# Homework Help: Two different demonstrations on the complex plane

1. Sep 3, 2011

### Telemachus

Hi there, I have to prove this two sentences . I think I've solved the first, but I'm quiet stuck with the second.

The first says:

1) Demonstrate that the equation of a line or a circumference in the complex plane can be written this way: $$\alpha z . \bar{z}+\beta z+\bar{\beta} \bar{z}+\gamma=0$$, with $$\alpha,\gamma\in{R},\beta\in{C}$$

So I called z and beta:

$$z=x+iy,\beta=u+iv$$

Then developing the products I get:
$$\alpha(x^2+y^2)+2(ux-vy)+\gamma=0$$

And making alpha equal zero I get the equation for a line, right? (for u and v fixed).

Then completing the square and reordering:

$$\left(x+\displaystyle\frac{u}{\alpha}\right )^2+\left(y-\displaystyle\frac{v}{\alpha}\right )^2=\displaystyle\frac{u^2}{\alpha^2}+\frac{v^2}{\alpha^2}-\frac{\gamma}{\alpha}$$

This is the equation for the circle, is this right?

In the other hand I got:

2) Prove that the geometrical place for the points $$z\in{C}$$ that verifies $$\left\|{\displaystyle\frac{z-1}{z+1}}\right\|=k$$ is a circumference ($$k\in{R},1\neq{k}>0$$).

I couldn't make much for this. I called z=x+iy again:

$$\left\|{\frac{z-1}{z+1}}\right\|=k\Rightarrow{\left\|{\frac{(x+iy-1)(x-iy+1)}{(x+iy+1)(x-iy+1)}}\right\|=k}\Rightarrow{\left\|{\frac{x^2+y^2+2iy-1}{(x+1)^2+y^2}}\right\|=k}$$

I don't know what to do from there.

Bye.

PS: sorry if it bothers you that I made two questions in the same topic, but I thought it was quiet trivial and it didn't deserved two different topics, but I can split it if necessary.

Last edited: Sep 3, 2011
2. Sep 3, 2011

### Dickfore

The thing in the denominator is real by construction (You multiplied it by the complex conjugate). What are the real and imaginary parts of the numerator?

3. Sep 3, 2011

### Telemachus

$$Re(w)=\frac{x^2+y^2-1}{(x+1)^2+y^2}$$
$$Im(w)=\frac{2y}{(x+1)^2+y^2}$$

BTW, is the first exercise right?

Last edited: Sep 3, 2011
4. Sep 3, 2011

### Dickfore

The first exercise is correct.

For the second one, a modulus of a complex number:
$$|w| = \sqrt{ \left( \mathrm{Re} w \right)^{2} + \left( \mathrm{Im} w \right)^{2}} = k$$
Square the equation, get rid off the denominators and simplify the expression. Write what you get.

5. Sep 3, 2011

Thanks :)

6. Sep 3, 2011

### Dickfore

Actually, it does not seem to correspond to a circumference. The condition

$$\left|\frac{z - 1}{z + 1}\right| = k \Leftrightarrow |z - 1| = k |z + 1|$$

means that the distance from the point $(1, 0)$ is $k$ times bigger than the distance from the point $(-1, 0)$. I am not sure if this is a circle in general.

7. Sep 3, 2011

### Dickfore

Actually, scratch what I was saying. Square the above:
$$|z - 1|^{2} = k^{2} |z + 1|^{2}$$

But, $| w |^{2} = \bar{w} w$, so:
$$(z - 1)(\bar{z} - 1) = k^{2} (z + 1) (\bar{z} + 1)$$

Simplify this and use part 1 of the problem.

8. Sep 3, 2011

### Telemachus

Thank you very much :)