Complex inequality proof [duplicate]
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This question already has an answer here:
Show that $left|frac{alpha - beta}{1-bar{alpha}beta}right| < 1$ when $|alpha|,|beta| < 1$
1 answer
If $z,sin mathbb{C}$ ( with norm less than $1$) then show that $left|frac{1-s/z}{1/z-overline{s}}right|<1$. I'm hoping for an algebraic proof but possibly also a proof by geometric means if possible ( even with a picture if possible).
complex-analysis complex-numbers
asked Jan 18 at 1:08
SquirtleSquirtle
4,2331741
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marked as duplicate by Martin R, Arnaud D., José Carlos Santos complex-analysis
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Jan 18 at 13:48
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Show that $left|frac{alpha - beta}{1-bar{alpha}beta}right| < 1$ when $|alpha|,|beta| < 1$
1 answer
If $z,sin mathbb{C}$ ( with norm less than $1$) then show that $left|frac{1-s/z}{1/z-overline{s}}right|<1$. I'm hoping for an algebraic proof but possibly also a proof by geometric means if possible ( even with a picture if possible).
complex-analysis complex-numbers
asked Jan 18 at 1:08
SquirtleSquirtle
4,2331741
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marked as duplicate by Martin R, Arnaud D., José Carlos Santos complex-analysis
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Jan 18 at 13:48
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
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An analytic way: Have a look at the automorphism group of the unit disk, and use the Schwarz lemma.
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– T. Bongers
Jan 18 at 1:22
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I don't see how to apply that theorem since $|f(0)|=|s|$ which isn't necessarily zero.
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– Squirtle
Jan 18 at 1:37
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You don't apply it to this function directly, but rather start by recognizing this as the value of an automorphism of $mathbb{D}$ evaluated at a point.
$endgroup$
– T. Bongers
Jan 18 at 2:11
2
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Possible duplicate of Show that $left|frac{alpha - beta}{1-bar{alpha}beta}right| < 1$ when $|alpha|,|beta| < 1$
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– Martin R
Jan 18 at 6:16
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Also: math.stackexchange.com/q/1630930/42969, math.stackexchange.com/q/342181/42969.
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– Martin R
Jan 18 at 6:17
add a comment |
$begingroup$
This question already has an answer here:
Show that $left|frac{alpha - beta}{1-bar{alpha}beta}right| < 1$ when $|alpha|,|beta| < 1$
1 answer
If $z,sin mathbb{C}$ ( with norm less than $1$) then show that $left|frac{1-s/z}{1/z-overline{s}}right|<1$. I'm hoping for an algebraic proof but possibly also a proof by geometric means if possible ( even with a picture if possible).
complex-analysis complex-numbers
asked Jan 18 at 1:08
SquirtleSquirtle
4,2331741
$endgroup$
This question already has an answer here:
Show that $left|frac{alpha - beta}{1-bar{alpha}beta}right| < 1$ when $|alpha|,|beta| < 1$
1 answer
If $z,sin mathbb{C}$ ( with norm less than $1$) then show that $left|frac{1-s/z}{1/z-overline{s}}right|<1$. I'm hoping for an algebraic proof but possibly also a proof by geometric means if possible ( even with a picture if possible).
This question already has an answer here:
Show that $left|frac{alpha - beta}{1-bar{alpha}beta}right| < 1$ when $|alpha|,|beta| < 1$
1 answer
complex-analysis complex-numbers
complex-analysis complex-numbers
asked Jan 18 at 1:08
SquirtleSquirtle
4,2331741
asked Jan 18 at 1:08
SquirtleSquirtle
4,2331741
edited Jan 18 at 1:18
Squirtle
asked Jan 18 at 1:08
SquirtleSquirtle
4,2331741
asked Jan 18 at 1:08
SquirtleSquirtle
4,2331741
asked Jan 18 at 1:08
SquirtleSquirtle
4,2331741
4,2331741
marked as duplicate by Martin R, Arnaud D., José Carlos Santos complex-analysis
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Jan 18 at 13:48
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Martin R, Arnaud D., José Carlos Santos complex-analysis
Users with the complex-analysis badge can single-handedly close complex-analysis questions as duplicates and reopen them as needed.
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Jan 18 at 13:48
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
$begingroup$
An analytic way: Have a look at the automorphism group of the unit disk, and use the Schwarz lemma.
$endgroup$
– T. Bongers
Jan 18 at 1:22
$begingroup$
I don't see how to apply that theorem since $|f(0)|=|s|$ which isn't necessarily zero.
$endgroup$
– Squirtle
Jan 18 at 1:37
$begingroup$
You don't apply it to this function directly, but rather start by recognizing this as the value of an automorphism of $mathbb{D}$ evaluated at a point.
$endgroup$
– T. Bongers
Jan 18 at 2:11
2
$begingroup$
Possible duplicate of Show that $left|frac{alpha - beta}{1-bar{alpha}beta}right| < 1$ when $|alpha|,|beta| < 1$
$endgroup$
– Martin R
Jan 18 at 6:16
$begingroup$
Also: math.stackexchange.com/q/1630930/42969, math.stackexchange.com/q/342181/42969.
$endgroup$
– Martin R
Jan 18 at 6:17
add a comment |
1
$begingroup$
An analytic way: Have a look at the automorphism group of the unit disk, and use the Schwarz lemma.
$endgroup$
– T. Bongers
Jan 18 at 1:22
$begingroup$
I don't see how to apply that theorem since $|f(0)|=|s|$ which isn't necessarily zero.
$endgroup$
– Squirtle
Jan 18 at 1:37
$begingroup$
You don't apply it to this function directly, but rather start by recognizing this as the value of an automorphism of $mathbb{D}$ evaluated at a point.
$endgroup$
– T. Bongers
Jan 18 at 2:11
2
$begingroup$
Possible duplicate of Show that $left|frac{alpha - beta}{1-bar{alpha}beta}right| < 1$ when $|alpha|,|beta| < 1$
$endgroup$
– Martin R
Jan 18 at 6:16
$begingroup$
Also: math.stackexchange.com/q/1630930/42969, math.stackexchange.com/q/342181/42969.
$endgroup$
– Martin R
Jan 18 at 6:17
1
1
$begingroup$
An analytic way: Have a look at the automorphism group of the unit disk, and use the Schwarz lemma.
$endgroup$
– T. Bongers
Jan 18 at 1:22
$begingroup$
An analytic way: Have a look at the automorphism group of the unit disk, and use the Schwarz lemma.
$endgroup$
– T. Bongers
Jan 18 at 1:22
$begingroup$
I don't see how to apply that theorem since $|f(0)|=|s|$ which isn't necessarily zero.
$endgroup$
– Squirtle
Jan 18 at 1:37
$begingroup$
I don't see how to apply that theorem since $|f(0)|=|s|$ which isn't necessarily zero.
$endgroup$
– Squirtle
Jan 18 at 1:37
$begingroup$
You don't apply it to this function directly, but rather start by recognizing this as the value of an automorphism of $mathbb{D}$ evaluated at a point.
$endgroup$
– T. Bongers
Jan 18 at 2:11
$begingroup$
You don't apply it to this function directly, but rather start by recognizing this as the value of an automorphism of $mathbb{D}$ evaluated at a point.
$endgroup$
– T. Bongers
Jan 18 at 2:11
2
2
$begingroup$
Possible duplicate of Show that $left|frac{alpha - beta}{1-bar{alpha}beta}right| < 1$ when $|alpha|,|beta| < 1$
$endgroup$
– Martin R
Jan 18 at 6:16
$begingroup$
Possible duplicate of Show that $left|frac{alpha - beta}{1-bar{alpha}beta}right| < 1$ when $|alpha|,|beta| < 1$
$endgroup$
– Martin R
Jan 18 at 6:16
$begingroup$
Also: math.stackexchange.com/q/1630930/42969, math.stackexchange.com/q/342181/42969.
$endgroup$
– Martin R
Jan 18 at 6:17
$begingroup$
Also: math.stackexchange.com/q/1630930/42969, math.stackexchange.com/q/342181/42969.
$endgroup$
– Martin R
Jan 18 at 6:17
add a comment |
1 Answer
1
active
oldest
votes
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Your inequality is equivalent to $|1 - s/z| < |1/z - bar{s}| iff |z - s|^2 < |1 - bar{s} z|^2$. Expanding yields
$$ z bar{z} + s bar{s} - z bar{s} - bar{z} s < 1 - z bar{z} s bar{s} - z bar{s} - s bar{z}$$
So we need to show $|z|^2 + |s|^2 < 1 + |s|^2|z|^2$. But this is equivalent to $0 < (1 - |z|^2) ( 1 - |s|^2)$ which is clearly true since $|z|, |s| < 1$.
edited Jan 18 at 6:18
Martin R
31k33561
answered Jan 18 at 1:26
0x5390x539
1,470518
$endgroup$
$begingroup$
Seems obvious now. Thanks!
$endgroup$
– Squirtle
Jan 18 at 1:46
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your inequality is equivalent to $|1 - s/z| < |1/z - bar{s}| iff |z - s|^2 < |1 - bar{s} z|^2$. Expanding yields
$$ z bar{z} + s bar{s} - z bar{s} - bar{z} s < 1 - z bar{z} s bar{s} - z bar{s} - s bar{z}$$
So we need to show $|z|^2 + |s|^2 < 1 + |s|^2|z|^2$. But this is equivalent to $0 < (1 - |z|^2) ( 1 - |s|^2)$ which is clearly true since $|z|, |s| < 1$.
edited Jan 18 at 6:18
Martin R
31k33561
answered Jan 18 at 1:26
0x5390x539
1,470518
$endgroup$
$begingroup$
Seems obvious now. Thanks!
$endgroup$
– Squirtle
Jan 18 at 1:46
add a comment |
$begingroup$
Your inequality is equivalent to $|1 - s/z| < |1/z - bar{s}| iff |z - s|^2 < |1 - bar{s} z|^2$. Expanding yields
$$ z bar{z} + s bar{s} - z bar{s} - bar{z} s < 1 - z bar{z} s bar{s} - z bar{s} - s bar{z}$$
So we need to show $|z|^2 + |s|^2 < 1 + |s|^2|z|^2$. But this is equivalent to $0 < (1 - |z|^2) ( 1 - |s|^2)$ which is clearly true since $|z|, |s| < 1$.
edited Jan 18 at 6:18
Martin R
31k33561
answered Jan 18 at 1:26
0x5390x539
1,470518
$endgroup$
$begingroup$
Seems obvious now. Thanks!
$endgroup$
– Squirtle
Jan 18 at 1:46
add a comment |
$begingroup$
Your inequality is equivalent to $|1 - s/z| < |1/z - bar{s}| iff |z - s|^2 < |1 - bar{s} z|^2$. Expanding yields
$$ z bar{z} + s bar{s} - z bar{s} - bar{z} s < 1 - z bar{z} s bar{s} - z bar{s} - s bar{z}$$
So we need to show $|z|^2 + |s|^2 < 1 + |s|^2|z|^2$. But this is equivalent to $0 < (1 - |z|^2) ( 1 - |s|^2)$ which is clearly true since $|z|, |s| < 1$.
edited Jan 18 at 6:18
Martin R
31k33561
answered Jan 18 at 1:26
0x5390x539
1,470518
$endgroup$
Your inequality is equivalent to $|1 - s/z| < |1/z - bar{s}| iff |z - s|^2 < |1 - bar{s} z|^2$. Expanding yields
$$ z bar{z} + s bar{s} - z bar{s} - bar{z} s < 1 - z bar{z} s bar{s} - z bar{s} - s bar{z}$$
So we need to show $|z|^2 + |s|^2 < 1 + |s|^2|z|^2$. But this is equivalent to $0 < (1 - |z|^2) ( 1 - |s|^2)$ which is clearly true since $|z|, |s| < 1$.
edited Jan 18 at 6:18
Martin R
31k33561
answered Jan 18 at 1:26
0x5390x539
1,470518
edited Jan 18 at 6:18
Martin R
31k33561
edited Jan 18 at 6:18
Martin R
31k33561
edited Jan 18 at 6:18
Martin R
31k33561
31k33561
answered Jan 18 at 1:26
0x5390x539
1,470518
answered Jan 18 at 1:26
0x5390x539
1,470518
answered Jan 18 at 1:26
0x5390x539
1,470518
1,470518
$begingroup$
Seems obvious now. Thanks!
$endgroup$
– Squirtle
Jan 18 at 1:46
add a comment |
$begingroup$
Seems obvious now. Thanks!
$endgroup$
– Squirtle
Jan 18 at 1:46
$begingroup$
Seems obvious now. Thanks!
$endgroup$
– Squirtle
Jan 18 at 1:46
$begingroup$
Seems obvious now. Thanks!
$endgroup$
– Squirtle
Jan 18 at 1:46
add a comment |
1
$begingroup$
An analytic way: Have a look at the automorphism group of the unit disk, and use the Schwarz lemma.
$endgroup$
– T. Bongers
Jan 18 at 1:22
$begingroup$
I don't see how to apply that theorem since $|f(0)|=|s|$ which isn't necessarily zero.
$endgroup$
– Squirtle
Jan 18 at 1:37
$begingroup$
You don't apply it to this function directly, but rather start by recognizing this as the value of an automorphism of $mathbb{D}$ evaluated at a point.
$endgroup$
– T. Bongers
Jan 18 at 2:11
2
$begingroup$
Possible duplicate of Show that $left|frac{alpha - beta}{1-bar{alpha}beta}right| < 1$ when $|alpha|,|beta| < 1$
$endgroup$
– Martin R
Jan 18 at 6:16
$begingroup$
Also: math.stackexchange.com/q/1630930/42969, math.stackexchange.com/q/342181/42969.
$endgroup$
– Martin R
Jan 18 at 6:17