Mixing
Show that $sin(z)$ is surjective even if $-pi/2 le x le pi/2.$
There are questions and answer in MSE regarding surjectivity of $sin(z)$. But they all prove that there is a $z in mathbb{C}$ such that ... not a proof that $z$ can be a proper subset of $mathbb{C}$.
The mapping $w=sin(z)$ maps the lines $x=c$ for $-pi/2 le c le pi/2$ to hyperbolas, $v$ axis and lines $u ge 1$ and $u le -1$. So it 'looks' that $w=sin(z)$ maps the strip $-pi/2 le x le pi/2$ to whole $mathbb{C}$ as it 'looks' that the hyperbolas sweeps entire plane as lines $x=c$ moves continuously in $-pi/2 le x le pi/2$. I am trying to prove these 'look's rigorously, that is for any $w in mathbb{C}$ there is a $z in$ the mentioned strip such that $w=sin(z)$ but I stuck no matter which way I am trying. Proving that the real part of $ sin^{-1} z =-i log[iz + (1 - z^2)^{frac12}]$ can always lie on $-pi/2 le x le pi/2$ for a suitable choice of $n$ is also impossible!
Is there any way to solve this at all? (algebraically not geometrically)
complex-analysis
asked Nov 22 '18 at 0:20
72D72D
540116
add a comment |
There are questions and answer in MSE regarding surjectivity of $sin(z)$. But they all prove that there is a $z in mathbb{C}$ such that ... not a proof that $z$ can be a proper subset of $mathbb{C}$.
The mapping $w=sin(z)$ maps the lines $x=c$ for $-pi/2 le c le pi/2$ to hyperbolas, $v$ axis and lines $u ge 1$ and $u le -1$. So it 'looks' that $w=sin(z)$ maps the strip $-pi/2 le x le pi/2$ to whole $mathbb{C}$ as it 'looks' that the hyperbolas sweeps entire plane as lines $x=c$ moves continuously in $-pi/2 le x le pi/2$. I am trying to prove these 'look's rigorously, that is for any $w in mathbb{C}$ there is a $z in$ the mentioned strip such that $w=sin(z)$ but I stuck no matter which way I am trying. Proving that the real part of $ sin^{-1} z =-i log[iz + (1 - z^2)^{frac12}]$ can always lie on $-pi/2 le x le pi/2$ for a suitable choice of $n$ is also impossible!
Is there any way to solve this at all? (algebraically not geometrically)
complex-analysis
asked Nov 22 '18 at 0:20
72D72D
540116
add a comment |
There are questions and answer in MSE regarding surjectivity of $sin(z)$. But they all prove that there is a $z in mathbb{C}$ such that ... not a proof that $z$ can be a proper subset of $mathbb{C}$.
The mapping $w=sin(z)$ maps the lines $x=c$ for $-pi/2 le c le pi/2$ to hyperbolas, $v$ axis and lines $u ge 1$ and $u le -1$. So it 'looks' that $w=sin(z)$ maps the strip $-pi/2 le x le pi/2$ to whole $mathbb{C}$ as it 'looks' that the hyperbolas sweeps entire plane as lines $x=c$ moves continuously in $-pi/2 le x le pi/2$. I am trying to prove these 'look's rigorously, that is for any $w in mathbb{C}$ there is a $z in$ the mentioned strip such that $w=sin(z)$ but I stuck no matter which way I am trying. Proving that the real part of $ sin^{-1} z =-i log[iz + (1 - z^2)^{frac12}]$ can always lie on $-pi/2 le x le pi/2$ for a suitable choice of $n$ is also impossible!
Is there any way to solve this at all? (algebraically not geometrically)
complex-analysis
asked Nov 22 '18 at 0:20
72D72D
540116
There are questions and answer in MSE regarding surjectivity of $sin(z)$. But they all prove that there is a $z in mathbb{C}$ such that ... not a proof that $z$ can be a proper subset of $mathbb{C}$.
The mapping $w=sin(z)$ maps the lines $x=c$ for $-pi/2 le c le pi/2$ to hyperbolas, $v$ axis and lines $u ge 1$ and $u le -1$. So it 'looks' that $w=sin(z)$ maps the strip $-pi/2 le x le pi/2$ to whole $mathbb{C}$ as it 'looks' that the hyperbolas sweeps entire plane as lines $x=c$ moves continuously in $-pi/2 le x le pi/2$. I am trying to prove these 'look's rigorously, that is for any $w in mathbb{C}$ there is a $z in$ the mentioned strip such that $w=sin(z)$ but I stuck no matter which way I am trying. Proving that the real part of $ sin^{-1} z =-i log[iz + (1 - z^2)^{frac12}]$ can always lie on $-pi/2 le x le pi/2$ for a suitable choice of $n$ is also impossible!
Is there any way to solve this at all? (algebraically not geometrically)
complex-analysis
complex-analysis
asked Nov 22 '18 at 0:20
72D72D
540116
asked Nov 22 '18 at 0:20
72D72D
540116
asked Nov 22 '18 at 0:20
72D72D
540116
asked Nov 22 '18 at 0:20
72D72D
540116
asked Nov 22 '18 at 0:20
72D72D
540116
540116
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
It is easy to see that $sin , z=c$ has a solution for any complex number $c$ (by solving a quadratic). By periodicity we may find a solution with $-pi leq Re (z) leq pi$. Using the fact that $sin(pi-z)=sin, z$ we can find a solution $c$ with $-pi /2 leq Re (z) leq pi /2$. [For example if $ Re, zin [pi /2, pi]$ then $Re(pi -z)in [-pi /2,0]$].
answered Nov 22 '18 at 0:41
Kavi Rama MurthyKavi Rama Murthy
51.8k32055
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%2f3008583%2fshow-that-sinz-is-surjective-even-if-pi-2-le-x-le-pi-2%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
It is easy to see that $sin , z=c$ has a solution for any complex number $c$ (by solving a quadratic). By periodicity we may find a solution with $-pi leq Re (z) leq pi$. Using the fact that $sin(pi-z)=sin, z$ we can find a solution $c$ with $-pi /2 leq Re (z) leq pi /2$. [For example if $ Re, zin [pi /2, pi]$ then $Re(pi -z)in [-pi /2,0]$].
answered Nov 22 '18 at 0:41
Kavi Rama MurthyKavi Rama Murthy
51.8k32055
add a comment |
It is easy to see that $sin , z=c$ has a solution for any complex number $c$ (by solving a quadratic). By periodicity we may find a solution with $-pi leq Re (z) leq pi$. Using the fact that $sin(pi-z)=sin, z$ we can find a solution $c$ with $-pi /2 leq Re (z) leq pi /2$. [For example if $ Re, zin [pi /2, pi]$ then $Re(pi -z)in [-pi /2,0]$].
answered Nov 22 '18 at 0:41
Kavi Rama MurthyKavi Rama Murthy
51.8k32055
add a comment |
It is easy to see that $sin , z=c$ has a solution for any complex number $c$ (by solving a quadratic). By periodicity we may find a solution with $-pi leq Re (z) leq pi$. Using the fact that $sin(pi-z)=sin, z$ we can find a solution $c$ with $-pi /2 leq Re (z) leq pi /2$. [For example if $ Re, zin [pi /2, pi]$ then $Re(pi -z)in [-pi /2,0]$].
answered Nov 22 '18 at 0:41
Kavi Rama MurthyKavi Rama Murthy
51.8k32055
It is easy to see that $sin , z=c$ has a solution for any complex number $c$ (by solving a quadratic). By periodicity we may find a solution with $-pi leq Re (z) leq pi$. Using the fact that $sin(pi-z)=sin, z$ we can find a solution $c$ with $-pi /2 leq Re (z) leq pi /2$. [For example if $ Re, zin [pi /2, pi]$ then $Re(pi -z)in [-pi /2,0]$].
answered Nov 22 '18 at 0:41
Kavi Rama MurthyKavi Rama Murthy
51.8k32055
edited Nov 22 '18 at 5:15
answered Nov 22 '18 at 0:41
Kavi Rama MurthyKavi Rama Murthy
51.8k32055
answered Nov 22 '18 at 0:41
Kavi Rama MurthyKavi Rama Murthy
51.8k32055
answered Nov 22 '18 at 0:41
Kavi Rama MurthyKavi Rama Murthy
51.8k32055
51.8k32055
add a comment |
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3008583%2fshow-that-sinz-is-surjective-even-if-pi-2-le-x-le-pi-2%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
