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- #1

- Thread starter dwsmith
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- Thread starter
- #1

- Jan 26, 2012

- 19

You solve that inequality..surely you can do that, no?$\displaystyle\sum_{n=0}^{\infty}\left(\frac{z}{z+1}\right)^n$

$z\in\mathbb{C}$

By the ratio test,

$\displaystyle\left|\frac{z}{z+1}\right|<1$

I am stuck at this part. How do I find the z such that some is convergence?

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- #3

Apparently I can't because I keep getting it wrong.You solve that inequality..surely you can do that, no?

- Jan 26, 2012

- 19

Ok, so we need to solve $|z|<|z+1|$ or $x^2+y^2<(x+1)^2+y^2$ or $x^2<(x+1)^2$ or $x>\frac{-1}{2}$.Apparently I can't because I keep getting it wrong.

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- #5

How did you go from this $|z|<|z+1|$ to this $x^2+y^2<(x+1)^2+y^2$??Ok, so we need to solve $|z|<|z+1|$ or $x^2+y^2<(x+1)^2+y^2$ or $x^2<(x+1)^2$ or $x>\frac{-1}{2}$.

- Jan 26, 2012

- 19

Let $z=x+iy$.How did you go from this $|z|<|z+1|$ to this $x^2+y^2<(x+1)^2+y^2$??

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- #7

I did but I don't see what happened to all the i's.Let $z=x+iy$.

- Jan 26, 2012

- 19

If $z=x+iy$ then $|z|=\sqrt{x^2+y^2}$.I did but I don't see what happened to all the i's.

Last edited:

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- #9

Shouldn't it be square rooted?If $z=x+iy$ then $|z|=x^2+y^2$.

- Jan 26, 2012

- 19

Yes, it should have been, and then I proceeded from there. Do you see how?Shouldn't it be square rooted?

P.S. Sorry if I sound terse, I'm just busy doing other things!