Mixing
Show that $zinmathbb{C}$ satisfying $argBig(dfrac{z-1}{z+1}Big)=dfrac{pi}{4}$ is a circle
$begingroup$
Show that the complex number $z$ satisfying the condition $argBig(dfrac{z-1}{z+1}Big)=dfrac{pi}{4}$ is a circle
I can prove it by expressing $z=x+iy$,
$$
argBig(frac{x-1+iy}{x+1+iy}Big)=arg(x-1+iy)-arg(x+1+iy)=frac{pi}{4}\
1=frac{frac{y}{x-1}-frac{y}{x+1}}{1+frac{y^2}{x^2-1}}=frac{xy+y-xy+y}{x^2-1+y^2}implies x^2-1+y^2=2y\implies x^2+y^2-2y-1=0
$$
But can I show that using geometry and the concept of Apollonius circle ?
complex-numbers circles
asked Jan 31 at 20:50
ss1729ss1729
2,05611124
$endgroup$
add a comment |
$begingroup$
Show that the complex number $z$ satisfying the condition $argBig(dfrac{z-1}{z+1}Big)=dfrac{pi}{4}$ is a circle
I can prove it by expressing $z=x+iy$,
$$
argBig(frac{x-1+iy}{x+1+iy}Big)=arg(x-1+iy)-arg(x+1+iy)=frac{pi}{4}\
1=frac{frac{y}{x-1}-frac{y}{x+1}}{1+frac{y^2}{x^2-1}}=frac{xy+y-xy+y}{x^2-1+y^2}implies x^2-1+y^2=2y\implies x^2+y^2-2y-1=0
$$
But can I show that using geometry and the concept of Apollonius circle ?
complex-numbers circles
asked Jan 31 at 20:50
ss1729ss1729
2,05611124
$endgroup$
$begingroup$
Just a remark: as far as I remember from middle-school geometry, the answer is only a circular arc.
$endgroup$
– Bernard
Jan 31 at 20:56
$begingroup$
Here is a related link.....
$endgroup$
– Eleven-Eleven
Jan 31 at 21:02
add a comment |
$begingroup$
Show that the complex number $z$ satisfying the condition $argBig(dfrac{z-1}{z+1}Big)=dfrac{pi}{4}$ is a circle
I can prove it by expressing $z=x+iy$,
$$
argBig(frac{x-1+iy}{x+1+iy}Big)=arg(x-1+iy)-arg(x+1+iy)=frac{pi}{4}\
1=frac{frac{y}{x-1}-frac{y}{x+1}}{1+frac{y^2}{x^2-1}}=frac{xy+y-xy+y}{x^2-1+y^2}implies x^2-1+y^2=2y\implies x^2+y^2-2y-1=0
$$
But can I show that using geometry and the concept of Apollonius circle ?
complex-numbers circles
asked Jan 31 at 20:50
ss1729ss1729
2,05611124
$endgroup$
Show that the complex number $z$ satisfying the condition $argBig(dfrac{z-1}{z+1}Big)=dfrac{pi}{4}$ is a circle
I can prove it by expressing $z=x+iy$,
$$
argBig(frac{x-1+iy}{x+1+iy}Big)=arg(x-1+iy)-arg(x+1+iy)=frac{pi}{4}\
1=frac{frac{y}{x-1}-frac{y}{x+1}}{1+frac{y^2}{x^2-1}}=frac{xy+y-xy+y}{x^2-1+y^2}implies x^2-1+y^2=2y\implies x^2+y^2-2y-1=0
$$
But can I show that using geometry and the concept of Apollonius circle ?
complex-numbers circles
complex-numbers circles
asked Jan 31 at 20:50
ss1729ss1729
2,05611124
asked Jan 31 at 20:50
ss1729ss1729
2,05611124
asked Jan 31 at 20:50
ss1729ss1729
2,05611124
asked Jan 31 at 20:50
ss1729ss1729
2,05611124
asked Jan 31 at 20:50
ss1729ss1729
2,05611124
2,05611124
$begingroup$
Just a remark: as far as I remember from middle-school geometry, the answer is only a circular arc.
$endgroup$
– Bernard
Jan 31 at 20:56
$begingroup$
Here is a related link.....
$endgroup$
– Eleven-Eleven
Jan 31 at 21:02
add a comment |
$begingroup$
Just a remark: as far as I remember from middle-school geometry, the answer is only a circular arc.
$endgroup$
– Bernard
Jan 31 at 20:56
$begingroup$
Here is a related link.....
$endgroup$
– Eleven-Eleven
Jan 31 at 21:02
$begingroup$
Just a remark: as far as I remember from middle-school geometry, the answer is only a circular arc.
$endgroup$
– Bernard
Jan 31 at 20:56
$begingroup$
Just a remark: as far as I remember from middle-school geometry, the answer is only a circular arc.
$endgroup$
– Bernard
Jan 31 at 20:56
$begingroup$
Here is a related link.....
$endgroup$
– Eleven-Eleven
Jan 31 at 21:02
$begingroup$
Here is a related link.....
$endgroup$
– Eleven-Eleven
Jan 31 at 21:02
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
${rm arg}left(frac{z-1}{z+1}right)$ is a measure of the angle between $overrightarrow{AM}$ and $overrightarrow{BM}$, where $A$ has affix $-1$ and $B$ has affix $1$. Find one point $M_0$ verifying this equation which is not $A$ or $B$. The circumcircle of triangle $M_0AB$, excluding the arc between $A$ and $B$, is the solution, because if $Omega$ is its center, for every $M$ on this circle, the angle between $overrightarrow{AM}$ and $overrightarrow{BM}$ is half the angle between $overrightarrow{Omega A}$ and $overrightarrow{Omega B}$, which is $fracpi2$.
answered Jan 31 at 21:00
Nicolas FRANCOISNicolas FRANCOIS
3,7921516
$endgroup$
add a comment |
$begingroup$
Let $M(z), A(-1)$ and $B(1)$.
$arg(frac{z-1}{z+1})=(overrightarrow{MA},overrightarrow{MB})=frac{pi}{4} [pi]$ is a circle $Gamma$ passing through $A$ and $B$ except those 2 points.
How to get this circle: let $Delta$ the perpendicular bisector of $[AB]$, and choose $T$ a point such that $(overrightarrow{AB},overrightarrow{AT})=frac{pi}{4}$, define $Delta’$ as the line perpendicular to $(AT)$ at $A$ and finally $Omega=DeltabigcapDelta’$. $Gamma$ is the circle centered a $Omega$ passing through $A$ or/and $B$.
$\$
If $arg(frac{z-1}{z+1})=(overrightarrow{MA},overrightarrow{MB})=frac{pi}{4} [2pi]$ $Gamma$ is the arc located in the half plan delimited by $(AB)$ and not containing the point $T$. This method is still valid if we change $frac{pi}{4}$ by any value $theta$
answered Jan 31 at 21:14
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,808212
$endgroup$
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%2f3095436%2fshow-that-z-in-mathbbc-satisfying-arg-big-dfracz-1z1-big-dfrac-pi%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
$begingroup$
${rm arg}left(frac{z-1}{z+1}right)$ is a measure of the angle between $overrightarrow{AM}$ and $overrightarrow{BM}$, where $A$ has affix $-1$ and $B$ has affix $1$. Find one point $M_0$ verifying this equation which is not $A$ or $B$. The circumcircle of triangle $M_0AB$, excluding the arc between $A$ and $B$, is the solution, because if $Omega$ is its center, for every $M$ on this circle, the angle between $overrightarrow{AM}$ and $overrightarrow{BM}$ is half the angle between $overrightarrow{Omega A}$ and $overrightarrow{Omega B}$, which is $fracpi2$.
answered Jan 31 at 21:00
Nicolas FRANCOISNicolas FRANCOIS
3,7921516
$endgroup$
add a comment |
$begingroup$
${rm arg}left(frac{z-1}{z+1}right)$ is a measure of the angle between $overrightarrow{AM}$ and $overrightarrow{BM}$, where $A$ has affix $-1$ and $B$ has affix $1$. Find one point $M_0$ verifying this equation which is not $A$ or $B$. The circumcircle of triangle $M_0AB$, excluding the arc between $A$ and $B$, is the solution, because if $Omega$ is its center, for every $M$ on this circle, the angle between $overrightarrow{AM}$ and $overrightarrow{BM}$ is half the angle between $overrightarrow{Omega A}$ and $overrightarrow{Omega B}$, which is $fracpi2$.
answered Jan 31 at 21:00
Nicolas FRANCOISNicolas FRANCOIS
3,7921516
$endgroup$
add a comment |
$begingroup$
${rm arg}left(frac{z-1}{z+1}right)$ is a measure of the angle between $overrightarrow{AM}$ and $overrightarrow{BM}$, where $A$ has affix $-1$ and $B$ has affix $1$. Find one point $M_0$ verifying this equation which is not $A$ or $B$. The circumcircle of triangle $M_0AB$, excluding the arc between $A$ and $B$, is the solution, because if $Omega$ is its center, for every $M$ on this circle, the angle between $overrightarrow{AM}$ and $overrightarrow{BM}$ is half the angle between $overrightarrow{Omega A}$ and $overrightarrow{Omega B}$, which is $fracpi2$.
answered Jan 31 at 21:00
Nicolas FRANCOISNicolas FRANCOIS
3,7921516
$endgroup$
${rm arg}left(frac{z-1}{z+1}right)$ is a measure of the angle between $overrightarrow{AM}$ and $overrightarrow{BM}$, where $A$ has affix $-1$ and $B$ has affix $1$. Find one point $M_0$ verifying this equation which is not $A$ or $B$. The circumcircle of triangle $M_0AB$, excluding the arc between $A$ and $B$, is the solution, because if $Omega$ is its center, for every $M$ on this circle, the angle between $overrightarrow{AM}$ and $overrightarrow{BM}$ is half the angle between $overrightarrow{Omega A}$ and $overrightarrow{Omega B}$, which is $fracpi2$.
answered Jan 31 at 21:00
Nicolas FRANCOISNicolas FRANCOIS
3,7921516
answered Jan 31 at 21:00
Nicolas FRANCOISNicolas FRANCOIS
3,7921516
answered Jan 31 at 21:00
Nicolas FRANCOISNicolas FRANCOIS
3,7921516
answered Jan 31 at 21:00
Nicolas FRANCOISNicolas FRANCOIS
3,7921516
3,7921516
add a comment |
add a comment |
$begingroup$
Let $M(z), A(-1)$ and $B(1)$.
$arg(frac{z-1}{z+1})=(overrightarrow{MA},overrightarrow{MB})=frac{pi}{4} [pi]$ is a circle $Gamma$ passing through $A$ and $B$ except those 2 points.
How to get this circle: let $Delta$ the perpendicular bisector of $[AB]$, and choose $T$ a point such that $(overrightarrow{AB},overrightarrow{AT})=frac{pi}{4}$, define $Delta’$ as the line perpendicular to $(AT)$ at $A$ and finally $Omega=DeltabigcapDelta’$. $Gamma$ is the circle centered a $Omega$ passing through $A$ or/and $B$.
$\$
If $arg(frac{z-1}{z+1})=(overrightarrow{MA},overrightarrow{MB})=frac{pi}{4} [2pi]$ $Gamma$ is the arc located in the half plan delimited by $(AB)$ and not containing the point $T$. This method is still valid if we change $frac{pi}{4}$ by any value $theta$
answered Jan 31 at 21:14
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,808212
$endgroup$
add a comment |
$begingroup$
Let $M(z), A(-1)$ and $B(1)$.
$arg(frac{z-1}{z+1})=(overrightarrow{MA},overrightarrow{MB})=frac{pi}{4} [pi]$ is a circle $Gamma$ passing through $A$ and $B$ except those 2 points.
How to get this circle: let $Delta$ the perpendicular bisector of $[AB]$, and choose $T$ a point such that $(overrightarrow{AB},overrightarrow{AT})=frac{pi}{4}$, define $Delta’$ as the line perpendicular to $(AT)$ at $A$ and finally $Omega=DeltabigcapDelta’$. $Gamma$ is the circle centered a $Omega$ passing through $A$ or/and $B$.
$\$
If $arg(frac{z-1}{z+1})=(overrightarrow{MA},overrightarrow{MB})=frac{pi}{4} [2pi]$ $Gamma$ is the arc located in the half plan delimited by $(AB)$ and not containing the point $T$. This method is still valid if we change $frac{pi}{4}$ by any value $theta$
answered Jan 31 at 21:14
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,808212
$endgroup$
add a comment |
$begingroup$
Let $M(z), A(-1)$ and $B(1)$.
$arg(frac{z-1}{z+1})=(overrightarrow{MA},overrightarrow{MB})=frac{pi}{4} [pi]$ is a circle $Gamma$ passing through $A$ and $B$ except those 2 points.
How to get this circle: let $Delta$ the perpendicular bisector of $[AB]$, and choose $T$ a point such that $(overrightarrow{AB},overrightarrow{AT})=frac{pi}{4}$, define $Delta’$ as the line perpendicular to $(AT)$ at $A$ and finally $Omega=DeltabigcapDelta’$. $Gamma$ is the circle centered a $Omega$ passing through $A$ or/and $B$.
$\$
If $arg(frac{z-1}{z+1})=(overrightarrow{MA},overrightarrow{MB})=frac{pi}{4} [2pi]$ $Gamma$ is the arc located in the half plan delimited by $(AB)$ and not containing the point $T$. This method is still valid if we change $frac{pi}{4}$ by any value $theta$
answered Jan 31 at 21:14
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,808212
$endgroup$
Let $M(z), A(-1)$ and $B(1)$.
$arg(frac{z-1}{z+1})=(overrightarrow{MA},overrightarrow{MB})=frac{pi}{4} [pi]$ is a circle $Gamma$ passing through $A$ and $B$ except those 2 points.
How to get this circle: let $Delta$ the perpendicular bisector of $[AB]$, and choose $T$ a point such that $(overrightarrow{AB},overrightarrow{AT})=frac{pi}{4}$, define $Delta’$ as the line perpendicular to $(AT)$ at $A$ and finally $Omega=DeltabigcapDelta’$. $Gamma$ is the circle centered a $Omega$ passing through $A$ or/and $B$.
$\$
If $arg(frac{z-1}{z+1})=(overrightarrow{MA},overrightarrow{MB})=frac{pi}{4} [2pi]$ $Gamma$ is the arc located in the half plan delimited by $(AB)$ and not containing the point $T$. This method is still valid if we change $frac{pi}{4}$ by any value $theta$
answered Jan 31 at 21:14
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,808212
edited Jan 31 at 21:29
answered Jan 31 at 21:14
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,808212
answered Jan 31 at 21:14
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,808212
answered Jan 31 at 21:14
HAMIDINE SOUMAREHAMIDINE SOUMARE
1,808212
1,808212
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%2f3095436%2fshow-that-z-in-mathbbc-satisfying-arg-big-dfracz-1z1-big-dfrac-pi%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
$begingroup$
Just a remark: as far as I remember from middle-school geometry, the answer is only a circular arc.
$endgroup$
– Bernard
Jan 31 at 20:56
$begingroup$
Here is a related link.....
$endgroup$
– Eleven-Eleven
Jan 31 at 21:02