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Post by mondo on May 29, 2023 7:30:11 GMT
I wonder how can we prove $g(z) < f(z)$ on $C$, author gives a hint $|z| > 1$ but I don't know how to use it.
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Post by Admin on May 29, 2023 9:32:41 GMT
Mondo
That's not what we are asked to prove according to my copy of the book.
Complex numbers have no order like real numbers, so for any two complex numbers $u$ and $v$ we can never write $u<v$.
Presumably you mean $|g(z)|<|f(z)|$?
Vasco
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Post by mondo on May 29, 2023 19:23:24 GMT
Mondo That's not what we are asked to prove according to my copy of the book. Complex numbers have no order like real numbers, so for any two complex numbers $u$ and $v$ we can never write $u<v$. Presumably you mean $|g(z)|<|f(z)|$? Vasco Yes, to quote the book exactly I should have written $|g(z)|<|f(z)|$. However, why the absolute value? I never said these are in order like in a real case. We just claim here that the first term is bigger than the sum of all the rest. That sum may contain negative numbers as well right? |
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Post by Admin on May 29, 2023 20:06:22 GMT
Mondo
A complex number cannot be greater or less than another complex number.
Vasco
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Post by Admin on May 29, 2023 20:18:54 GMT
Mondo
Also complex numbers are not positive or negative. These words have no meaning for complex numbers.
Vasco
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Post by mondo on May 29, 2023 20:23:07 GMT
Mondo A complex number cannot be greater or less than another complex number. Vasco Ok I see, so that's why we compare their lengths. This now becomes a problem of proving that the length of the first term $z^n$ is bigger than the sum of all the rest $Az^{n-1} + Bz^{n-2} + ... + E$ but still how to prove it? |
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Post by Admin on May 29, 2023 20:36:00 GMT
Mondo A complex number cannot be greater or less than another complex number. Vasco Ok I see, so that's why we compare their lengths. This now becomes a problem of proving that the length of the first term $z^n$ is bigger than the sum of all the rest $Az^{n-1} + Bz^{n-2} + ... + E$ but still how to prove it? Mondo No that is not correct. If $u$,$v$,$w$ are complex numbers then $|u+v+w|\neq |u|+|v|+|w|$ Vasco |
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Post by mondo on May 29, 2023 21:04:06 GMT
Ok I see, so that's why we compare their lengths. This now becomes a problem of proving that the length of the first term $z^n$ is bigger than the sum of all the rest $Az^{n-1} + Bz^{n-2} + ... + E$ but still how to prove it? Mondo No that is not correct. If $u$,$v$,$w$ are complex numbers then $|u+v+w|\neq |u|+|v|+|w|$ Vasco I agree, what needs to be proven is $|f(z)| > |Az^{n-1} + Bz^{n-2} + ... + E|$. right? How can we use the fact "|z| > 1" to prove it? |
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Post by mondo on May 30, 2023 1:34:09 GMT
Here is my attempt: $g(z)$ is the sum with the formula $\frac{E - z^n}{1-z}$ and since both $E$ and $1$ are neglectable in comparison to $z^n$ we can simplify this sum to $\frac{z^n}{z}$ and this is smaller than $z^n$ hence proving $|g(z)| < |f(z)|$.
The hint that author have about $|z| > 1$ was needed to conclude $\frac{z^n}{z}$ < $z^n$.
Do you agree Vasco?
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Post by Admin on May 31, 2023 13:28:57 GMT
Mondo
You seem to be comparing complex numbers again. I thought we had agreed that this does not make sense.
Vasco
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Post by Admin on May 31, 2023 19:03:58 GMT
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Post by mondo on Jun 1, 2023 7:25:44 GMT
Thank you for your proposed solution Vasco.
Right, I have to get rid of this tendency..
As to your solution, I agree with the first step, $|g(z)| = |Az^{n-1} + Bz^{n-2} + ... + E| \le |A||z^{n-1}| + |B||z^{n-2} + ... + |E|$ - triangle inequality
But I don't know how can you say $(|A| + |B| + ... + |E|)|z|^{n-1} \le |z|^n = f(z)$, where does it come from?
And what is $|z|$ - is that a radius of the circle?
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Post by Admin on Jun 1, 2023 7:47:37 GMT
Mondo
Here is a hint: if $|z|>1$ then all powers of it are $>1$ and so if we replace all the lower powers by $|z|^{n-1}$ then the right hand side of the inequality gets bigger and so the inequality is still true.
Vasco
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Post by Admin on Jun 1, 2023 7:51:28 GMT
Mondo
We are considering all the points $z$ to lie on an origin centred circle which has a radius $>1$.
Vasco
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Post by mondo on Jun 1, 2023 7:56:36 GMT
Mondo Here is a hint: if $|z|>1$ then all powers of it are $>1$ and so if we replace all the lower powers by $|z|^{n-1}$ then the right hand side of the inequality gets bigger and so the inequality is still true. Vasco This makes sense to me but you are saying something different in the last inequality. You say thay the sum of $(|A| + |B| + ... + |E|)z^{n-1}$ is less than $z^{n}$ alone, how can that be? |
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