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What would $arg(\frac{z-4i}{z+2i}) = \frac{\pi}{4}$ look like on an Argand diagram and why?

I don't know how to check my answer because Desmos does not support complex numbers.

PhysicsMathsLove
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We have a Möbius transformation

$$z'=\frac{z-4i}{z+2i}$$

which maps circles to circles or straight lines, and straight lines to circles or straight lines. We are looking for the shape of the set of complex numbers which will be mapped to a straight line of slope $1$. We suspect that the shape is that of a circle.

Let $z=a+ib.$ Then

$$\frac{a+ib-4i}{a+ib+2i}=\frac{(a+ib-4i)(a-ib-2i)}{(a+i(b+2))(a-i(b+2))}=$$

$$=\frac{a^2+b^2-2b-8-i6a}{a^2+(b+2)^2}.$$

The argument is $\frac{\pi}4$. So,

$$\operatorname {atan}\left(-\frac{6a}{a^2+b^2-2b-8}\right)=\frac{\pi}4.$$

Taking the tangent of both sides we get $$-\frac{6a}{a^2+b^2-2b-8}=1.$$

Thus

$$a^2+b^2-2b+6a-8=(a+3)^2+(b-1)^2=18.$$

This is a circle centered at $-3+i$ and of radius $\sqrt{18}.$

EDIT

Here is the plot of the locus of the $z$'s sought:

enter image description here

So, negative $a$'s are considered.

zoli
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  • do you mean where $a<0$ also? in fact, is it $a<0$ or $a \leq 0$ ? – PhysicsMathsLove Mar 20 '17 at 21:24
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    @Tilly: $-3-\sqrt{18}\le a\le -3+\sqrt{18}.$ – zoli Mar 20 '17 at 22:50
  • @zoli i'm working on a similiar question and was just wondering, wouldn't the loci be restricted to Q2 and Q3? My book has an angle subtended by the chord (here it'd be (0,4) and (0,-2)). the angle to the circumference would be = pi/4, and the angle subtended to the circumference to the right of the y-axis is >pi/4, so this isn't part of the solution is it? – Noobcoder Apr 05 '20 at 10:26