- #1

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z in C.

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

Thanks.

- Thread starter fauboca
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- #1

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z in C.

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

Thanks.

- #2

vela

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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.

- #3

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Ratio Test

- #4

vela

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You need to explain what you did in more detail.

- #5

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By the ratio test,You need to explain what you did in more detail.

[tex]\left|\frac{z}{z+1}\right|<1[/tex]

Is this correct now?

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

- #6

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Your inequality is true if and only if

[tex]\left|\frac{z}{z+1}\right|^2 < 1[/tex]

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?

- #7

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Why did you square it?Your inequality is true if and only if

[tex]\left|\frac{z}{z+1}\right|^2 < 1[/tex]

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?

- #8

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Because you can then simplify [itex]|z + 1|^2[/itex]. If you're not sure how, try writing it as z + 1 times its complex conjugate.Why did you square it?

- #9

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I don't understand what you mean.Because you can then simplify [itex]|z + 1|^2[/itex]. If you're not sure how, try writing it as z + 1 times its complex conjugate.

z + 1 times it conjugate is

[tex]z^2 - 1[/tex]

- #10

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No, that's z + 1 times z - 1. The conjugate of z + 1 is not z - 1.I don't understand what you mean.

z + 1 times it conjugate is

[tex]z^2 - 1[/tex]

- #11

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So am I supposed to get [itex]\bar{z}> -1[/itex]??No, that's z + 1 times z - 1. The conjugate of z + 1 is not z - 1.

- #12

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No. That doesn't even make any sense, as [itex]\bar{z}[/itex] is a complex number.So am I supposed to get [itex]\bar{z}> -1[/itex]??

[tex]|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[/tex]

So what does your inequality become?

Hint: you can simplify [itex]z + \bar{z}[/itex].

- #13

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I defined [itex]z = x + yi[/itex] and then did what you said. But we had [itex]|z| < |z+1|[/itex] That is how I obtained my solution.No. That doesn't even make any sense, as [itex]\bar{z}[/itex] is a complex number.

[tex]|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[/tex]

So what does your inequality become?

Hint: you can simplify [itex]z + \bar{z}[/itex].

- #14

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But you didn't obtain a correct solution, or at least you haven't posted it here.I defined [itex]z = x + yi[/itex] and then did what you said. But we had [itex]|z| < |z+1|[/itex] That is how I obtained my solution.

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)|

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