Perturbation of complex square root function
up vote
1
down vote
favorite
Let $z$ and $w$ be complex numbers. Than it is claimed that
$$
|sqrt{z+w} - sqrt{z}| leq frac{|w|}{sqrt{|z| + |w|}}
$$
or
$$
|sqrt{z+w} + sqrt{z}| leq frac{|w|}{sqrt{|z| + |w|}}
$$
I am hoping someone can show me how to perform such a perturbation analysis in complex analysis. I am able to prove this for real numbers via Taylor expansion.
complex-analysis inequality perturbation-theory
asked Nov 23 at 17:26
JohnKnoxV
500113
add a comment |
up vote
1
down vote
favorite
Let $z$ and $w$ be complex numbers. Than it is claimed that
$$
|sqrt{z+w} - sqrt{z}| leq frac{|w|}{sqrt{|z| + |w|}}
$$
or
$$
|sqrt{z+w} + sqrt{z}| leq frac{|w|}{sqrt{|z| + |w|}}
$$
I am hoping someone can show me how to perform such a perturbation analysis in complex analysis. I am able to prove this for real numbers via Taylor expansion.
complex-analysis inequality perturbation-theory
asked Nov 23 at 17:26
JohnKnoxV
500113
You also have to specify which branch of the complex square root is chosen (there are two choices). If you are thinking of the “main branch” (which maps into the right half plane) then this looks wrong for $z = -1 -iepsilon$, $z+w= -1 + iepsilon$.
– Martin R
Nov 23 at 17:47
Yes, but the previous conjecture was false. I'm adding the "or" condition here.
– JohnKnoxV
Nov 23 at 18:44
Sorry, I've fixed a typo. The two inequalities differ by a "+"
– JohnKnoxV
Nov 23 at 18:44
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $z$ and $w$ be complex numbers. Than it is claimed that
$$
|sqrt{z+w} - sqrt{z}| leq frac{|w|}{sqrt{|z| + |w|}}
$$
or
$$
|sqrt{z+w} + sqrt{z}| leq frac{|w|}{sqrt{|z| + |w|}}
$$
I am hoping someone can show me how to perform such a perturbation analysis in complex analysis. I am able to prove this for real numbers via Taylor expansion.
complex-analysis inequality perturbation-theory
asked Nov 23 at 17:26
JohnKnoxV
500113
Let $z$ and $w$ be complex numbers. Than it is claimed that
$$
|sqrt{z+w} - sqrt{z}| leq frac{|w|}{sqrt{|z| + |w|}}
$$
or
$$
|sqrt{z+w} + sqrt{z}| leq frac{|w|}{sqrt{|z| + |w|}}
$$
I am hoping someone can show me how to perform such a perturbation analysis in complex analysis. I am able to prove this for real numbers via Taylor expansion.
complex-analysis inequality perturbation-theory
complex-analysis inequality perturbation-theory
asked Nov 23 at 17:26
JohnKnoxV
500113
asked Nov 23 at 17:26
JohnKnoxV
500113
edited Nov 23 at 18:43
asked Nov 23 at 17:26
JohnKnoxV
500113
asked Nov 23 at 17:26
JohnKnoxV
500113
asked Nov 23 at 17:26
JohnKnoxV
500113
500113
You also have to specify which branch of the complex square root is chosen (there are two choices). If you are thinking of the “main branch” (which maps into the right half plane) then this looks wrong for $z = -1 -iepsilon$, $z+w= -1 + iepsilon$.
– Martin R
Nov 23 at 17:47
Yes, but the previous conjecture was false. I'm adding the "or" condition here.
– JohnKnoxV
Nov 23 at 18:44
Sorry, I've fixed a typo. The two inequalities differ by a "+"
– JohnKnoxV
Nov 23 at 18:44
add a comment |
You also have to specify which branch of the complex square root is chosen (there are two choices). If you are thinking of the “main branch” (which maps into the right half plane) then this looks wrong for $z = -1 -iepsilon$, $z+w= -1 + iepsilon$.
– Martin R
Nov 23 at 17:47
Yes, but the previous conjecture was false. I'm adding the "or" condition here.
– JohnKnoxV
Nov 23 at 18:44
Sorry, I've fixed a typo. The two inequalities differ by a "+"
– JohnKnoxV
Nov 23 at 18:44
You also have to specify which branch of the complex square root is chosen (there are two choices). If you are thinking of the “main branch” (which maps into the right half plane) then this looks wrong for $z = -1 -iepsilon$, $z+w= -1 + iepsilon$.
– Martin R
Nov 23 at 17:47
You also have to specify which branch of the complex square root is chosen (there are two choices). If you are thinking of the “main branch” (which maps into the right half plane) then this looks wrong for $z = -1 -iepsilon$, $z+w= -1 + iepsilon$.
– Martin R
Nov 23 at 17:47
Yes, but the previous conjecture was false. I'm adding the "or" condition here.
– JohnKnoxV
Nov 23 at 18:44
Yes, but the previous conjecture was false. I'm adding the "or" condition here.
– JohnKnoxV
Nov 23 at 18:44
Sorry, I've fixed a typo. The two inequalities differ by a "+"
– JohnKnoxV
Nov 23 at 18:44
Sorry, I've fixed a typo. The two inequalities differ by a "+"
– JohnKnoxV
Nov 23 at 18:44
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
This claim is absolutely wrong. Choose $w=-1,z=1$ therefore$$|sqrt{z+w}-sqrt z|=1\|sqrt{z+w}+sqrt z|=1$$while $${|w|over sqrt{|w|+|z|}}={1oversqrt 2}$$but $$1notle {1over sqrt 2}$$
answered Nov 27 at 21:52
Mostafa Ayaz
13.3k3836
Oh wow, you're right.
– JohnKnoxV
Nov 28 at 1:12
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
This claim is absolutely wrong. Choose $w=-1,z=1$ therefore$$|sqrt{z+w}-sqrt z|=1\|sqrt{z+w}+sqrt z|=1$$while $${|w|over sqrt{|w|+|z|}}={1oversqrt 2}$$but $$1notle {1over sqrt 2}$$
answered Nov 27 at 21:52
Mostafa Ayaz
13.3k3836
Oh wow, you're right.
– JohnKnoxV
Nov 28 at 1:12
add a comment |
up vote
1
down vote
accepted
This claim is absolutely wrong. Choose $w=-1,z=1$ therefore$$|sqrt{z+w}-sqrt z|=1\|sqrt{z+w}+sqrt z|=1$$while $${|w|over sqrt{|w|+|z|}}={1oversqrt 2}$$but $$1notle {1over sqrt 2}$$
answered Nov 27 at 21:52
Mostafa Ayaz
13.3k3836
Oh wow, you're right.
– JohnKnoxV
Nov 28 at 1:12
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
This claim is absolutely wrong. Choose $w=-1,z=1$ therefore$$|sqrt{z+w}-sqrt z|=1\|sqrt{z+w}+sqrt z|=1$$while $${|w|over sqrt{|w|+|z|}}={1oversqrt 2}$$but $$1notle {1over sqrt 2}$$
answered Nov 27 at 21:52
Mostafa Ayaz
13.3k3836
This claim is absolutely wrong. Choose $w=-1,z=1$ therefore$$|sqrt{z+w}-sqrt z|=1\|sqrt{z+w}+sqrt z|=1$$while $${|w|over sqrt{|w|+|z|}}={1oversqrt 2}$$but $$1notle {1over sqrt 2}$$
answered Nov 27 at 21:52
Mostafa Ayaz
13.3k3836
answered Nov 27 at 21:52
Mostafa Ayaz
13.3k3836
answered Nov 27 at 21:52
Mostafa Ayaz
13.3k3836
answered Nov 27 at 21:52
Mostafa Ayaz
13.3k3836
13.3k3836
Oh wow, you're right.
– JohnKnoxV
Nov 28 at 1:12
add a comment |
Oh wow, you're right.
– JohnKnoxV
Nov 28 at 1:12
Oh wow, you're right.
– JohnKnoxV
Nov 28 at 1:12
Oh wow, you're right.
– JohnKnoxV
Nov 28 at 1:12
add a comment |
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You also have to specify which branch of the complex square root is chosen (there are two choices). If you are thinking of the “main branch” (which maps into the right half plane) then this looks wrong for $z = -1 -iepsilon$, $z+w= -1 + iepsilon$.
– Martin R
Nov 23 at 17:47
Yes, but the previous conjecture was false. I'm adding the "or" condition here.
– JohnKnoxV
Nov 23 at 18:44
Sorry, I've fixed a typo. The two inequalities differ by a "+"
– JohnKnoxV
Nov 23 at 18:44