Hi there, I have to prove this two sentences . I think I've solved the first, but I'm quiet stuck with the second.(adsbygoogle = window.adsbygoogle || []).push({});

The first says:

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

So I called z and beta:

[tex]z=x+iy,\beta=u+iv[/tex]

Then developing the products I get:

[tex]\alpha(x^2+y^2)+2(ux-vy)+\gamma=0[/tex]

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

Then completing the square and reordering:

[tex]\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}[/tex]

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 [tex]z\in{C}[/tex] that verifies [tex] \left\|{\displaystyle\frac{z-1}{z+1}}\right\|=k[/tex] is a circumference ([tex]k\in{R},1\neq{k}>0[/tex]).

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

[tex] \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}[/tex]

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.

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# Two different demonstrations on the complex plane

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