Convergence of $\sum_{n=0}^{\infty}\left(\frac{z}{z+1}\right)^n (z\in C)

[fauboca]

\sum_{n=0}^{\infty}\left(\frac{z}{z+1}\right)^n

z in C.

This only converges for z>\frac{-1}{2}, correct?

Thanks.

 

[vela]

No. For one thing, z is complex, so it doesn't make sense to say z > -1/2. In any case, show us how you reached this conclusion.

 

[fauboca]

No. For one thing, z is complex, so it doesn't make sense to say z > -1/2. In any case, show us how you reached this conclusion.

Ratio Test

 

[vela]

You need to explain what you did in more detail.

 

[fauboca]

You need to explain what you did in more detail.

By the ratio test,

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

Is this correct now?

If so, how do I find the appropriate z for which it converges?

 

[jbunniii]

Your inequality is true if and only if

\left|\frac{z}{z+1}\right|^2 < 1

You can rearrange this and simplify to find a simple inequality for z.

Another way: your inequality is true if and only if

|z| < |z + 1|

which can be written more suggestively as

|z - 0| < |z + 1|

What does this mean geometrically?

 

[fauboca]

Your inequality is true if and only if

\left|\frac{z}{z+1}\right|^2 &lt; 1

You can rearrange this and simplify to find a simple inequality for z.

Another way: your inequality is true if and only if

|z| < |z + 1|

What does this mean geometrically?

Why did you square it?

 

[jbunniii]

Why did you square it?

Because you can then simplify |z + 1|^2. If you're not sure how, try writing it as z + 1 times its complex conjugate.

 

[fauboca]

Because you can then simplify |z + 1|^2. If you're not sure how, try writing it as z + 1 times its complex conjugate.

I don't understand what you mean.

z + 1 times it conjugate is

z^2 - 1

 

[jbunniii]

I don't understand what you mean.

z + 1 times it conjugate is

z^2 - 1

No, that's z + 1 times z - 1. The conjugate of z + 1 is not z - 1.

 

[fauboca]

No, that's z + 1 times z - 1. The conjugate of z + 1 is not z - 1.

So am I supposed to get \bar{z}&gt; -1??

 

[jbunniii]

So am I supposed to get \bar{z}&gt; -1??

No. That doesn't even make any sense, as \bar{z} is a complex number.

|z+1|^2 = (z+1)(\overline{z + 1}) = (z + 1)(\bar{z} + 1) = z\bar{z} + z + \bar{z} + 1 = |z|^2 + z + \bar{z} + 1

So what does your inequality become?

Hint: you can simplify z + \bar{z}.

 

[fauboca]

No. That doesn't even make any sense, as \bar{z} is a complex number.

|z+1|^2 = (z+1)(\overline{z + 1}) = (z + 1)(\bar{z} + 1) = z\bar{z} + z + \bar{z} + 1 = |z|^2 + z + \bar{z} + 1

So what does your inequality become?

Hint: you can simplify z + \bar{z}.

I defined z = x + yi and then did what you said. But we had |z| &lt; |z+1| That is how I obtained my solution.

 

[jbunniii]

I defined z = x + yi and then did what you said. But we had |z| &lt; |z+1| That is how I obtained my solution.

But you didn't obtain a correct solution, or at least you haven't posted it here.

OK, forget the squaring for now and let's focus on this inequality:

|z| < |z + 1|

What does this mean geometrically?

I will write it more suggestively as follows:

|z - 0| < |z - (-1)|

What values of z satisfy this inequality? It's a certain region of the complex plane. What region?

This kind of thing is often easier to answer if you first determine what region is specified by the EQUALITY:

|z - 0| = |z - (-1)|