Proving if $|f(z)|leqfrac2R^2+1$ is true
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Suppose that $$f(z)=frace^-zz^2+1.$$
Is it true that $$|f(z)|leqfrac2R^2+1$$ for all $z$ such that $Re(z)geq 0$ and $|z|>R$?
My attempt:
consider the triangle inequality beginalign
|z^2+1|&leq |z|^2+1 \
frac1&geq frac1 \
&leqfrac1R^2+1 \
frac&leq frac \
&=frac1e^Re(z)frac1R^2+1 \
&leqfrac1R^2+1
endalign
While my working suggests the statement is incorrect, I wonder if the logic from my second to third line is correct? Any advice would be greatly appreciated.
complex-analysis proof-verification
asked Aug 29 at 4:26
Bell
826314
add a comment |Â
up vote
1
down vote
favorite
Suppose that $$f(z)=frace^-zz^2+1.$$
Is it true that $$|f(z)|leqfrac2R^2+1$$ for all $z$ such that $Re(z)geq 0$ and $|z|>R$?
My attempt:
consider the triangle inequality beginalign
|z^2+1|&leq |z|^2+1 \
frac1&geq frac1 \
&leqfrac1R^2+1 \
frac&leq frac \
&=frac1e^Re(z)frac1R^2+1 \
&leqfrac1R^2+1
endalign
While my working suggests the statement is incorrect, I wonder if the logic from my second to third line is correct? Any advice would be greatly appreciated.
complex-analysis proof-verification
asked Aug 29 at 4:26
Bell
826314
No, from the second to the third line is not correct. Knowing that $b$ is less than both $a$ and $c$ tells you nothing about the relationship between $a$ and $c$.
– Theo Bendit
Aug 29 at 4:38
You may write $|z^2+1|ge |z|^2-1$. Then $displaystyle frac1le frac1$
– Empty
Aug 29 at 5:10
Yep. By the reverse triangle angle inequality. Thanks!
– Bell
Aug 29 at 5:59
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Suppose that $$f(z)=frace^-zz^2+1.$$
Is it true that $$|f(z)|leqfrac2R^2+1$$ for all $z$ such that $Re(z)geq 0$ and $|z|>R$?
My attempt:
consider the triangle inequality beginalign
|z^2+1|&leq |z|^2+1 \
frac1&geq frac1 \
&leqfrac1R^2+1 \
frac&leq frac \
&=frac1e^Re(z)frac1R^2+1 \
&leqfrac1R^2+1
endalign
While my working suggests the statement is incorrect, I wonder if the logic from my second to third line is correct? Any advice would be greatly appreciated.
complex-analysis proof-verification
asked Aug 29 at 4:26
Bell
826314
Suppose that $$f(z)=frace^-zz^2+1.$$
Is it true that $$|f(z)|leqfrac2R^2+1$$ for all $z$ such that $Re(z)geq 0$ and $|z|>R$?
My attempt:
consider the triangle inequality beginalign
|z^2+1|&leq |z|^2+1 \
frac1&geq frac1 \
&leqfrac1R^2+1 \
frac&leq frac \
&=frac1e^Re(z)frac1R^2+1 \
&leqfrac1R^2+1
endalign
While my working suggests the statement is incorrect, I wonder if the logic from my second to third line is correct? Any advice would be greatly appreciated.
complex-analysis proof-verification
asked Aug 29 at 4:26
Bell
826314
edited Aug 29 at 4:34
asked Aug 29 at 4:26
Bell
826314
asked Aug 29 at 4:26
Bell
826314
asked Aug 29 at 4:26
Bell
826314
826314
No, from the second to the third line is not correct. Knowing that $b$ is less than both $a$ and $c$ tells you nothing about the relationship between $a$ and $c$.
– Theo Bendit
Aug 29 at 4:38
You may write $|z^2+1|ge |z|^2-1$. Then $displaystyle frac1le frac1$
– Empty
Aug 29 at 5:10
Yep. By the reverse triangle angle inequality. Thanks!
– Bell
Aug 29 at 5:59
add a comment |Â
No, from the second to the third line is not correct. Knowing that $b$ is less than both $a$ and $c$ tells you nothing about the relationship between $a$ and $c$.
– Theo Bendit
Aug 29 at 4:38
You may write $|z^2+1|ge |z|^2-1$. Then $displaystyle frac1le frac1$
– Empty
Aug 29 at 5:10
Yep. By the reverse triangle angle inequality. Thanks!
– Bell
Aug 29 at 5:59
No, from the second to the third line is not correct. Knowing that $b$ is less than both $a$ and $c$ tells you nothing about the relationship between $a$ and $c$.
– Theo Bendit
Aug 29 at 4:38
No, from the second to the third line is not correct. Knowing that $b$ is less than both $a$ and $c$ tells you nothing about the relationship between $a$ and $c$.
– Theo Bendit
Aug 29 at 4:38
You may write $|z^2+1|ge |z|^2-1$. Then $displaystyle frac1le frac1$
– Empty
Aug 29 at 5:10
You may write $|z^2+1|ge |z|^2-1$. Then $displaystyle frac1le frac1$
– Empty
Aug 29 at 5:10
Yep. By the reverse triangle angle inequality. Thanks!
– Bell
Aug 29 at 5:59
Yep. By the reverse triangle angle inequality. Thanks!
– Bell
Aug 29 at 5:59
add a comment |Â
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No, from the second to the third line is not correct. Knowing that $b$ is less than both $a$ and $c$ tells you nothing about the relationship between $a$ and $c$.
– Theo Bendit
Aug 29 at 4:38
You may write $|z^2+1|ge |z|^2-1$. Then $displaystyle frac1le frac1$
– Empty
Aug 29 at 5:10
Yep. By the reverse triangle angle inequality. Thanks!
– Bell
Aug 29 at 5:59