What is the proper way of introducing a pair of invertible complex functions $exp$ and $log$?
$begingroup$
I need to introduce a pair of invertible complex functions $exp$ and $log$ with the following properties:
$A$ being a branch (or strip?) of $mathbb{C} backslash {0}$:
$forall a in A quad exp(log(a)) = log(exp(a)) = a$
$forall a in mathbb{R}^* quad log(-|a|) = log(|a|) + ipi$
What is the proper way of introducing these two functions? I am especially concerned about the proper enunciation of the functions’ domains and codomains.
Furthermore, is there a proper way of introducing the “continuation” of the first property to a subset containing 0?
I am somewhat familiar with complex logarithms, but I still struggle with the proper definition of domains and codomains using branches (strips?). I would like the definitions to be as precise and unambiguous as can be.
complex-numbers
asked Jan 9 at 16:09
ismaelismael
274216
$endgroup$
add a comment |
$begingroup$
I need to introduce a pair of invertible complex functions $exp$ and $log$ with the following properties:
$A$ being a branch (or strip?) of $mathbb{C} backslash {0}$:
$forall a in A quad exp(log(a)) = log(exp(a)) = a$
$forall a in mathbb{R}^* quad log(-|a|) = log(|a|) + ipi$
What is the proper way of introducing these two functions? I am especially concerned about the proper enunciation of the functions’ domains and codomains.
Furthermore, is there a proper way of introducing the “continuation” of the first property to a subset containing 0?
I am somewhat familiar with complex logarithms, but I still struggle with the proper definition of domains and codomains using branches (strips?). I would like the definitions to be as precise and unambiguous as can be.
complex-numbers
asked Jan 9 at 16:09
ismaelismael
274216
$endgroup$
1
$begingroup$
I am afraid that your functions cannot coincide with the usual exponential and logarithm. Because exponential is not injective on $mathbb Csetminus{0}$, and so it cannot have an inverse and the first property cannot hold. What you can do is considering the restriction of the exponential to a strip ${z=x+iy : yin [a, a+2pi)}$. This function is a bijection onto $mathbb Csetminus{0}$.
$endgroup$
– Giuseppe Negro
Jan 9 at 16:16
$begingroup$
@GiuseppeNegro I believe this is what I am looking for. Would you mind writing the full definitions in the form $f : A longrightarrow B$ for both $exp$ and $log$ as an answer to the question? Thanks.
$endgroup$
– ismael
Jan 9 at 16:19
1
$begingroup$
I disagree with the downvote, which, moreover, is not accompanied by a motivating comment. I'll see if I can convert my comment into an answer.
$endgroup$
– Giuseppe Negro
Jan 9 at 16:50
$begingroup$
@GiuseppeNegro Thank you! I edited the original question in order to clarify things.
$endgroup$
– ismael
Jan 9 at 16:51
$begingroup$
This is a great case of definition question.
$endgroup$
– ismael
Jan 10 at 15:32
add a comment |
$begingroup$
I need to introduce a pair of invertible complex functions $exp$ and $log$ with the following properties:
$A$ being a branch (or strip?) of $mathbb{C} backslash {0}$:
$forall a in A quad exp(log(a)) = log(exp(a)) = a$
$forall a in mathbb{R}^* quad log(-|a|) = log(|a|) + ipi$
What is the proper way of introducing these two functions? I am especially concerned about the proper enunciation of the functions’ domains and codomains.
Furthermore, is there a proper way of introducing the “continuation” of the first property to a subset containing 0?
I am somewhat familiar with complex logarithms, but I still struggle with the proper definition of domains and codomains using branches (strips?). I would like the definitions to be as precise and unambiguous as can be.
complex-numbers
asked Jan 9 at 16:09
ismaelismael
274216
$endgroup$
I need to introduce a pair of invertible complex functions $exp$ and $log$ with the following properties:
$A$ being a branch (or strip?) of $mathbb{C} backslash {0}$:
$forall a in A quad exp(log(a)) = log(exp(a)) = a$
$forall a in mathbb{R}^* quad log(-|a|) = log(|a|) + ipi$
What is the proper way of introducing these two functions? I am especially concerned about the proper enunciation of the functions’ domains and codomains.
Furthermore, is there a proper way of introducing the “continuation” of the first property to a subset containing 0?
I am somewhat familiar with complex logarithms, but I still struggle with the proper definition of domains and codomains using branches (strips?). I would like the definitions to be as precise and unambiguous as can be.
complex-numbers
complex-numbers
asked Jan 9 at 16:09
ismaelismael
274216
asked Jan 9 at 16:09
ismaelismael
274216
edited Jan 9 at 17:03
ismael
asked Jan 9 at 16:09
ismaelismael
274216
asked Jan 9 at 16:09
ismaelismael
274216
asked Jan 9 at 16:09
ismaelismael
274216
274216
1
$begingroup$
I am afraid that your functions cannot coincide with the usual exponential and logarithm. Because exponential is not injective on $mathbb Csetminus{0}$, and so it cannot have an inverse and the first property cannot hold. What you can do is considering the restriction of the exponential to a strip ${z=x+iy : yin [a, a+2pi)}$. This function is a bijection onto $mathbb Csetminus{0}$.
$endgroup$
– Giuseppe Negro
Jan 9 at 16:16
$begingroup$
@GiuseppeNegro I believe this is what I am looking for. Would you mind writing the full definitions in the form $f : A longrightarrow B$ for both $exp$ and $log$ as an answer to the question? Thanks.
$endgroup$
– ismael
Jan 9 at 16:19
1
$begingroup$
I disagree with the downvote, which, moreover, is not accompanied by a motivating comment. I'll see if I can convert my comment into an answer.
$endgroup$
– Giuseppe Negro
Jan 9 at 16:50
$begingroup$
@GiuseppeNegro Thank you! I edited the original question in order to clarify things.
$endgroup$
– ismael
Jan 9 at 16:51
$begingroup$
This is a great case of definition question.
$endgroup$
– ismael
Jan 10 at 15:32
add a comment |
1
$begingroup$
I am afraid that your functions cannot coincide with the usual exponential and logarithm. Because exponential is not injective on $mathbb Csetminus{0}$, and so it cannot have an inverse and the first property cannot hold. What you can do is considering the restriction of the exponential to a strip ${z=x+iy : yin [a, a+2pi)}$. This function is a bijection onto $mathbb Csetminus{0}$.
$endgroup$
– Giuseppe Negro
Jan 9 at 16:16
$begingroup$
@GiuseppeNegro I believe this is what I am looking for. Would you mind writing the full definitions in the form $f : A longrightarrow B$ for both $exp$ and $log$ as an answer to the question? Thanks.
$endgroup$
– ismael
Jan 9 at 16:19
1
$begingroup$
I disagree with the downvote, which, moreover, is not accompanied by a motivating comment. I'll see if I can convert my comment into an answer.
$endgroup$
– Giuseppe Negro
Jan 9 at 16:50
$begingroup$
@GiuseppeNegro Thank you! I edited the original question in order to clarify things.
$endgroup$
– ismael
Jan 9 at 16:51
$begingroup$
This is a great case of definition question.
$endgroup$
– ismael
Jan 10 at 15:32
1
1
$begingroup$
I am afraid that your functions cannot coincide with the usual exponential and logarithm. Because exponential is not injective on $mathbb Csetminus{0}$, and so it cannot have an inverse and the first property cannot hold. What you can do is considering the restriction of the exponential to a strip ${z=x+iy : yin [a, a+2pi)}$. This function is a bijection onto $mathbb Csetminus{0}$.
$endgroup$
– Giuseppe Negro
Jan 9 at 16:16
$begingroup$
I am afraid that your functions cannot coincide with the usual exponential and logarithm. Because exponential is not injective on $mathbb Csetminus{0}$, and so it cannot have an inverse and the first property cannot hold. What you can do is considering the restriction of the exponential to a strip ${z=x+iy : yin [a, a+2pi)}$. This function is a bijection onto $mathbb Csetminus{0}$.
$endgroup$
– Giuseppe Negro
Jan 9 at 16:16
$begingroup$
@GiuseppeNegro I believe this is what I am looking for. Would you mind writing the full definitions in the form $f : A longrightarrow B$ for both $exp$ and $log$ as an answer to the question? Thanks.
$endgroup$
– ismael
Jan 9 at 16:19
$begingroup$
@GiuseppeNegro I believe this is what I am looking for. Would you mind writing the full definitions in the form $f : A longrightarrow B$ for both $exp$ and $log$ as an answer to the question? Thanks.
$endgroup$
– ismael
Jan 9 at 16:19
1
1
$begingroup$
I disagree with the downvote, which, moreover, is not accompanied by a motivating comment. I'll see if I can convert my comment into an answer.
$endgroup$
– Giuseppe Negro
Jan 9 at 16:50
$begingroup$
I disagree with the downvote, which, moreover, is not accompanied by a motivating comment. I'll see if I can convert my comment into an answer.
$endgroup$
– Giuseppe Negro
Jan 9 at 16:50
$begingroup$
@GiuseppeNegro Thank you! I edited the original question in order to clarify things.
$endgroup$
– ismael
Jan 9 at 16:51
$begingroup$
@GiuseppeNegro Thank you! I edited the original question in order to clarify things.
$endgroup$
– ismael
Jan 9 at 16:51
$begingroup$
This is a great case of definition question.
$endgroup$
– ismael
Jan 10 at 15:32
$begingroup$
This is a great case of definition question.
$endgroup$
– ismael
Jan 10 at 15:32
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For the first property to hold, the exponential must be restricted to a subset where it is injective. Now, the exponential is injective only on "horizontal strips"
$$
A_lambda:={x+iy : yin [lambda, lambda+2pi)}, $$
where $lambdain mathbb R$. This is a consequence of Euler's formula $$e^{x+iy}=e^x(cos y + i sin y).$$
For each fixed $lambda$, since $exp$ is a bijection of $A_lambda $ onto $mathbb Csetminus{0}$, there is an inverse function $$log_lambda colon mathbb Csetminus{0} to A_lambda.$$ WARNING: This function is discontinuous on the half-line $${re^{ilambda} : rge 0}.$$
For all $lambdainmathbb R$, it holds that
$$log_lambda(-z)=log_lambda(z)pm ipi,$$
where the sign is chosen in such a way that $log_lambda(z)pm ipi$ stays in $A_lambda$.
answered Jan 10 at 10:06
Giuseppe NegroGiuseppe Negro
16.9k330123
$endgroup$
$begingroup$
Thank you! This is exactly what I was looking for.
$endgroup$
– ismael
Jan 10 at 14:49
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
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%2f3067621%2fwhat-is-the-proper-way-of-introducing-a-pair-of-invertible-complex-functions-e%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
$begingroup$
For the first property to hold, the exponential must be restricted to a subset where it is injective. Now, the exponential is injective only on "horizontal strips"
$$
A_lambda:={x+iy : yin [lambda, lambda+2pi)}, $$
where $lambdain mathbb R$. This is a consequence of Euler's formula $$e^{x+iy}=e^x(cos y + i sin y).$$
For each fixed $lambda$, since $exp$ is a bijection of $A_lambda $ onto $mathbb Csetminus{0}$, there is an inverse function $$log_lambda colon mathbb Csetminus{0} to A_lambda.$$ WARNING: This function is discontinuous on the half-line $${re^{ilambda} : rge 0}.$$
For all $lambdainmathbb R$, it holds that
$$log_lambda(-z)=log_lambda(z)pm ipi,$$
where the sign is chosen in such a way that $log_lambda(z)pm ipi$ stays in $A_lambda$.
answered Jan 10 at 10:06
Giuseppe NegroGiuseppe Negro
16.9k330123
$endgroup$
$begingroup$
Thank you! This is exactly what I was looking for.
$endgroup$
– ismael
Jan 10 at 14:49
add a comment |
$begingroup$
For the first property to hold, the exponential must be restricted to a subset where it is injective. Now, the exponential is injective only on "horizontal strips"
$$
A_lambda:={x+iy : yin [lambda, lambda+2pi)}, $$
where $lambdain mathbb R$. This is a consequence of Euler's formula $$e^{x+iy}=e^x(cos y + i sin y).$$
For each fixed $lambda$, since $exp$ is a bijection of $A_lambda $ onto $mathbb Csetminus{0}$, there is an inverse function $$log_lambda colon mathbb Csetminus{0} to A_lambda.$$ WARNING: This function is discontinuous on the half-line $${re^{ilambda} : rge 0}.$$
For all $lambdainmathbb R$, it holds that
$$log_lambda(-z)=log_lambda(z)pm ipi,$$
where the sign is chosen in such a way that $log_lambda(z)pm ipi$ stays in $A_lambda$.
answered Jan 10 at 10:06
Giuseppe NegroGiuseppe Negro
16.9k330123
$endgroup$
$begingroup$
Thank you! This is exactly what I was looking for.
$endgroup$
– ismael
Jan 10 at 14:49
add a comment |
$begingroup$
For the first property to hold, the exponential must be restricted to a subset where it is injective. Now, the exponential is injective only on "horizontal strips"
$$
A_lambda:={x+iy : yin [lambda, lambda+2pi)}, $$
where $lambdain mathbb R$. This is a consequence of Euler's formula $$e^{x+iy}=e^x(cos y + i sin y).$$
For each fixed $lambda$, since $exp$ is a bijection of $A_lambda $ onto $mathbb Csetminus{0}$, there is an inverse function $$log_lambda colon mathbb Csetminus{0} to A_lambda.$$ WARNING: This function is discontinuous on the half-line $${re^{ilambda} : rge 0}.$$
For all $lambdainmathbb R$, it holds that
$$log_lambda(-z)=log_lambda(z)pm ipi,$$
where the sign is chosen in such a way that $log_lambda(z)pm ipi$ stays in $A_lambda$.
answered Jan 10 at 10:06
Giuseppe NegroGiuseppe Negro
16.9k330123
$endgroup$
For the first property to hold, the exponential must be restricted to a subset where it is injective. Now, the exponential is injective only on "horizontal strips"
$$
A_lambda:={x+iy : yin [lambda, lambda+2pi)}, $$
where $lambdain mathbb R$. This is a consequence of Euler's formula $$e^{x+iy}=e^x(cos y + i sin y).$$
For each fixed $lambda$, since $exp$ is a bijection of $A_lambda $ onto $mathbb Csetminus{0}$, there is an inverse function $$log_lambda colon mathbb Csetminus{0} to A_lambda.$$ WARNING: This function is discontinuous on the half-line $${re^{ilambda} : rge 0}.$$
For all $lambdainmathbb R$, it holds that
$$log_lambda(-z)=log_lambda(z)pm ipi,$$
where the sign is chosen in such a way that $log_lambda(z)pm ipi$ stays in $A_lambda$.
answered Jan 10 at 10:06
Giuseppe NegroGiuseppe Negro
16.9k330123
answered Jan 10 at 10:06
Giuseppe NegroGiuseppe Negro
16.9k330123
answered Jan 10 at 10:06
Giuseppe NegroGiuseppe Negro
16.9k330123
answered Jan 10 at 10:06
Giuseppe NegroGiuseppe Negro
16.9k330123
16.9k330123
$begingroup$
Thank you! This is exactly what I was looking for.
$endgroup$
– ismael
Jan 10 at 14:49
add a comment |
$begingroup$
Thank you! This is exactly what I was looking for.
$endgroup$
– ismael
Jan 10 at 14:49
$begingroup$
Thank you! This is exactly what I was looking for.
$endgroup$
– ismael
Jan 10 at 14:49
$begingroup$
Thank you! This is exactly what I was looking for.
$endgroup$
– ismael
Jan 10 at 14:49
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%2f3067621%2fwhat-is-the-proper-way-of-introducing-a-pair-of-invertible-complex-functions-e%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
1
$begingroup$
I am afraid that your functions cannot coincide with the usual exponential and logarithm. Because exponential is not injective on $mathbb Csetminus{0}$, and so it cannot have an inverse and the first property cannot hold. What you can do is considering the restriction of the exponential to a strip ${z=x+iy : yin [a, a+2pi)}$. This function is a bijection onto $mathbb Csetminus{0}$.
$endgroup$
– Giuseppe Negro
Jan 9 at 16:16
$begingroup$
@GiuseppeNegro I believe this is what I am looking for. Would you mind writing the full definitions in the form $f : A longrightarrow B$ for both $exp$ and $log$ as an answer to the question? Thanks.
$endgroup$
– ismael
Jan 9 at 16:19
1
$begingroup$
I disagree with the downvote, which, moreover, is not accompanied by a motivating comment. I'll see if I can convert my comment into an answer.
$endgroup$
– Giuseppe Negro
Jan 9 at 16:50
$begingroup$
@GiuseppeNegro Thank you! I edited the original question in order to clarify things.
$endgroup$
– ismael
Jan 9 at 16:51
$begingroup$
This is a great case of definition question.
$endgroup$
– ismael
Jan 10 at 15:32