Find the value of Given function?
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0
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Let, $f(z)$ be analytics in $|z| le 1$ and $|f(z) |le 1$ with $f(0) = frac{1 +i}{sqrt 2}$ . Then find the value of $f(i) -f(1)$
I Thinks it will be $0$ by Liouville theorem.
Is Its correct ??
complex-analysis
add a comment |
up vote
0
down vote
favorite
Let, $f(z)$ be analytics in $|z| le 1$ and $|f(z) |le 1$ with $f(0) = frac{1 +i}{sqrt 2}$ . Then find the value of $f(i) -f(1)$
I Thinks it will be $0$ by Liouville theorem.
Is Its correct ??
complex-analysis
1
Since Liouville's theorem is for entire functions, I don't see how you plan to apply it here.
– José Carlos Santos
Nov 19 at 15:14
Okkkss sir @jose Carlos santos
– santosh
Nov 19 at 15:17
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let, $f(z)$ be analytics in $|z| le 1$ and $|f(z) |le 1$ with $f(0) = frac{1 +i}{sqrt 2}$ . Then find the value of $f(i) -f(1)$
I Thinks it will be $0$ by Liouville theorem.
Is Its correct ??
complex-analysis
Let, $f(z)$ be analytics in $|z| le 1$ and $|f(z) |le 1$ with $f(0) = frac{1 +i}{sqrt 2}$ . Then find the value of $f(i) -f(1)$
I Thinks it will be $0$ by Liouville theorem.
Is Its correct ??
complex-analysis
complex-analysis
edited Nov 19 at 17:43
Empty
8,04742358
8,04742358
asked Nov 19 at 15:11
santosh
898
898
1
Since Liouville's theorem is for entire functions, I don't see how you plan to apply it here.
– José Carlos Santos
Nov 19 at 15:14
Okkkss sir @jose Carlos santos
– santosh
Nov 19 at 15:17
add a comment |
1
Since Liouville's theorem is for entire functions, I don't see how you plan to apply it here.
– José Carlos Santos
Nov 19 at 15:14
Okkkss sir @jose Carlos santos
– santosh
Nov 19 at 15:17
1
1
Since Liouville's theorem is for entire functions, I don't see how you plan to apply it here.
– José Carlos Santos
Nov 19 at 15:14
Since Liouville's theorem is for entire functions, I don't see how you plan to apply it here.
– José Carlos Santos
Nov 19 at 15:14
Okkkss sir @jose Carlos santos
– santosh
Nov 19 at 15:17
Okkkss sir @jose Carlos santos
– santosh
Nov 19 at 15:17
add a comment |
1 Answer
1
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oldest
votes
up vote
2
down vote
accepted
Hint : $|f(0)|=1$. Use Maximum Modulus Theorem, which states that "A non constant analytic function in a domain $D$ attains maximum value on it's boundary $partial D$.
Then answer will be 0 na??
– santosh
Nov 19 at 15:17
1
@santosh Yeah !
– Empty
Nov 19 at 15:18
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Hint : $|f(0)|=1$. Use Maximum Modulus Theorem, which states that "A non constant analytic function in a domain $D$ attains maximum value on it's boundary $partial D$.
Then answer will be 0 na??
– santosh
Nov 19 at 15:17
1
@santosh Yeah !
– Empty
Nov 19 at 15:18
add a comment |
up vote
2
down vote
accepted
Hint : $|f(0)|=1$. Use Maximum Modulus Theorem, which states that "A non constant analytic function in a domain $D$ attains maximum value on it's boundary $partial D$.
Then answer will be 0 na??
– santosh
Nov 19 at 15:17
1
@santosh Yeah !
– Empty
Nov 19 at 15:18
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Hint : $|f(0)|=1$. Use Maximum Modulus Theorem, which states that "A non constant analytic function in a domain $D$ attains maximum value on it's boundary $partial D$.
Hint : $|f(0)|=1$. Use Maximum Modulus Theorem, which states that "A non constant analytic function in a domain $D$ attains maximum value on it's boundary $partial D$.
edited Nov 19 at 15:17
answered Nov 19 at 15:14
Empty
8,04742358
8,04742358
Then answer will be 0 na??
– santosh
Nov 19 at 15:17
1
@santosh Yeah !
– Empty
Nov 19 at 15:18
add a comment |
Then answer will be 0 na??
– santosh
Nov 19 at 15:17
1
@santosh Yeah !
– Empty
Nov 19 at 15:18
Then answer will be 0 na??
– santosh
Nov 19 at 15:17
Then answer will be 0 na??
– santosh
Nov 19 at 15:17
1
1
@santosh Yeah !
– Empty
Nov 19 at 15:18
@santosh Yeah !
– Empty
Nov 19 at 15:18
add a comment |
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Since Liouville's theorem is for entire functions, I don't see how you plan to apply it here.
– José Carlos Santos
Nov 19 at 15:14
Okkkss sir @jose Carlos santos
– santosh
Nov 19 at 15:17