z-z_2)=\frac{\pi}{4}$, the difference would be $-\frac{\pi}2$

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