Mixing
Prove that for $|z| < R$, we have $frac{|z|}{|z^2-n^2|} leq frac{R}{n^2-R^2}$
Prove that for $|z| < R$, we have $frac{|z|}{|z^2-n^2|} leq frac{R}{n^2-R^2}$, where $n$ is an integer
greater than $R$.
This is from page 168, Lang's Complex Analysis. I am having trouble verifying this. Certainly, we have that $|n-z| geq n - |z| geq n-R$, but since $|z| geq |z+n|-n notgeq R$ I am not sure what to do.
complex-analysis inequality
asked Nov 20 '18 at 5:34
Cute Brownie
995416
add a comment |
Prove that for $|z| < R$, we have $frac{|z|}{|z^2-n^2|} leq frac{R}{n^2-R^2}$, where $n$ is an integer
greater than $R$.
This is from page 168, Lang's Complex Analysis. I am having trouble verifying this. Certainly, we have that $|n-z| geq n - |z| geq n-R$, but since $|z| geq |z+n|-n notgeq R$ I am not sure what to do.
complex-analysis inequality
asked Nov 20 '18 at 5:34
Cute Brownie
995416
That's wrong if $n le R$.
– Martin R
Nov 20 '18 at 5:46
add a comment |
Prove that for $|z| < R$, we have $frac{|z|}{|z^2-n^2|} leq frac{R}{n^2-R^2}$, where $n$ is an integer
greater than $R$.
This is from page 168, Lang's Complex Analysis. I am having trouble verifying this. Certainly, we have that $|n-z| geq n - |z| geq n-R$, but since $|z| geq |z+n|-n notgeq R$ I am not sure what to do.
complex-analysis inequality
asked Nov 20 '18 at 5:34
Cute Brownie
995416
Prove that for $|z| < R$, we have $frac{|z|}{|z^2-n^2|} leq frac{R}{n^2-R^2}$, where $n$ is an integer
greater than $R$.
This is from page 168, Lang's Complex Analysis. I am having trouble verifying this. Certainly, we have that $|n-z| geq n - |z| geq n-R$, but since $|z| geq |z+n|-n notgeq R$ I am not sure what to do.
complex-analysis inequality
complex-analysis inequality
asked Nov 20 '18 at 5:34
Cute Brownie
995416
asked Nov 20 '18 at 5:34
Cute Brownie
995416
edited Nov 20 '18 at 6:15
asked Nov 20 '18 at 5:34
Cute Brownie
995416
asked Nov 20 '18 at 5:34
Cute Brownie
995416
asked Nov 20 '18 at 5:34
Cute Brownie
995416
995416
That's wrong if $n le R$.
– Martin R
Nov 20 '18 at 5:46
add a comment |
That's wrong if $n le R$.
– Martin R
Nov 20 '18 at 5:46
That's wrong if $n le R$.
– Martin R
Nov 20 '18 at 5:46
That's wrong if $n le R$.
– Martin R
Nov 20 '18 at 5:46
add a comment |
1 Answer
1
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oldest
votes
Hint:
begin{align}
|z^2-n^2|geq left|n^2-|z|^2right| geq n^2-R^2
end{align}
answered Nov 20 '18 at 5:37
Jacky Chong
17.7k21128
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Hint:
begin{align}
|z^2-n^2|geq left|n^2-|z|^2right| geq n^2-R^2
end{align}
answered Nov 20 '18 at 5:37
Jacky Chong
17.7k21128
add a comment |
Hint:
begin{align}
|z^2-n^2|geq left|n^2-|z|^2right| geq n^2-R^2
end{align}
answered Nov 20 '18 at 5:37
Jacky Chong
17.7k21128
add a comment |
Hint:
begin{align}
|z^2-n^2|geq left|n^2-|z|^2right| geq n^2-R^2
end{align}
answered Nov 20 '18 at 5:37
Jacky Chong
17.7k21128
Hint:
begin{align}
|z^2-n^2|geq left|n^2-|z|^2right| geq n^2-R^2
end{align}
answered Nov 20 '18 at 5:37
Jacky Chong
17.7k21128
answered Nov 20 '18 at 5:37
Jacky Chong
17.7k21128
answered Nov 20 '18 at 5:37
Jacky Chong
17.7k21128
answered Nov 20 '18 at 5:37
Jacky Chong
17.7k21128
17.7k21128
add a comment |
add a comment |
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That's wrong if $n le R$.
– Martin R
Nov 20 '18 at 5:46