# Homework Help: Loci of Complex Equation

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1. Jan 15, 2017

### Alettix

1. The problem statement, all variables and given/known data
Consider the relation $|\frac{z-i}{z*-i}| = \lambda$ where z = x + yi
a) For $\lambda = 1$ show that the locus is a line in the complex plane and find its equation.
b) What is the locus when $\lambda = 0$?
c) Show that for all other positive $\lambda$ the locus may be written as $zz* + bz* + b*z + c = 0$ where $c$ is a real number. Find conditions on complex #b# and real #c# for which the expression describes a
i) Point
ii) Circle
iii) Line

2. Relevant equations
$\frac{m}{n} = \frac{mn*}{|n|^2}$
if $m = k + ip$, $|m| = \sqrt{k^2+p^2}$

3. The attempt at a solution
I tried the following on the first part:
$1 = |\frac{z-i}{z*-i}| = |\frac{x+i(y-1)}{x-i(y+1)}| = |\frac{(x+i(y-1))\cdot (x+i(y+1))}{x^2+(y+1)^2}| = |\frac{x^2-y^2-2xyi}{x^2+(y+1)^2}|$
but from here I am stuck, because this does not resemble the equation of a line at all.

Thanks for all help and hints! :)

2. Jan 15, 2017

### Staff: Mentor

For the first part, with $\lambda = 1$, you have $|\frac{z-i}{\bar{z}-i}| = 1$, or equivalently, $|z - i| = |\bar{z}-i|$. Simplify this equation by replacing z and $\bar{z}$, and by taking magnitudes on each side.

3. Jan 15, 2017

### Alettix

Right! That seems easier! So then we have:
$x^2 + (y-1)^2 = x^2 + (y+1)^2$
$y^2 -2y +1 = y^2 + 2y +1$
$y = 0$
Thus the loci of all z must be the real axis?

For part b) then: $|z-i| = 0$ requires z=i, thus the loci is a single point?

in c) I have the following:
$\sqrt{(z-i)(z*+i)} = \lambda \sqrt{(z*-i)(z+i)}$ yielding:
$(\lambda^2 -1)z*z - (\lambda^2+1)i z + (\lambda^2+1)iz* + (\lambda^2-1) = 0$
but I am not sure how I should derive the conditions for the different loci.
Inserting $z = x +yi$ in $z^*z + bz^* +b^*z + c = 0$ and noting $b = -b^*$ and $c=1$ simply yields: $x^2 + y^2 + 1 = 0$ which is the equation of a parabola...

Last edited: Jan 15, 2017
4. Jan 15, 2017

### Staff: Mentor

Yes on both a) and b).
I haven't looked at c) yet, but will do so now.

5. Jan 15, 2017

### haruspex

Check that step
Umm.. No it isn't.

6. Jan 16, 2017

### Alettix

Well: $z^*z + bz^* +b^*z + c = x^2+y^2+bx-by+b^*x+b^*y +c = 0$
noting $b = -b^*$ and $c=1$ gives $x^2+y^2+c=0$
which is $x^2+y^2+1=0$
What am I doing wrong?
Sorry, I meant hyperbola!

7. Jan 16, 2017

### SammyS

Staff Emeritus
Why did you choose $\ c=1\,?$

8. Jan 16, 2017

### haruspex

You are cancelling a zb with a z*b.
Not that either.

9. Jan 16, 2017

### Alettix

I didn't choose it, it followed from $(\lambda^2 -1)z*z - (\lambda^2+1)i z + (\lambda^2+1)iz* + (\lambda^2-1) = 0$ assuming $\lambda \ne 1$

10. Jan 16, 2017

### Alettix

It should be: $x^2 + y^2 - 2bi y + 1 = x^2 +y^2 + 2|b|y + 1 = 0$ which can be written as $x^2 + (y+|b|)^2 = |b|^2 - 1$. Which is the equation of a circle. This becomes a point if $|b|^2 = 1$. But how can this be made a line?

11. Jan 16, 2017

### haruspex

To arrive at this equation you made an assumption about λ, remember?

12. Jan 16, 2017

### Ray Vickson

Use geometry! $|z-i|$ is the distance of the point $(x,y)$ from the point $(0,1)$ in the Cartesian plane. $|z^*-i| = |(z^*-i)^*| = |z+i|$ is the distance from $(x,y)$ to $(0,-1)$.

So, for (a), you want all the points $(x,y)$ that are the same distance to the two points $(0,1)$ and $(0,-1)$.
For (b) you want $d[(x,y) \to (0,1)]= 0\, d[(x,y) \to (0,-1)] = 0$. For (c) you want $d[(x,y) \to (0,1)] = \lambda d[(x,y) \to (0,-1)]$, or
$d[(x,y) \to (0,1)]^2 = \lambda^2 d[(x,y) \to (0,-1)]^2.$ Expand this equation out in terms of powers of $x$ and $y$, to see what you get.

13. Jan 17, 2017

### Alettix

Yes, I assumed that $\lambda >1$. This also means that I cannot insert $lambda = 1$ (which gave a line) into my $z^*z + bz^* +b^*z + c = 0$ equation, because in that case $b$ diverges. Therefore I am not really sure how to find the conditions of a line.

14. Jan 17, 2017

### haruspex

Sorry, I overlooked that it was asking for conditions on b and c, so is specifically in the context of λ≠1. As you found, it can never be a line, though it does tend to a line in the limit as b tends to infinity.

15. Jan 18, 2017

### Alettix

Got it! Thank you very much!