2
$\begingroup$

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

$\endgroup$
1
  • 1
    $\begingroup$ Draw the semi - circle counter clockwise from $z_{1}$ to $z_{2}$ $\endgroup$
    – acat3
    Commented Aug 9, 2021 at 6:27

1 Answer 1

3
$\begingroup$

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 $-\frac{\pi}2$

$\endgroup$
4
  • $\begingroup$ Hi @SiongThyeGoh, I have edited the question because I still have a doubt. I'd be glad if you could address that also. Thanks. $\endgroup$
    – MangoPizza
    Commented Aug 10, 2021 at 14:59
  • $\begingroup$ we should compute in terms of modulo $2\pi$. $\endgroup$ Commented Aug 10, 2021 at 16:04
  • $\begingroup$ Why can we do that? Isn't this this contradictory with what the range of the argument is $(\pi, \pi]$? $\endgroup$
    – MangoPizza
    Commented Aug 10, 2021 at 17:27
  • $\begingroup$ note that $arg(z) \in \{Arg(z) + 2n\pi, n \in \mathbb{Z} \}$ as stated in wikipedia $\endgroup$ Commented Aug 10, 2021 at 23:14

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .