Find $lim _ { z rightarrow z _ { 0 } } frac { f ( z ) } { g ( z ) }$ with f and g holomorphic and $z_0$ a...
$begingroup$
Problem : Let $f$ and $g$ holomorphic within the neighbourhood of $z_0$. Knowing that $z_0$ is a zero of order $k$ of $f$, and a zero of order $l$ of $g$ with $l>k$
My Answer : Since f is holomorphic in a neighbourhood of $z_0$ we have :
$$
begin{aligned} f ( z ) & = f left( z _ { 0 } right) + f ^ { prime } left( z _ { 0 } right) left( z - z _ { 0 } right) + frac { f ^ { prime prime } left( z _ { 0 } right) } { 2 ! } left( z - z _ { 0 } right) ^ { 2 } + cdots \ & = left( z - z _ { 0 } right) ^ { l } left[ frac { f ^ { ( l ) } left( z _ { 0 } right) } { l ! } + sum _ { n = l + 1 } ^ { + infty } frac { f ^ { ( n ) } left( z _ { 0 } right) } { n ! } left( z - z _ { 0 } right) ^ { n - l } right] \ & = left( z - z _ { 0 } right) ^ { l } frac { f ^ { ( l ) } left( z _ { 0 } right) } { l ! } + left( z - z _ { 0 } right) ^ { l } F ( z ) end{aligned}
$$
In the same way we obtain :
$$
begin{aligned} g ( z ) & = left( z - z _ { 0 } right) ^ { l } left[ frac { g ^ { ( l ) } left( z _ { 0 } right) } { l ! } + sum _ { n = l + 1 } ^ { + infty } frac { g ^ { ( n ) } left( z _ { 0 } right) } { n ! } left( z - z _ { 0 } right) ^ { n - l } right] \ & = left( z - z _ { 0 } right) ^ { l } frac { g ^ { ( l ) } left( z _ { 0 } right) } { l ! } + left( z - z _ { 0 } right) ^ { l } G ( z ) end{aligned}
$$
My issue : To conclude I would like to write the following steps :
$$
lim _ { z rightarrow z _ { 0 } } left[ frac { f ( z ) } { g ( z ) } right] = lim _ { z rightarrow z _ { 0 } } left[ frac { frac { f ^ { ( l ) } left( z _ { 0 } right) } { l ! } + F ( z ) } { frac { g ^ { ( l ) } left( z _ { 0 } right) } { l ! } + G ( z ) } right] = frac { f ^ { ( l ) } left( z _ { 0 } right) } { g ^ { ( l ) } left( z _ { 0 } right) }
$$
- Since $z_0$ is a zero of order $l$ of $g$ I know that ${ g ^ { ( l ) } left( z _ { 0 } right) } neq 0$.
$G(z_0)$ and $F(z_0)$ need to be equal to $0$ but I'm not sure how to get this property. Where this property would come from?
complex-analysis complex-numbers holomorphic-functions
edited Dec 29 '18 at 12:11
Oscar Lanzi
13.7k12136
asked Dec 29 '18 at 11:47
NotAbelianGroupNotAbelianGroup
20911
$endgroup$
add a comment |
$begingroup$
Problem : Let $f$ and $g$ holomorphic within the neighbourhood of $z_0$. Knowing that $z_0$ is a zero of order $k$ of $f$, and a zero of order $l$ of $g$ with $l>k$
My Answer : Since f is holomorphic in a neighbourhood of $z_0$ we have :
$$
begin{aligned} f ( z ) & = f left( z _ { 0 } right) + f ^ { prime } left( z _ { 0 } right) left( z - z _ { 0 } right) + frac { f ^ { prime prime } left( z _ { 0 } right) } { 2 ! } left( z - z _ { 0 } right) ^ { 2 } + cdots \ & = left( z - z _ { 0 } right) ^ { l } left[ frac { f ^ { ( l ) } left( z _ { 0 } right) } { l ! } + sum _ { n = l + 1 } ^ { + infty } frac { f ^ { ( n ) } left( z _ { 0 } right) } { n ! } left( z - z _ { 0 } right) ^ { n - l } right] \ & = left( z - z _ { 0 } right) ^ { l } frac { f ^ { ( l ) } left( z _ { 0 } right) } { l ! } + left( z - z _ { 0 } right) ^ { l } F ( z ) end{aligned}
$$
In the same way we obtain :
$$
begin{aligned} g ( z ) & = left( z - z _ { 0 } right) ^ { l } left[ frac { g ^ { ( l ) } left( z _ { 0 } right) } { l ! } + sum _ { n = l + 1 } ^ { + infty } frac { g ^ { ( n ) } left( z _ { 0 } right) } { n ! } left( z - z _ { 0 } right) ^ { n - l } right] \ & = left( z - z _ { 0 } right) ^ { l } frac { g ^ { ( l ) } left( z _ { 0 } right) } { l ! } + left( z - z _ { 0 } right) ^ { l } G ( z ) end{aligned}
$$
My issue : To conclude I would like to write the following steps :
$$
lim _ { z rightarrow z _ { 0 } } left[ frac { f ( z ) } { g ( z ) } right] = lim _ { z rightarrow z _ { 0 } } left[ frac { frac { f ^ { ( l ) } left( z _ { 0 } right) } { l ! } + F ( z ) } { frac { g ^ { ( l ) } left( z _ { 0 } right) } { l ! } + G ( z ) } right] = frac { f ^ { ( l ) } left( z _ { 0 } right) } { g ^ { ( l ) } left( z _ { 0 } right) }
$$
- Since $z_0$ is a zero of order $l$ of $g$ I know that ${ g ^ { ( l ) } left( z _ { 0 } right) } neq 0$.
$G(z_0)$ and $F(z_0)$ need to be equal to $0$ but I'm not sure how to get this property. Where this property would come from?
complex-analysis complex-numbers holomorphic-functions
edited Dec 29 '18 at 12:11
Oscar Lanzi
13.7k12136
asked Dec 29 '18 at 11:47
NotAbelianGroupNotAbelianGroup
20911
$endgroup$
$begingroup$
But you have an explicit formula for $F(z)$ and $G(z)$, right? So you just need to plug in $z=z_0$ in these formulas.
$endgroup$
– Did
Dec 29 '18 at 11:53
add a comment |
$begingroup$
Problem : Let $f$ and $g$ holomorphic within the neighbourhood of $z_0$. Knowing that $z_0$ is a zero of order $k$ of $f$, and a zero of order $l$ of $g$ with $l>k$
My Answer : Since f is holomorphic in a neighbourhood of $z_0$ we have :
$$
begin{aligned} f ( z ) & = f left( z _ { 0 } right) + f ^ { prime } left( z _ { 0 } right) left( z - z _ { 0 } right) + frac { f ^ { prime prime } left( z _ { 0 } right) } { 2 ! } left( z - z _ { 0 } right) ^ { 2 } + cdots \ & = left( z - z _ { 0 } right) ^ { l } left[ frac { f ^ { ( l ) } left( z _ { 0 } right) } { l ! } + sum _ { n = l + 1 } ^ { + infty } frac { f ^ { ( n ) } left( z _ { 0 } right) } { n ! } left( z - z _ { 0 } right) ^ { n - l } right] \ & = left( z - z _ { 0 } right) ^ { l } frac { f ^ { ( l ) } left( z _ { 0 } right) } { l ! } + left( z - z _ { 0 } right) ^ { l } F ( z ) end{aligned}
$$
In the same way we obtain :
$$
begin{aligned} g ( z ) & = left( z - z _ { 0 } right) ^ { l } left[ frac { g ^ { ( l ) } left( z _ { 0 } right) } { l ! } + sum _ { n = l + 1 } ^ { + infty } frac { g ^ { ( n ) } left( z _ { 0 } right) } { n ! } left( z - z _ { 0 } right) ^ { n - l } right] \ & = left( z - z _ { 0 } right) ^ { l } frac { g ^ { ( l ) } left( z _ { 0 } right) } { l ! } + left( z - z _ { 0 } right) ^ { l } G ( z ) end{aligned}
$$
My issue : To conclude I would like to write the following steps :
$$
lim _ { z rightarrow z _ { 0 } } left[ frac { f ( z ) } { g ( z ) } right] = lim _ { z rightarrow z _ { 0 } } left[ frac { frac { f ^ { ( l ) } left( z _ { 0 } right) } { l ! } + F ( z ) } { frac { g ^ { ( l ) } left( z _ { 0 } right) } { l ! } + G ( z ) } right] = frac { f ^ { ( l ) } left( z _ { 0 } right) } { g ^ { ( l ) } left( z _ { 0 } right) }
$$
- Since $z_0$ is a zero of order $l$ of $g$ I know that ${ g ^ { ( l ) } left( z _ { 0 } right) } neq 0$.
$G(z_0)$ and $F(z_0)$ need to be equal to $0$ but I'm not sure how to get this property. Where this property would come from?
complex-analysis complex-numbers holomorphic-functions
edited Dec 29 '18 at 12:11
Oscar Lanzi
13.7k12136
asked Dec 29 '18 at 11:47
NotAbelianGroupNotAbelianGroup
20911
$endgroup$
Problem : Let $f$ and $g$ holomorphic within the neighbourhood of $z_0$. Knowing that $z_0$ is a zero of order $k$ of $f$, and a zero of order $l$ of $g$ with $l>k$
My Answer : Since f is holomorphic in a neighbourhood of $z_0$ we have :
$$
begin{aligned} f ( z ) & = f left( z _ { 0 } right) + f ^ { prime } left( z _ { 0 } right) left( z - z _ { 0 } right) + frac { f ^ { prime prime } left( z _ { 0 } right) } { 2 ! } left( z - z _ { 0 } right) ^ { 2 } + cdots \ & = left( z - z _ { 0 } right) ^ { l } left[ frac { f ^ { ( l ) } left( z _ { 0 } right) } { l ! } + sum _ { n = l + 1 } ^ { + infty } frac { f ^ { ( n ) } left( z _ { 0 } right) } { n ! } left( z - z _ { 0 } right) ^ { n - l } right] \ & = left( z - z _ { 0 } right) ^ { l } frac { f ^ { ( l ) } left( z _ { 0 } right) } { l ! } + left( z - z _ { 0 } right) ^ { l } F ( z ) end{aligned}
$$
In the same way we obtain :
$$
begin{aligned} g ( z ) & = left( z - z _ { 0 } right) ^ { l } left[ frac { g ^ { ( l ) } left( z _ { 0 } right) } { l ! } + sum _ { n = l + 1 } ^ { + infty } frac { g ^ { ( n ) } left( z _ { 0 } right) } { n ! } left( z - z _ { 0 } right) ^ { n - l } right] \ & = left( z - z _ { 0 } right) ^ { l } frac { g ^ { ( l ) } left( z _ { 0 } right) } { l ! } + left( z - z _ { 0 } right) ^ { l } G ( z ) end{aligned}
$$
My issue : To conclude I would like to write the following steps :
$$
lim _ { z rightarrow z _ { 0 } } left[ frac { f ( z ) } { g ( z ) } right] = lim _ { z rightarrow z _ { 0 } } left[ frac { frac { f ^ { ( l ) } left( z _ { 0 } right) } { l ! } + F ( z ) } { frac { g ^ { ( l ) } left( z _ { 0 } right) } { l ! } + G ( z ) } right] = frac { f ^ { ( l ) } left( z _ { 0 } right) } { g ^ { ( l ) } left( z _ { 0 } right) }
$$
- Since $z_0$ is a zero of order $l$ of $g$ I know that ${ g ^ { ( l ) } left( z _ { 0 } right) } neq 0$.
$G(z_0)$ and $F(z_0)$ need to be equal to $0$ but I'm not sure how to get this property. Where this property would come from?
complex-analysis complex-numbers holomorphic-functions
complex-analysis complex-numbers holomorphic-functions
edited Dec 29 '18 at 12:11
Oscar Lanzi
13.7k12136
asked Dec 29 '18 at 11:47
NotAbelianGroupNotAbelianGroup
20911
edited Dec 29 '18 at 12:11
Oscar Lanzi
13.7k12136
asked Dec 29 '18 at 11:47
NotAbelianGroupNotAbelianGroup
20911
edited Dec 29 '18 at 12:11
Oscar Lanzi
13.7k12136
edited Dec 29 '18 at 12:11
Oscar Lanzi
13.7k12136
edited Dec 29 '18 at 12:11
Oscar Lanzi
13.7k12136
13.7k12136
asked Dec 29 '18 at 11:47
NotAbelianGroupNotAbelianGroup
20911
asked Dec 29 '18 at 11:47
NotAbelianGroupNotAbelianGroup
20911
asked Dec 29 '18 at 11:47
NotAbelianGroupNotAbelianGroup
20911
20911
$begingroup$
But you have an explicit formula for $F(z)$ and $G(z)$, right? So you just need to plug in $z=z_0$ in these formulas.
$endgroup$
– Did
Dec 29 '18 at 11:53
add a comment |
$begingroup$
But you have an explicit formula for $F(z)$ and $G(z)$, right? So you just need to plug in $z=z_0$ in these formulas.
$endgroup$
– Did
Dec 29 '18 at 11:53
$begingroup$
But you have an explicit formula for $F(z)$ and $G(z)$, right? So you just need to plug in $z=z_0$ in these formulas.
$endgroup$
– Did
Dec 29 '18 at 11:53
$begingroup$
But you have an explicit formula for $F(z)$ and $G(z)$, right? So you just need to plug in $z=z_0$ in these formulas.
$endgroup$
– Did
Dec 29 '18 at 11:53
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
A better approach: we can write $f(z)=(z-z_0)^{l} f_1(z)$ where $f_1$ is anlytic in some disk around $z_0$ with $f_1(z_0) neq 0$ and $g(z)=(z-z_0)^{k} g_1(z)$ where $g_1$ is analytic in some disk around $z_0$ and $g_1(z_0) neq 0$. This immediately gives the result since $frac {f(z)} {g(z)} to 0$ if $l >k$ and $frac {f(z)} {g(z)} to frac {f_1(z_0)} {g_1(z_0)}$ if $l=k$. Can you identify the limit in the second case by differentiating the functions $l$ times and setting $z=z_0$?
answered Dec 29 '18 at 12:03
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
$endgroup$
$begingroup$
Thank for your answer. Would the limit in the second case be $frac { f_1 ^ { ( l ) } left( z _ { 0 } right) } { g_1 ^ { ( l ) } left( z _ { 0 } right) }$? However I don't seem to understand how your second case relate to my problem. In my exercise $l>k$, so is the solution $0$ since the $l$-th derivative of $f$ equal to $0$? $$lim _ { z rightarrow z _ { 0 } } left[ frac { f ( z ) } { g ( z ) } right] = frac { f ^ { ( l ) } left( z _ { 0 } right) } { g ^ { ( l ) } left( z _ { 0 } right)} = 0$$
$endgroup$
– NotAbelianGroup
Dec 29 '18 at 13:38
add a comment |
Your Answer
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3055763%2ffind-lim-z-rightarrow-z-0-frac-f-z-g-z-with-f-a%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
A better approach: we can write $f(z)=(z-z_0)^{l} f_1(z)$ where $f_1$ is anlytic in some disk around $z_0$ with $f_1(z_0) neq 0$ and $g(z)=(z-z_0)^{k} g_1(z)$ where $g_1$ is analytic in some disk around $z_0$ and $g_1(z_0) neq 0$. This immediately gives the result since $frac {f(z)} {g(z)} to 0$ if $l >k$ and $frac {f(z)} {g(z)} to frac {f_1(z_0)} {g_1(z_0)}$ if $l=k$. Can you identify the limit in the second case by differentiating the functions $l$ times and setting $z=z_0$?
answered Dec 29 '18 at 12:03
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
$endgroup$
$begingroup$
Thank for your answer. Would the limit in the second case be $frac { f_1 ^ { ( l ) } left( z _ { 0 } right) } { g_1 ^ { ( l ) } left( z _ { 0 } right) }$? However I don't seem to understand how your second case relate to my problem. In my exercise $l>k$, so is the solution $0$ since the $l$-th derivative of $f$ equal to $0$? $$lim _ { z rightarrow z _ { 0 } } left[ frac { f ( z ) } { g ( z ) } right] = frac { f ^ { ( l ) } left( z _ { 0 } right) } { g ^ { ( l ) } left( z _ { 0 } right)} = 0$$
$endgroup$
– NotAbelianGroup
Dec 29 '18 at 13:38
add a comment |
$begingroup$
A better approach: we can write $f(z)=(z-z_0)^{l} f_1(z)$ where $f_1$ is anlytic in some disk around $z_0$ with $f_1(z_0) neq 0$ and $g(z)=(z-z_0)^{k} g_1(z)$ where $g_1$ is analytic in some disk around $z_0$ and $g_1(z_0) neq 0$. This immediately gives the result since $frac {f(z)} {g(z)} to 0$ if $l >k$ and $frac {f(z)} {g(z)} to frac {f_1(z_0)} {g_1(z_0)}$ if $l=k$. Can you identify the limit in the second case by differentiating the functions $l$ times and setting $z=z_0$?
answered Dec 29 '18 at 12:03
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
$endgroup$
$begingroup$
Thank for your answer. Would the limit in the second case be $frac { f_1 ^ { ( l ) } left( z _ { 0 } right) } { g_1 ^ { ( l ) } left( z _ { 0 } right) }$? However I don't seem to understand how your second case relate to my problem. In my exercise $l>k$, so is the solution $0$ since the $l$-th derivative of $f$ equal to $0$? $$lim _ { z rightarrow z _ { 0 } } left[ frac { f ( z ) } { g ( z ) } right] = frac { f ^ { ( l ) } left( z _ { 0 } right) } { g ^ { ( l ) } left( z _ { 0 } right)} = 0$$
$endgroup$
– NotAbelianGroup
Dec 29 '18 at 13:38
add a comment |
$begingroup$
A better approach: we can write $f(z)=(z-z_0)^{l} f_1(z)$ where $f_1$ is anlytic in some disk around $z_0$ with $f_1(z_0) neq 0$ and $g(z)=(z-z_0)^{k} g_1(z)$ where $g_1$ is analytic in some disk around $z_0$ and $g_1(z_0) neq 0$. This immediately gives the result since $frac {f(z)} {g(z)} to 0$ if $l >k$ and $frac {f(z)} {g(z)} to frac {f_1(z_0)} {g_1(z_0)}$ if $l=k$. Can you identify the limit in the second case by differentiating the functions $l$ times and setting $z=z_0$?
answered Dec 29 '18 at 12:03
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
$endgroup$
A better approach: we can write $f(z)=(z-z_0)^{l} f_1(z)$ where $f_1$ is anlytic in some disk around $z_0$ with $f_1(z_0) neq 0$ and $g(z)=(z-z_0)^{k} g_1(z)$ where $g_1$ is analytic in some disk around $z_0$ and $g_1(z_0) neq 0$. This immediately gives the result since $frac {f(z)} {g(z)} to 0$ if $l >k$ and $frac {f(z)} {g(z)} to frac {f_1(z_0)} {g_1(z_0)}$ if $l=k$. Can you identify the limit in the second case by differentiating the functions $l$ times and setting $z=z_0$?
answered Dec 29 '18 at 12:03
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
answered Dec 29 '18 at 12:03
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
answered Dec 29 '18 at 12:03
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
answered Dec 29 '18 at 12:03
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
74.9k53270
$begingroup$
Thank for your answer. Would the limit in the second case be $frac { f_1 ^ { ( l ) } left( z _ { 0 } right) } { g_1 ^ { ( l ) } left( z _ { 0 } right) }$? However I don't seem to understand how your second case relate to my problem. In my exercise $l>k$, so is the solution $0$ since the $l$-th derivative of $f$ equal to $0$? $$lim _ { z rightarrow z _ { 0 } } left[ frac { f ( z ) } { g ( z ) } right] = frac { f ^ { ( l ) } left( z _ { 0 } right) } { g ^ { ( l ) } left( z _ { 0 } right)} = 0$$
$endgroup$
– NotAbelianGroup
Dec 29 '18 at 13:38
add a comment |
$begingroup$
Thank for your answer. Would the limit in the second case be $frac { f_1 ^ { ( l ) } left( z _ { 0 } right) } { g_1 ^ { ( l ) } left( z _ { 0 } right) }$? However I don't seem to understand how your second case relate to my problem. In my exercise $l>k$, so is the solution $0$ since the $l$-th derivative of $f$ equal to $0$? $$lim _ { z rightarrow z _ { 0 } } left[ frac { f ( z ) } { g ( z ) } right] = frac { f ^ { ( l ) } left( z _ { 0 } right) } { g ^ { ( l ) } left( z _ { 0 } right)} = 0$$
$endgroup$
– NotAbelianGroup
Dec 29 '18 at 13:38
$begingroup$
Thank for your answer. Would the limit in the second case be $frac { f_1 ^ { ( l ) } left( z _ { 0 } right) } { g_1 ^ { ( l ) } left( z _ { 0 } right) }$? However I don't seem to understand how your second case relate to my problem. In my exercise $l>k$, so is the solution $0$ since the $l$-th derivative of $f$ equal to $0$? $$lim _ { z rightarrow z _ { 0 } } left[ frac { f ( z ) } { g ( z ) } right] = frac { f ^ { ( l ) } left( z _ { 0 } right) } { g ^ { ( l ) } left( z _ { 0 } right)} = 0$$
$endgroup$
– NotAbelianGroup
Dec 29 '18 at 13:38
$begingroup$
Thank for your answer. Would the limit in the second case be $frac { f_1 ^ { ( l ) } left( z _ { 0 } right) } { g_1 ^ { ( l ) } left( z _ { 0 } right) }$? However I don't seem to understand how your second case relate to my problem. In my exercise $l>k$, so is the solution $0$ since the $l$-th derivative of $f$ equal to $0$? $$lim _ { z rightarrow z _ { 0 } } left[ frac { f ( z ) } { g ( z ) } right] = frac { f ^ { ( l ) } left( z _ { 0 } right) } { g ^ { ( l ) } left( z _ { 0 } right)} = 0$$
$endgroup$
– NotAbelianGroup
Dec 29 '18 at 13:38
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3055763%2ffind-lim-z-rightarrow-z-0-frac-f-z-g-z-with-f-a%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
But you have an explicit formula for $F(z)$ and $G(z)$, right? So you just need to plug in $z=z_0$ in these formulas.
$endgroup$
– Did
Dec 29 '18 at 11:53