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

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:

or this:

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

• Draw the semi - circle counter clockwise from $z_{1}$ to $z_{2}$ Commented Aug 9, 2021 at 6:27

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 $$-\frac{\pi}2$$

• Hi @SiongThyeGoh, I have edited the question because I still have a doubt. I'd be glad if you could address that also. Thanks. Commented Aug 10, 2021 at 14:59
• we should compute in terms of modulo $2\pi$. Commented Aug 10, 2021 at 16:04
• Why can we do that? Isn't this this contradictory with what the range of the argument is $(\pi, \pi]$? Commented Aug 10, 2021 at 17:27
• note that $arg(z) \in \{Arg(z) + 2n\pi, n \in \mathbb{Z} \}$ as stated in wikipedia Commented Aug 10, 2021 at 23:14