# Math Help - Need help calculating convergence, series.

1. ## Need help calculating convergence, series.

I have an exercise where I'm supposed to find convergence if any and if there are show how it's... constituted. The series:

$$\displaystyle \sum_{n=0}^{\infty} \left(\frac{z-1}{1-2z}\right)^n$
$

I'd like help when it comes to ways of solving it. I solve it as a real problem, getting the zeros on the real axis. Then I use these to find the zeros in the img-plane.

$$\displaystyle \frac{\left|(\frac{z-1}{1-2z})^{n+1}\right|}{\left|(\frac{z-1}{1-2z})^n\right|}$
$
= $$\left|(\frac{z-1}{1-2z})^{1}\right|$
$
= $$\displaystyle \frac{\left|z-1\right|}{\left|1-2z\right|} $ For what z is the fraction smaller than 1? $\[ \displaystyle \frac{\left|z-1\right|}{\left|1-2z\right|} < 1$$

I get $0 < z < \frac{2}{3}$ - treating z as a real variable, so centre on x-axis should be $\frac{1}{3}$

Now to the imaginary part.
$$\displaystyle \frac{\left|(x+iy)-1\right|}{\left|1-2(x+iy)\right|} < 1$$
which gives $$\displaystyle \frac{\left|(\frac{1}{3}+iy)-1\right|}{\left|1-2(\frac{1}{3}+iy)\right|} < 1$$

I get $y = \pm \frac{i}{3}$ - which goes nicely with the real values.

Centre = $\frac{1}{3}$
Radius = $\frac{1}{3}$

First on all, is this right? And is there a more solid way of doing this?

2. That is the right answer. I would do it this way, using the fact that $|z|^2 = z\overline{z}$, where $\overline{z}$ is the complex conjugate of z. Then $\frac{|z-1|}{|1-2z|} = 1$ is equivalent to $(z-1)(\overline{z}-1) = (1-2z)(1-2\overline{z})$. Multiply that out and rearrange it as $3z\overline{z}-(z+\overline{z}) = 0$. Now "complete the square" (except that it's not a square, but the method is the same) to write this as $(z-\frac13)(\overline{z}-\frac13) = \frac19$. Finally, take the square root to get $|z-\frac13| = \frac13$, which is the equation of a circle centred at 1/3, with radius 1/3.

3. Ahh... Conjugate. Your method is so simple that it's out right beautiful!

Thank you very much!