Complex loci problem

**Stonehambey** · July 28th 2009, 06:14 AM

Hi,

I know this locus is a circle, but only because I recognise the form, I cannot see how to show it algebraically.

$\arg(z-2)-\arg(z-2i)=\frac{3\pi}{4}$

I would like to find the centre and radius (or its equation).

The second part of the question asks to show that $\frac{|z-4|}{|z-1|}$ is constant on this circle, but I don't even know what it's asking for that one!

Any nudges in the correct direction would be great, not asking for a full solution, just some help getting started.

Cheers

**Opalg** · July 28th 2009, 08:12 AM

Originally Posted by Stonehambey

I know this locus is a circle, but only because I recognise the form, I cannot see how to show it algebraically.

$\arg(z-2)\mathrel{\color{red}-}\arg(z-2i)=\frac{3\pi}{4}$

I would like to find the centre and radius (or its equation).

Notice the typo: the first = should have been – (otherwise the question doesn't make sense).

The difference between the arguments of two complex numbers is the argument of their quotient. So $\arg(z-2)-\arg(z-2i) = \arg\Bigl(\frac{z-2}{z-2i}\Bigr)$ . You want that argument to be $\tfrac{3\pi}4$ .

Now a complex number $w=u+iv$ has argument $\tfrac{3\pi}4$ if $u=-v$ (and $v>0$ ).

So let $z=x+iy$ . Then $\frac{z-2}{z-2i} = \frac{x-2+iy}{x+(y-2)i}$ . Find the real and imaginary parts of that number in the usual way (by rationalising the denominator) and write down the condition that the real part is the negative of the imaginary part. That will give you the equation of a circle. If you want extra credit, you could point out that the locus is in fact only part of that circle (because of the condition $v>0$ in the previous paragraph, which says that the imaginary part of $\frac{z-2}{z-2i}$ should be positive).

Originally Posted by Stonehambey

The second part of the question asks to show that $\frac{|z-4|}{|z-1|}$ is constant on this circle, but I don't even know what it's asking for that one!

If $\frac{|z-4|}{|z-1|} = k$ (constant) then $|z-4|^2 = k^2|z-1|^2$ . Putting $z=x+iy$ again, that becomes $(x-4)^2+y^2 = k^2\bigl((x-1)^2+y^2\bigr)$ , which is the equation of a circle. Compare it with the equation of the circle that you got in the first part of the question, and see if there is a value of k that makes those two equations the same.

**Stonehambey** · July 28th 2009, 10:59 AM

Thanks! I did a bit of hunting around the web and found a method which involved taking the tangent of both sides and then using the identity for tan(A-B), I will try both methods and see if I get the same equation for the circle.

When going the tangent route, can the additional condition be obtained by noting that if $\left(\frac{z-2}{z-2i}\right)$ has an argument of $\frac{3\pi}{4}$ then it must be in the second quadrant, so it's real part negative and imaginary part positive?

**Opalg** · July 28th 2009, 12:30 PM

Originally Posted by Stonehambey

When going the tangent route, can the additional condition be obtained by noting that if $\left(\frac{z-2}{z-2i}\right)$ has an argument of $\frac{3\pi}{4}$ then it must be in the second quadrant, so it's real part negative and imaginary part positive?

Yes, that's exactly the reason.

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