holomorphy domain
$begingroup$
could someone please help me with this exercise?
Let $$f_n=frac{logleft(z-frac{1}{n}right)}{log^frac{3}{2}(n)left(z^2+nright)}$$
Where $log(z)$ is the natural branch of the function logarithm.
Find the holomorphy domain of $f(z)=sumlimits_{n=2}^{infty} f_n$.
I tried to use the fact that a series of holomorphic functions witch converge nomally to a function $f$ implies the holomorphy of $f$ but i can't find a majoration.
I also would be happy to recive other approaches to this kind of exercises.
complex-analysis
edited Jan 7 at 21:26
Fabio
1029
asked Jan 7 at 20:06
Elia OrsiElia Orsi
13
$endgroup$
add a comment |
$begingroup$
could someone please help me with this exercise?
Let $$f_n=frac{logleft(z-frac{1}{n}right)}{log^frac{3}{2}(n)left(z^2+nright)}$$
Where $log(z)$ is the natural branch of the function logarithm.
Find the holomorphy domain of $f(z)=sumlimits_{n=2}^{infty} f_n$.
I tried to use the fact that a series of holomorphic functions witch converge nomally to a function $f$ implies the holomorphy of $f$ but i can't find a majoration.
I also would be happy to recive other approaches to this kind of exercises.
complex-analysis
edited Jan 7 at 21:26
Fabio
1029
asked Jan 7 at 20:06
Elia OrsiElia Orsi
13
$endgroup$
add a comment |
$begingroup$
could someone please help me with this exercise?
Let $$f_n=frac{logleft(z-frac{1}{n}right)}{log^frac{3}{2}(n)left(z^2+nright)}$$
Where $log(z)$ is the natural branch of the function logarithm.
Find the holomorphy domain of $f(z)=sumlimits_{n=2}^{infty} f_n$.
I tried to use the fact that a series of holomorphic functions witch converge nomally to a function $f$ implies the holomorphy of $f$ but i can't find a majoration.
I also would be happy to recive other approaches to this kind of exercises.
complex-analysis
edited Jan 7 at 21:26
Fabio
1029
asked Jan 7 at 20:06
Elia OrsiElia Orsi
13
$endgroup$
could someone please help me with this exercise?
Let $$f_n=frac{logleft(z-frac{1}{n}right)}{log^frac{3}{2}(n)left(z^2+nright)}$$
Where $log(z)$ is the natural branch of the function logarithm.
Find the holomorphy domain of $f(z)=sumlimits_{n=2}^{infty} f_n$.
I tried to use the fact that a series of holomorphic functions witch converge nomally to a function $f$ implies the holomorphy of $f$ but i can't find a majoration.
I also would be happy to recive other approaches to this kind of exercises.
complex-analysis
complex-analysis
edited Jan 7 at 21:26
Fabio
1029
asked Jan 7 at 20:06
Elia OrsiElia Orsi
13
edited Jan 7 at 21:26
Fabio
1029
asked Jan 7 at 20:06
Elia OrsiElia Orsi
13
edited Jan 7 at 21:26
Fabio
1029
edited Jan 7 at 21:26
Fabio
1029
edited Jan 7 at 21:26
Fabio
1029
1029
asked Jan 7 at 20:06
Elia OrsiElia Orsi
13
asked Jan 7 at 20:06
Elia OrsiElia Orsi
13
asked Jan 7 at 20:06
Elia OrsiElia Orsi
13
13
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint: For each $n$ determine where $f_n$ is well defined. This will be a set of the form $mathbb Csetminus E_n.$ The set where $sum f_n$ makes sense can be no larger than $mathbb Csetminus (cup ,E_n).$
answered Jan 7 at 21:07
zhw.zhw.
71.9k43075
$endgroup$
$begingroup$
Thank you, $E_n$ should be {$zin mathbb{C}:zneq isqrt{n},$ $zneq-isqrt{n}$ } $cup$ $ ${$zin mathbb{C}: Re(z)leq frac{1}{n}$ $and $ $ Img(z)=0$}.
$endgroup$
– Elia Orsi
Jan 7 at 22:10
$begingroup$
I would write $E_n = {isqrt n,-isqrt n}cup (-infty,1/n].$
$endgroup$
– zhw.
Jan 7 at 23:26
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Hint: For each $n$ determine where $f_n$ is well defined. This will be a set of the form $mathbb Csetminus E_n.$ The set where $sum f_n$ makes sense can be no larger than $mathbb Csetminus (cup ,E_n).$
answered Jan 7 at 21:07
zhw.zhw.
71.9k43075
$endgroup$
$begingroup$
Thank you, $E_n$ should be {$zin mathbb{C}:zneq isqrt{n},$ $zneq-isqrt{n}$ } $cup$ $ ${$zin mathbb{C}: Re(z)leq frac{1}{n}$ $and $ $ Img(z)=0$}.
$endgroup$
– Elia Orsi
Jan 7 at 22:10
$begingroup$
I would write $E_n = {isqrt n,-isqrt n}cup (-infty,1/n].$
$endgroup$
– zhw.
Jan 7 at 23:26
add a comment |
$begingroup$
Hint: For each $n$ determine where $f_n$ is well defined. This will be a set of the form $mathbb Csetminus E_n.$ The set where $sum f_n$ makes sense can be no larger than $mathbb Csetminus (cup ,E_n).$
answered Jan 7 at 21:07
zhw.zhw.
71.9k43075
$endgroup$
$begingroup$
Thank you, $E_n$ should be {$zin mathbb{C}:zneq isqrt{n},$ $zneq-isqrt{n}$ } $cup$ $ ${$zin mathbb{C}: Re(z)leq frac{1}{n}$ $and $ $ Img(z)=0$}.
$endgroup$
– Elia Orsi
Jan 7 at 22:10
$begingroup$
I would write $E_n = {isqrt n,-isqrt n}cup (-infty,1/n].$
$endgroup$
– zhw.
Jan 7 at 23:26
add a comment |
$begingroup$
Hint: For each $n$ determine where $f_n$ is well defined. This will be a set of the form $mathbb Csetminus E_n.$ The set where $sum f_n$ makes sense can be no larger than $mathbb Csetminus (cup ,E_n).$
answered Jan 7 at 21:07
zhw.zhw.
71.9k43075
$endgroup$
Hint: For each $n$ determine where $f_n$ is well defined. This will be a set of the form $mathbb Csetminus E_n.$ The set where $sum f_n$ makes sense can be no larger than $mathbb Csetminus (cup ,E_n).$
answered Jan 7 at 21:07
zhw.zhw.
71.9k43075
answered Jan 7 at 21:07
zhw.zhw.
71.9k43075
answered Jan 7 at 21:07
zhw.zhw.
71.9k43075
answered Jan 7 at 21:07
zhw.zhw.
71.9k43075
71.9k43075
$begingroup$
Thank you, $E_n$ should be {$zin mathbb{C}:zneq isqrt{n},$ $zneq-isqrt{n}$ } $cup$ $ ${$zin mathbb{C}: Re(z)leq frac{1}{n}$ $and $ $ Img(z)=0$}.
$endgroup$
– Elia Orsi
Jan 7 at 22:10
$begingroup$
I would write $E_n = {isqrt n,-isqrt n}cup (-infty,1/n].$
$endgroup$
– zhw.
Jan 7 at 23:26
add a comment |
$begingroup$
Thank you, $E_n$ should be {$zin mathbb{C}:zneq isqrt{n},$ $zneq-isqrt{n}$ } $cup$ $ ${$zin mathbb{C}: Re(z)leq frac{1}{n}$ $and $ $ Img(z)=0$}.
$endgroup$
– Elia Orsi
Jan 7 at 22:10
$begingroup$
I would write $E_n = {isqrt n,-isqrt n}cup (-infty,1/n].$
$endgroup$
– zhw.
Jan 7 at 23:26
$begingroup$
Thank you, $E_n$ should be {$zin mathbb{C}:zneq isqrt{n},$ $zneq-isqrt{n}$ } $cup$ $ ${$zin mathbb{C}: Re(z)leq frac{1}{n}$ $and $ $ Img(z)=0$}.
$endgroup$
– Elia Orsi
Jan 7 at 22:10
$begingroup$
Thank you, $E_n$ should be {$zin mathbb{C}:zneq isqrt{n},$ $zneq-isqrt{n}$ } $cup$ $ ${$zin mathbb{C}: Re(z)leq frac{1}{n}$ $and $ $ Img(z)=0$}.
$endgroup$
– Elia Orsi
Jan 7 at 22:10
$begingroup$
I would write $E_n = {isqrt n,-isqrt n}cup (-infty,1/n].$
$endgroup$
– zhw.
Jan 7 at 23:26
$begingroup$
I would write $E_n = {isqrt n,-isqrt n}cup (-infty,1/n].$
$endgroup$
– zhw.
Jan 7 at 23:26
add a comment |
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