Mixing
Plotting Reciprocal of a Complex inequality
$begingroup$
We have a complex inequality $|z-r|leq{sqrt{r}}$. (${r>0}$ is a real constant number and $z$ is a variable complex number). Now we want to find the plot of $f(z) = 1/z$ where $z$ are the answers of above inequality. i.e. we want to find plot of: $|1/z-r|leq sqrt{r}$ $~~~$
I know that the first one is the plot of a circle with Radius $=sqrt r$ and center is $r + 0 i$. (All the point inside and on the perimeter of circle).
But I don't know what to do with the reciprocal of $z$ in $|1/z-r|leq sqrt{r}$ .
complex-numbers graphing-functions
asked Nov 28 '18 at 20:10
amir naamir na
1466
$endgroup$
add a comment |
$begingroup$
We have a complex inequality $|z-r|leq{sqrt{r}}$. (${r>0}$ is a real constant number and $z$ is a variable complex number). Now we want to find the plot of $f(z) = 1/z$ where $z$ are the answers of above inequality. i.e. we want to find plot of: $|1/z-r|leq sqrt{r}$ $~~~$
I know that the first one is the plot of a circle with Radius $=sqrt r$ and center is $r + 0 i$. (All the point inside and on the perimeter of circle).
But I don't know what to do with the reciprocal of $z$ in $|1/z-r|leq sqrt{r}$ .
complex-numbers graphing-functions
asked Nov 28 '18 at 20:10
amir naamir na
1466
$endgroup$
add a comment |
$begingroup$
We have a complex inequality $|z-r|leq{sqrt{r}}$. (${r>0}$ is a real constant number and $z$ is a variable complex number). Now we want to find the plot of $f(z) = 1/z$ where $z$ are the answers of above inequality. i.e. we want to find plot of: $|1/z-r|leq sqrt{r}$ $~~~$
I know that the first one is the plot of a circle with Radius $=sqrt r$ and center is $r + 0 i$. (All the point inside and on the perimeter of circle).
But I don't know what to do with the reciprocal of $z$ in $|1/z-r|leq sqrt{r}$ .
complex-numbers graphing-functions
asked Nov 28 '18 at 20:10
amir naamir na
1466
$endgroup$
We have a complex inequality $|z-r|leq{sqrt{r}}$. (${r>0}$ is a real constant number and $z$ is a variable complex number). Now we want to find the plot of $f(z) = 1/z$ where $z$ are the answers of above inequality. i.e. we want to find plot of: $|1/z-r|leq sqrt{r}$ $~~~$
I know that the first one is the plot of a circle with Radius $=sqrt r$ and center is $r + 0 i$. (All the point inside and on the perimeter of circle).
But I don't know what to do with the reciprocal of $z$ in $|1/z-r|leq sqrt{r}$ .
complex-numbers graphing-functions
complex-numbers graphing-functions
asked Nov 28 '18 at 20:10
amir naamir na
1466
asked Nov 28 '18 at 20:10
amir naamir na
1466
asked Nov 28 '18 at 20:10
amir naamir na
1466
asked Nov 28 '18 at 20:10
amir naamir na
1466
asked Nov 28 '18 at 20:10
amir naamir na
1466
1466
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let $D$ be the unit disc, then your original inequality is satisfied by is the set
$$
r + zsqrt{r}, quad z in D
$$
and the inverse would be
$$
frac{1}{r+zsqrt{r}}, quad z in D.
$$
answered Nov 28 '18 at 20:16
gt6989bgt6989b
34.5k22456
$endgroup$
$begingroup$
Can you say what will be the shape of $frac{1}{r+zsqrt{r}}, quad z in D$ if we graph it? The question is maybe "what is reciprocal of a circle?".
$endgroup$
– amir na
Nov 28 '18 at 20:19
$begingroup$
@amirna why don't you do some work on this problem as well? Map out the image of the 4 extremes of the circle under the transformation and perhaps you will get some idea
$endgroup$
– gt6989b
Nov 28 '18 at 20:21
add a comment |
$begingroup$
Recall that if a point in the plane is represented by $z=x+iy$ for $x,yin mathbb{R}$, then you have
$$
frac{1}{z}=frac{bar{z}}{|z|^2}
quad implies quad
frac{1}{z-z_0}=frac{overline{z-z_0}}{|z-z_0|^2}.
$$
Take $z_0=x_0+iy_0$. This tells you that if you have $|z-z_0|le R$ for real number $R>0$, then $frac{1}{z-z_0}$ yields another circle for $overline{z-z_0}$ scaled by $|z-z_0|^2le R^2$. This is can be seen from
$$
frac{1}{z-z_0}
=frac{bar{z}-bar{z_0}}{|z-z_0|^2}
ge frac{bar{z}-bar{z_0}}{R^2}.
$$
In your case, you have
$$
left|frac{1}{z-r}right|
=left|frac{bar{z}-r}{|z-r|^2}right|
=frac{|bar{z}-r|}{|z-r|^2}
ge frac{|bar{z}-r|}{r}
=frac{sqrt{r}}{r}
=frac{1}{sqrt{r}}.
$$
What you want is the plot of the circle described by $left|frac{bar{z}-r}{r}right|le frac{1}{sqrt{r}}.$ That's it. But then you have to take into account the change of angle $theta$ when you write $z=sqrt{r}e^{itheta}$ then $bar{z}=sqrt{r}e^{-itheta}$.
answered Jan 18 at 16:54
M.D.M.D.
549
$endgroup$
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%2f3017637%2fplotting-reciprocal-of-a-complex-inequality%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $D$ be the unit disc, then your original inequality is satisfied by is the set
$$
r + zsqrt{r}, quad z in D
$$
and the inverse would be
$$
frac{1}{r+zsqrt{r}}, quad z in D.
$$
answered Nov 28 '18 at 20:16
gt6989bgt6989b
34.5k22456
$endgroup$
$begingroup$
Can you say what will be the shape of $frac{1}{r+zsqrt{r}}, quad z in D$ if we graph it? The question is maybe "what is reciprocal of a circle?".
$endgroup$
– amir na
Nov 28 '18 at 20:19
$begingroup$
@amirna why don't you do some work on this problem as well? Map out the image of the 4 extremes of the circle under the transformation and perhaps you will get some idea
$endgroup$
– gt6989b
Nov 28 '18 at 20:21
add a comment |
$begingroup$
Let $D$ be the unit disc, then your original inequality is satisfied by is the set
$$
r + zsqrt{r}, quad z in D
$$
and the inverse would be
$$
frac{1}{r+zsqrt{r}}, quad z in D.
$$
answered Nov 28 '18 at 20:16
gt6989bgt6989b
34.5k22456
$endgroup$
$begingroup$
Can you say what will be the shape of $frac{1}{r+zsqrt{r}}, quad z in D$ if we graph it? The question is maybe "what is reciprocal of a circle?".
$endgroup$
– amir na
Nov 28 '18 at 20:19
$begingroup$
@amirna why don't you do some work on this problem as well? Map out the image of the 4 extremes of the circle under the transformation and perhaps you will get some idea
$endgroup$
– gt6989b
Nov 28 '18 at 20:21
add a comment |
$begingroup$
Let $D$ be the unit disc, then your original inequality is satisfied by is the set
$$
r + zsqrt{r}, quad z in D
$$
and the inverse would be
$$
frac{1}{r+zsqrt{r}}, quad z in D.
$$
answered Nov 28 '18 at 20:16
gt6989bgt6989b
34.5k22456
$endgroup$
Let $D$ be the unit disc, then your original inequality is satisfied by is the set
$$
r + zsqrt{r}, quad z in D
$$
and the inverse would be
$$
frac{1}{r+zsqrt{r}}, quad z in D.
$$
answered Nov 28 '18 at 20:16
gt6989bgt6989b
34.5k22456
answered Nov 28 '18 at 20:16
gt6989bgt6989b
34.5k22456
answered Nov 28 '18 at 20:16
gt6989bgt6989b
34.5k22456
answered Nov 28 '18 at 20:16
gt6989bgt6989b
34.5k22456
34.5k22456
$begingroup$
Can you say what will be the shape of $frac{1}{r+zsqrt{r}}, quad z in D$ if we graph it? The question is maybe "what is reciprocal of a circle?".
$endgroup$
– amir na
Nov 28 '18 at 20:19
$begingroup$
@amirna why don't you do some work on this problem as well? Map out the image of the 4 extremes of the circle under the transformation and perhaps you will get some idea
$endgroup$
– gt6989b
Nov 28 '18 at 20:21
add a comment |
$begingroup$
Can you say what will be the shape of $frac{1}{r+zsqrt{r}}, quad z in D$ if we graph it? The question is maybe "what is reciprocal of a circle?".
$endgroup$
– amir na
Nov 28 '18 at 20:19
$begingroup$
@amirna why don't you do some work on this problem as well? Map out the image of the 4 extremes of the circle under the transformation and perhaps you will get some idea
$endgroup$
– gt6989b
Nov 28 '18 at 20:21
$begingroup$
Can you say what will be the shape of $frac{1}{r+zsqrt{r}}, quad z in D$ if we graph it? The question is maybe "what is reciprocal of a circle?".
$endgroup$
– amir na
Nov 28 '18 at 20:19
$begingroup$
Can you say what will be the shape of $frac{1}{r+zsqrt{r}}, quad z in D$ if we graph it? The question is maybe "what is reciprocal of a circle?".
$endgroup$
– amir na
Nov 28 '18 at 20:19
$begingroup$
@amirna why don't you do some work on this problem as well? Map out the image of the 4 extremes of the circle under the transformation and perhaps you will get some idea
$endgroup$
– gt6989b
Nov 28 '18 at 20:21
$begingroup$
@amirna why don't you do some work on this problem as well? Map out the image of the 4 extremes of the circle under the transformation and perhaps you will get some idea
$endgroup$
– gt6989b
Nov 28 '18 at 20:21
add a comment |
$begingroup$
Recall that if a point in the plane is represented by $z=x+iy$ for $x,yin mathbb{R}$, then you have
$$
frac{1}{z}=frac{bar{z}}{|z|^2}
quad implies quad
frac{1}{z-z_0}=frac{overline{z-z_0}}{|z-z_0|^2}.
$$
Take $z_0=x_0+iy_0$. This tells you that if you have $|z-z_0|le R$ for real number $R>0$, then $frac{1}{z-z_0}$ yields another circle for $overline{z-z_0}$ scaled by $|z-z_0|^2le R^2$. This is can be seen from
$$
frac{1}{z-z_0}
=frac{bar{z}-bar{z_0}}{|z-z_0|^2}
ge frac{bar{z}-bar{z_0}}{R^2}.
$$
In your case, you have
$$
left|frac{1}{z-r}right|
=left|frac{bar{z}-r}{|z-r|^2}right|
=frac{|bar{z}-r|}{|z-r|^2}
ge frac{|bar{z}-r|}{r}
=frac{sqrt{r}}{r}
=frac{1}{sqrt{r}}.
$$
What you want is the plot of the circle described by $left|frac{bar{z}-r}{r}right|le frac{1}{sqrt{r}}.$ That's it. But then you have to take into account the change of angle $theta$ when you write $z=sqrt{r}e^{itheta}$ then $bar{z}=sqrt{r}e^{-itheta}$.
answered Jan 18 at 16:54
M.D.M.D.
549
$endgroup$
add a comment |
$begingroup$
Recall that if a point in the plane is represented by $z=x+iy$ for $x,yin mathbb{R}$, then you have
$$
frac{1}{z}=frac{bar{z}}{|z|^2}
quad implies quad
frac{1}{z-z_0}=frac{overline{z-z_0}}{|z-z_0|^2}.
$$
Take $z_0=x_0+iy_0$. This tells you that if you have $|z-z_0|le R$ for real number $R>0$, then $frac{1}{z-z_0}$ yields another circle for $overline{z-z_0}$ scaled by $|z-z_0|^2le R^2$. This is can be seen from
$$
frac{1}{z-z_0}
=frac{bar{z}-bar{z_0}}{|z-z_0|^2}
ge frac{bar{z}-bar{z_0}}{R^2}.
$$
In your case, you have
$$
left|frac{1}{z-r}right|
=left|frac{bar{z}-r}{|z-r|^2}right|
=frac{|bar{z}-r|}{|z-r|^2}
ge frac{|bar{z}-r|}{r}
=frac{sqrt{r}}{r}
=frac{1}{sqrt{r}}.
$$
What you want is the plot of the circle described by $left|frac{bar{z}-r}{r}right|le frac{1}{sqrt{r}}.$ That's it. But then you have to take into account the change of angle $theta$ when you write $z=sqrt{r}e^{itheta}$ then $bar{z}=sqrt{r}e^{-itheta}$.
answered Jan 18 at 16:54
M.D.M.D.
549
$endgroup$
add a comment |
$begingroup$
Recall that if a point in the plane is represented by $z=x+iy$ for $x,yin mathbb{R}$, then you have
$$
frac{1}{z}=frac{bar{z}}{|z|^2}
quad implies quad
frac{1}{z-z_0}=frac{overline{z-z_0}}{|z-z_0|^2}.
$$
Take $z_0=x_0+iy_0$. This tells you that if you have $|z-z_0|le R$ for real number $R>0$, then $frac{1}{z-z_0}$ yields another circle for $overline{z-z_0}$ scaled by $|z-z_0|^2le R^2$. This is can be seen from
$$
frac{1}{z-z_0}
=frac{bar{z}-bar{z_0}}{|z-z_0|^2}
ge frac{bar{z}-bar{z_0}}{R^2}.
$$
In your case, you have
$$
left|frac{1}{z-r}right|
=left|frac{bar{z}-r}{|z-r|^2}right|
=frac{|bar{z}-r|}{|z-r|^2}
ge frac{|bar{z}-r|}{r}
=frac{sqrt{r}}{r}
=frac{1}{sqrt{r}}.
$$
What you want is the plot of the circle described by $left|frac{bar{z}-r}{r}right|le frac{1}{sqrt{r}}.$ That's it. But then you have to take into account the change of angle $theta$ when you write $z=sqrt{r}e^{itheta}$ then $bar{z}=sqrt{r}e^{-itheta}$.
answered Jan 18 at 16:54
M.D.M.D.
549
$endgroup$
Recall that if a point in the plane is represented by $z=x+iy$ for $x,yin mathbb{R}$, then you have
$$
frac{1}{z}=frac{bar{z}}{|z|^2}
quad implies quad
frac{1}{z-z_0}=frac{overline{z-z_0}}{|z-z_0|^2}.
$$
Take $z_0=x_0+iy_0$. This tells you that if you have $|z-z_0|le R$ for real number $R>0$, then $frac{1}{z-z_0}$ yields another circle for $overline{z-z_0}$ scaled by $|z-z_0|^2le R^2$. This is can be seen from
$$
frac{1}{z-z_0}
=frac{bar{z}-bar{z_0}}{|z-z_0|^2}
ge frac{bar{z}-bar{z_0}}{R^2}.
$$
In your case, you have
$$
left|frac{1}{z-r}right|
=left|frac{bar{z}-r}{|z-r|^2}right|
=frac{|bar{z}-r|}{|z-r|^2}
ge frac{|bar{z}-r|}{r}
=frac{sqrt{r}}{r}
=frac{1}{sqrt{r}}.
$$
What you want is the plot of the circle described by $left|frac{bar{z}-r}{r}right|le frac{1}{sqrt{r}}.$ That's it. But then you have to take into account the change of angle $theta$ when you write $z=sqrt{r}e^{itheta}$ then $bar{z}=sqrt{r}e^{-itheta}$.
answered Jan 18 at 16:54
M.D.M.D.
549
edited Jan 19 at 16:15
answered Jan 18 at 16:54
M.D.M.D.
549
answered Jan 18 at 16:54
M.D.M.D.
549
answered Jan 18 at 16:54
M.D.M.D.
549
549
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.
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%2f3017637%2fplotting-reciprocal-of-a-complex-inequality%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
