How to find value of an analytic function at a particular value given some conditions? [closed]

I've come across a question from gate 2018 paper.Not able to come up with solution,can somebody give some idea to solve this.

Let f : C → C be an entire function with f(0) = 1, f(1) = 2 and f′(0) = 0. If there exists
M > 0 such that |f′′(z)| ≤ M for all z ∈ C, then f(2) =
(A) 2 (B) 5 (C) 2 + 5i (D) 5 + 2i

Thanks in advance.

asked Dec 18 '18 at 4:03

sgn

274

closed as off-topic by mrtaurho, José Carlos Santos, Cesareo, Brahadeesh, user10354138 Dec 18 '18 at 12:02

This question appears to be off-topic. The users who voted to close gave this specific reason:

"This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – mrtaurho, José Carlos Santos, Cesareo, Brahadeesh, user10354138

If this question can be reworded to fit the rules in the help center, please edit the question.

1

$begingroup$
If you had the condition $|f(z)|le M$ for all $M$, then $f$ would be a constant, by Liouville. But you don't have that, you have $|f''(z)|le M$. What does this tell you about $f$?
$endgroup$
– Lord Shark the Unknown
Dec 18 '18 at 4:12

add a comment |

I've come across a question from gate 2018 paper.Not able to come up with solution,can somebody give some idea to solve this.

Let f : C → C be an entire function with f(0) = 1, f(1) = 2 and f′(0) = 0. If there exists
M > 0 such that |f′′(z)| ≤ M for all z ∈ C, then f(2) =
(A) 2 (B) 5 (C) 2 + 5i (D) 5 + 2i

Thanks in advance.

asked Dec 18 '18 at 4:03

sgn

274

closed as off-topic by mrtaurho, José Carlos Santos, Cesareo, Brahadeesh, user10354138 Dec 18 '18 at 12:02

This question appears to be off-topic. The users who voted to close gave this specific reason:

"This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – mrtaurho, José Carlos Santos, Cesareo, Brahadeesh, user10354138

If this question can be reworded to fit the rules in the help center, please edit the question.

1

$begingroup$
If you had the condition $|f(z)|le M$ for all $M$, then $f$ would be a constant, by Liouville. But you don't have that, you have $|f''(z)|le M$. What does this tell you about $f$?
$endgroup$
– Lord Shark the Unknown
Dec 18 '18 at 4:12

add a comment |

I've come across a question from gate 2018 paper.Not able to come up with solution,can somebody give some idea to solve this.

Let f : C → C be an entire function with f(0) = 1, f(1) = 2 and f′(0) = 0. If there exists
M > 0 such that |f′′(z)| ≤ M for all z ∈ C, then f(2) =
(A) 2 (B) 5 (C) 2 + 5i (D) 5 + 2i

Thanks in advance.

asked Dec 18 '18 at 4:03

sgn

274

I've come across a question from gate 2018 paper.Not able to come up with solution,can somebody give some idea to solve this.

Let f : C → C be an entire function with f(0) = 1, f(1) = 2 and f′(0) = 0. If there exists
M > 0 such that |f′′(z)| ≤ M for all z ∈ C, then f(2) =
(A) 2 (B) 5 (C) 2 + 5i (D) 5 + 2i

Thanks in advance.

complex-analysis

asked Dec 18 '18 at 4:03

sgn

274

asked Dec 18 '18 at 4:03

sgn

274

asked Dec 18 '18 at 4:03

sgn

274

asked Dec 18 '18 at 4:03

sgn

274

asked Dec 18 '18 at 4:03

sgn

274

closed as off-topic by mrtaurho, José Carlos Santos, Cesareo, Brahadeesh, user10354138 Dec 18 '18 at 12:02

This question appears to be off-topic. The users who voted to close gave this specific reason:

"This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – mrtaurho, José Carlos Santos, Cesareo, Brahadeesh, user10354138

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by mrtaurho, José Carlos Santos, Cesareo, Brahadeesh, user10354138 Dec 18 '18 at 12:02

This question appears to be off-topic. The users who voted to close gave this specific reason:

"This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – mrtaurho, José Carlos Santos, Cesareo, Brahadeesh, user10354138

If this question can be reworded to fit the rules in the help center, please edit the question.

1

$begingroup$
If you had the condition $|f(z)|le M$ for all $M$, then $f$ would be a constant, by Liouville. But you don't have that, you have $|f''(z)|le M$. What does this tell you about $f$?
$endgroup$
– Lord Shark the Unknown
Dec 18 '18 at 4:12

add a comment |

1

$begingroup$
If you had the condition $|f(z)|le M$ for all $M$, then $f$ would be a constant, by Liouville. But you don't have that, you have $|f''(z)|le M$. What does this tell you about $f$?
$endgroup$
– Lord Shark the Unknown
Dec 18 '18 at 4:12

If you had the condition $|f(z)|le M$ for all $M$, then $f$ would be a constant, by Liouville. But you don't have that, you have $|f''(z)|le M$. What does this tell you about $f$?

– Lord Shark the Unknown
Dec 18 '18 at 4:12

add a comment |

1 Answer
1

active

oldest

votes

Since $f(z)$ is entire, so is $f''(z)$; since $vert f''(z) vert < M$, by Liouville's theorem $f''(z)$ is constant:

$f''(z) = a in Bbb C; tag 1$

then we have

$f(z) = dfrac{1}{2}az^2 + bz + c, ; b, c in Bbb C; tag 2$

thus

$f'(z) = az + b; tag 3$

we now use the given data

$f(0) = 1, ; f(1) = 2, ; f'(0) = 0 tag 4$

and find that

$c = 1, dfrac{1}{2}a + b + c = 2, ; b = 0; tag 5$

together these give

$a = 2, tag 6$

whence

$f(z) = z^2 + 1; tag 7$

thus,

$f(2) = 5, tag 8$

and so (B) is correct.

answered Dec 18 '18 at 4:14

Robert Lewis

45k22964

add a comment |

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

Since $f(z)$ is entire, so is $f''(z)$; since $vert f''(z) vert < M$, by Liouville's theorem $f''(z)$ is constant:

$f''(z) = a in Bbb C; tag 1$

then we have

$f(z) = dfrac{1}{2}az^2 + bz + c, ; b, c in Bbb C; tag 2$

thus

$f'(z) = az + b; tag 3$

we now use the given data

$f(0) = 1, ; f(1) = 2, ; f'(0) = 0 tag 4$

and find that

$c = 1, dfrac{1}{2}a + b + c = 2, ; b = 0; tag 5$

together these give

$a = 2, tag 6$

whence

$f(z) = z^2 + 1; tag 7$

thus,

$f(2) = 5, tag 8$

and so (B) is correct.

answered Dec 18 '18 at 4:14

Robert Lewis

45k22964

add a comment |

Since $f(z)$ is entire, so is $f''(z)$; since $vert f''(z) vert < M$, by Liouville's theorem $f''(z)$ is constant:

$f''(z) = a in Bbb C; tag 1$

then we have

$f(z) = dfrac{1}{2}az^2 + bz + c, ; b, c in Bbb C; tag 2$

thus

$f'(z) = az + b; tag 3$

we now use the given data

$f(0) = 1, ; f(1) = 2, ; f'(0) = 0 tag 4$

and find that

$c = 1, dfrac{1}{2}a + b + c = 2, ; b = 0; tag 5$

together these give

$a = 2, tag 6$

whence

$f(z) = z^2 + 1; tag 7$

thus,

$f(2) = 5, tag 8$

and so (B) is correct.

answered Dec 18 '18 at 4:14

Robert Lewis

45k22964

add a comment |

Since $f(z)$ is entire, so is $f''(z)$; since $vert f''(z) vert < M$, by Liouville's theorem $f''(z)$ is constant:

$f''(z) = a in Bbb C; tag 1$

then we have

$f(z) = dfrac{1}{2}az^2 + bz + c, ; b, c in Bbb C; tag 2$

thus

$f'(z) = az + b; tag 3$

we now use the given data

$f(0) = 1, ; f(1) = 2, ; f'(0) = 0 tag 4$

and find that

$c = 1, dfrac{1}{2}a + b + c = 2, ; b = 0; tag 5$

together these give

$a = 2, tag 6$

whence

$f(z) = z^2 + 1; tag 7$

thus,

$f(2) = 5, tag 8$

and so (B) is correct.

answered Dec 18 '18 at 4:14

Robert Lewis

45k22964

Since $f(z)$ is entire, so is $f''(z)$; since $vert f''(z) vert < M$, by Liouville's theorem $f''(z)$ is constant:

$f''(z) = a in Bbb C; tag 1$

then we have

$f(z) = dfrac{1}{2}az^2 + bz + c, ; b, c in Bbb C; tag 2$

thus

$f'(z) = az + b; tag 3$

we now use the given data

$f(0) = 1, ; f(1) = 2, ; f'(0) = 0 tag 4$

and find that

$c = 1, dfrac{1}{2}a + b + c = 2, ; b = 0; tag 5$

together these give

$a = 2, tag 6$

whence

$f(z) = z^2 + 1; tag 7$

thus,

$f(2) = 5, tag 8$

and so (B) is correct.

answered Dec 18 '18 at 4:14

Robert Lewis

45k22964

answered Dec 18 '18 at 4:14

Robert Lewis

45k22964

answered Dec 18 '18 at 4:14

Robert Lewis

45k22964

answered Dec 18 '18 at 4:14

Robert Lewis

45k22964

add a comment |

This page is only for reference, If you need detailed information, please check here

搜尋此網誌

Xrfgtjtk