# Complex locus

1. Dec 9, 2008

### Mentallic

1. The problem statement, all variables and given/known data
Given $$Z=\frac{z-2}{z}$$, if $$|z|=1$$ prove that the locus of Z is another circle whose centre and radius must be determined. Also describe the direction of Z as z describes the unit circle in an anticlockwise direction.

2. Relevant equations
$$z=x+iy$$

3. The attempt at a solution
$$Z=\frac{x+iy-2}{x+iy}(\frac{x-iy}{x-iy})$$

expanded and simplified: $$\frac{x^2-2x+y^2+i2y}{x^2+y^2}$$

I think since $$|z|=1$$ then $$x^2+y^2=1$$?

This leaves $$-2x+i2y$$ and I am completely lost at this point...

2. Dec 9, 2008

### gabbagabbahey

Doesn't it leave you with $$Z=(1-2x)+(2y)i$$?....What does that say about $$|Z|$$?

3. Dec 10, 2008

### Mentallic

Ahh yes that was a typo.

$$\frac{x^2-2x+y^2+i2y}{x^2+y^2}$$

$$|z|=1, x^2+y^2=1$$

therefore simplified to: $$Z=1-2x+i2y$$

$$|Z|=x^2+y^2=(1-2x)^2+(2y)^2$$

Rearranged to: $$|Z|=4(x-\frac{1}{2})^2+4y^2$$

This is in the form of a circle (in cartesian form) but I don't know what it has to do with complex numbers, since on the argand diagram $$|z|=r$$ is a circle radius r.

4. Dec 10, 2008

### gabbagabbahey

For a circle centered at 0+0i, yes. Since |Z| is not of this form, you know it's center must be at a different point in the complex plane. If its center is at the point $\alpha=a+bi$, then you know $|Z-\alpha|=r$....correct?

So...what is $|Z-\alpha|$? In particular, what is $|Z-1|$?

5. Dec 10, 2008

### Mentallic

Ahh $$|Z-a|=r$$ would be a circle radius r centred at (a,bi)

So $$|Z-1|=r$$ is a circle centre (1,0i) radius r.

But how can $$|Z|=4(x-\frac{1}{2})^2+4y^2$$ be transformed into this form? r is real number that is more than 0 is it not?

6. Dec 10, 2008

### gabbagabbahey

Well, what is Z-1? So what is |Z-1|?

7. Dec 10, 2008

### Mentallic

I'm starting to lose the plot here
Are we talking about z or Z now? Remember from the question that $$Z=\frac{z-1}{z}$$ where $$z=x+iy$$

8. Dec 10, 2008

### gabbagabbahey

We're talking about capital Z now. If its Locus is a circle with radius r, centered at the point $\alpha=a+bi$ in the complex plane, then $|Z-\alpha|=r$

Do you understand that part?

....Now, you've already calculated that $Z=(1-2x)+(2y)i$, so what is $Z-1$?

9. Dec 10, 2008

### Mentallic

Yes, but I just hope I can apply it to the question.

Well, if $$Z=1-2x+i2y$$ then $$Z-1=-2x+i2y$$ (1)
but I also calculated it from scratch ~

$$Z-1=\frac{z-1}{z}-1$$

$$=\frac{z-1-z}{z}$$

$$=\frac{-1}{z}$$

$$=\frac{-1}{x+iy}(\frac{x-iy}{x-iy})$$

$$=\frac{-x+iy}{x^2+y^2}$$

therefore, $$Z-1=-x+iy$$ (2)

which is not the same to (1)?

10. Dec 10, 2008

### gabbagabbahey

That's because $$Z=\frac{z-2}{z}$$ not (z-1)/z.

11. Dec 10, 2008

### Mentallic

oh lol how embarrassing

ok $$Z-1=-2x+i2y$$

hence, $$|Z-1|=(-2x)^2+(2y)^2=4x^2+4y^2$$

um.. I'm taking a stab at this but does this mean $$|Z-1|=4(x^2+y^2)=4$$?

In which case we have a circle centre (1,0i) radius 4?

12. Dec 10, 2008

### gabbagabbahey

Looks good to me

That's why I chose alpha=1, to get Z-alpha into the above form.

Edit- shouldn't it be $|Z-1|=\sqrt{4(x^2+y^2)}=2$ ?

13. Dec 10, 2008

### Mentallic

Awesome! But what method or how did your intuition tell you to use alpha=1?

For this part I'm unsure what it is asking and how to find it mathematically. i.e. I am completely stumped.

14. Dec 10, 2008

### Mentallic

Ahh yes. For all the questions I've been provided to date, it has been with |z|=1. Totally forgot about the pythagorean method lol.

15. Dec 10, 2008

### gabbagabbahey

Well, I can't explain why my intuition was what it was, but I can give you a logical step by step reasoning for choosing alpha=1:

(1)Start by assuming a general alpha $\alpha=a+bi$
(2)Calculate $$|Z-\alpha|=\sqrt{[(1-2x-a)+(2y-b)i][(1-2x-a)-(2y-b)i]}=\sqrt{(1-2x-a)^2+(2y-b)^2}$$
(3) You're told to show that the Locus of Z is a circle, so you know you want $|Z-\alpha|$ to be some constant radius 'r'; and the only restriction you have regarding x and y is that x^2+y^2=1, so you need to choose a and b in such a way that $|Z-\alpha|=f(x^2+y^2)$. Choosing a=1 and b=0 clearly accomplishes that.

I think you need to express both z and Z in polar form. If you travel around the circle |z|=1 counter clockwise, is the phase of z increasing or decreasing? What happens to the phase of Z as the phase of z increases?

16. Dec 10, 2008

### Mentallic

That works by inspection, but I'll have to see how 'inspecting' works for the more difficult ones.

Sorry I've never learnt or even heard of the term 'phase'. It could just be that you are using different terminology to what our class is accustomed to.

gabbagabbahey I really appreciate your help. If you are capable of continuing with it for this next part, then please, I would love to have a taste of your wisdom if however, you aren't sure about what is required, then all you need to do is say so.

I graphed these circles in the cartesian plane. Maybe it would make more sense by visualising it?

17. Dec 10, 2008

### gabbagabbahey

Well, I assume you have at least been taught that any complex number z can be represented not only in the form x+yi, but also in polar form: $z=|z|e^{i\phi}$?

$\phi$ is called the phase angle of z. What happens to $\phi$ as you go counterclockwise around the unit circle?

18. Dec 10, 2008

### Mentallic

No sorry, we've only been taught $$z=x+iy=rcis\theta$$.

I might have to wait till we have class again then. Thank you so much for your help gabbagabbahey!

19. Dec 10, 2008

### gabbagabbahey

cis(theta) is another way of writing $e^{i\theta}[/tex]....Using you definition of polar form, theta is what I am calling the phase angle. What happens to theta as you go counterclockwise around the unit circle? Does it increase or decrease? If $$z=cis(\theta)$, what is [tex]Z$$?

20. Dec 10, 2008

### Mentallic

Thats a funny way of expressing the same thing

Well theta obviously increases as you go counter-clockwise around the unit circle. For Z, I'm not sure if I calculate theta from the centre of the circle it defines (1,0i) or from the origin, but nonetheless theta still increases as we travel counterclockwise.