plotting on argand diagram $\arg

Question

I'm a bit stuck on trying to plot loci of this complex number argument:

$\arg \frac{z}{z-3i} = \frac{\pi}{2}$

I know the chord lies on (0,0) and (0,3), and know that the loci is on the semi-circle, but I cannot for the life of me, figure out if the loci lies in quadrant 1 or quadrant 2. If it's a semi circle, does it lie in both quadrants? i.e. is the loci a circle centred (0,1.5)?

I've not been able to successfully manage to draw an arbitrary example either, because the pi/2 is throwing me off. Intuition is telling me the difference between the two principle angles will equal $\frac{\pi}{2}$ regardless of whether I consider quadrant 1 or 2.

Any help greatly appreciated!

Answer 1

Just to add an additional resource I found useful: https://www.nagwa.com/en/explainers/279138216750/

This explains how to come up with a loci very well, and simply too!

Answer 2

Well, we know that the $\arg$ function is given by:

$$ \arg\left(\underline{\text{z}}\right):=\begin{cases} 0\space,\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\Re\left(\underline{\text{z}}\right)\ge0\space\wedge\space\Im\left(\underline{\text{z}}\right)=0\\ \\ \frac{\pi}{2}\space,\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\Re\left(\underline{\text{z}}\right)=0\space\wedge\space\Im\left(\underline{\text{z}}\right)>0\\ \\ \pi\space,\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\Re\left(\underline{\text{z}}\right)<0\space\wedge\space\Im\left(\underline{\text{z}}\right)=0\\ \\ \frac{3\pi}{2}\space,\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\Re\left(\underline{\text{z}}\right)=0\space\wedge\space\Im\left(\underline{\text{z}}\right)<0\\ \\ \arctan\left(\frac{\Im\left(\underline{\text{z}}\right)}{\Re\left(\underline{\text{z}}\right)}\right)\space,\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\Re\left(\underline{\text{z}}\right)>0\space\wedge\space\Im\left(\underline{\text{z}}\right)>0\\ \\ \frac{\pi}{2}+\arctan\left(\frac{\left|\Re\left(\underline{\text{z}}\right)\right|}{\Im\left(\underline{\text{z}}\right)}\right)\space,\space\space\space\space\space\Re\left(\underline{\text{z}}\right)<0\space\wedge\space\Im\left(\underline{\text{z}}\right)>0\\ \\ \pi+\arctan\left(\frac{\left|\Im\left(\underline{\text{z}}\right)\right|}{\left|\Re\left(\underline{\text{z}}\right)\right|}\right)\space,\space\space\space\space\space\space\Re\left(\underline{\text{z}}\right)<0\space\wedge\space\Im\left(\underline{\text{z}}\right)<0\\ \\ \frac{3\pi}{2}+\arctan\left(\frac{\Re\left(\underline{\text{z}}\right)}{\left|\Im\left(\underline{\text{z}}\right)\right|}\right)\space,\space\space\space\space\space\Re\left(\underline{\text{z}}\right)>0\space\wedge\space\Im\left(\underline{\text{z}}\right)<0 \end{cases}\tag1 $$

Now, in your case we have:

$$\arg\left(\frac{\underline{\text{z}}}{\underline{\text{z}}-3i}\right)=\frac{\pi}{2}\tag2$$

This means that:

$$ \begin{cases} \Re\left(\frac{\underline{\text{z}}}{\underline{\text{z}}-3i}\right)=0\\ \\ \Im\left(\frac{\underline{\text{z}}}{\underline{\text{z}}-3i}\right)>0 \end{cases}\tag3 $$

So:

$$ \begin{cases} 0<\Re\left(\underline{\text{z}}\right)\le\frac{3}{2}\\ \\ \Im\left(\underline{\text{z}}\right)=\frac{3\pm\sqrt{9-4\Re\left(\underline{\text{z}}\right)^2}}{2} \end{cases}\tag4 $$

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EDIT

Just look at:

$$\Re\left(\frac{\underline{\text{z}}}{\underline{\text{z}}-3i}\right)=1+\frac{3\left(\Im\left(\underline{\text{z}}\right)-3\right)}{\Re\left(\underline{\text{z}}\right)^2+\left(\Im\left(\underline{\text{z}}\right)-3\right)^2}=0\tag5$$ $$\Im\left(\frac{\underline{\text{z}}}{\underline{\text{z}}-3i}\right)=\frac{3\Re\left(\underline{\text{z}}\right)}{\Re\left(\underline{\text{z}}\right)^2+\left(\Im\left(\underline{\text{z}}\right)-3\right)^2}>0\tag6$$