Strange way to define an ellipse in the complex plane












2












$begingroup$


I'm asked to prove that



$|z-i||z+i|=2$



defines an ellipse in the plane.



I have tried replacing $z = x+iy $ in the previous equation and brute forcing the result to no avail. Considering that $ |z-i||z+i| = |z^2+1| $ eases the algebra a bit but didn't help me that much.



Edit: I know it seems like there's a missing plus sign: $|z-i| + |z+i| = 2$ in the question, but that's what the exercise says.



In fact, the next exercise wants us to prove that $ |z-1||z+i| = 2$ defines a line in the complex plane.






complex-numbers conic-sections







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    I think there's a mistake. There's supposed to be a plus sign between the two absolute values, not multiplication .
    $endgroup$
    – zokomoko
    Aug 24 '17 at 5:37










  • $begingroup$
    I agree with you. That's definitely not an ellipse, although it looks really close to one.
    $endgroup$
    – Jairo Peña
    Aug 24 '17 at 6:00
















2












$begingroup$


I'm asked to prove that



$|z-i||z+i|=2$



defines an ellipse in the plane.



I have tried replacing $z = x+iy $ in the previous equation and brute forcing the result to no avail. Considering that $ |z-i||z+i| = |z^2+1| $ eases the algebra a bit but didn't help me that much.



Edit: I know it seems like there's a missing plus sign: $|z-i| + |z+i| = 2$ in the question, but that's what the exercise says.



In fact, the next exercise wants us to prove that $ |z-1||z+i| = 2$ defines a line in the complex plane.






complex-numbers conic-sections







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    I think there's a mistake. There's supposed to be a plus sign between the two absolute values, not multiplication .
    $endgroup$
    – zokomoko
    Aug 24 '17 at 5:37










  • $begingroup$
    I agree with you. That's definitely not an ellipse, although it looks really close to one.
    $endgroup$
    – Jairo Peña
    Aug 24 '17 at 6:00














2












2








2





$begingroup$


I'm asked to prove that



$|z-i||z+i|=2$



defines an ellipse in the plane.



I have tried replacing $z = x+iy $ in the previous equation and brute forcing the result to no avail. Considering that $ |z-i||z+i| = |z^2+1| $ eases the algebra a bit but didn't help me that much.



Edit: I know it seems like there's a missing plus sign: $|z-i| + |z+i| = 2$ in the question, but that's what the exercise says.



In fact, the next exercise wants us to prove that $ |z-1||z+i| = 2$ defines a line in the complex plane.






complex-numbers conic-sections







share|cite|improve this question











$endgroup$




I'm asked to prove that



$|z-i||z+i|=2$



defines an ellipse in the plane.



I have tried replacing $z = x+iy $ in the previous equation and brute forcing the result to no avail. Considering that $ |z-i||z+i| = |z^2+1| $ eases the algebra a bit but didn't help me that much.



Edit: I know it seems like there's a missing plus sign: $|z-i| + |z+i| = 2$ in the question, but that's what the exercise says.



In fact, the next exercise wants us to prove that $ |z-1||z+i| = 2$ defines a line in the complex plane.






complex-numbers conic-sections




complex-numbers conic-sections






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 10 at 10:23









José Carlos Santos

155k22124227




155k22124227










asked Aug 24 '17 at 5:33









Jairo PeñaJairo Peña

135




135








  • 2




    $begingroup$
    I think there's a mistake. There's supposed to be a plus sign between the two absolute values, not multiplication .
    $endgroup$
    – zokomoko
    Aug 24 '17 at 5:37










  • $begingroup$
    I agree with you. That's definitely not an ellipse, although it looks really close to one.
    $endgroup$
    – Jairo Peña
    Aug 24 '17 at 6:00














  • 2




    $begingroup$
    I think there's a mistake. There's supposed to be a plus sign between the two absolute values, not multiplication .
    $endgroup$
    – zokomoko
    Aug 24 '17 at 5:37










  • $begingroup$
    I agree with you. That's definitely not an ellipse, although it looks really close to one.
    $endgroup$
    – Jairo Peña
    Aug 24 '17 at 6:00








2




2




$begingroup$
I think there's a mistake. There's supposed to be a plus sign between the two absolute values, not multiplication .
$endgroup$
– zokomoko
Aug 24 '17 at 5:37




$begingroup$
I think there's a mistake. There's supposed to be a plus sign between the two absolute values, not multiplication .
$endgroup$
– zokomoko
Aug 24 '17 at 5:37












$begingroup$
I agree with you. That's definitely not an ellipse, although it looks really close to one.
$endgroup$
– Jairo Peña
Aug 24 '17 at 6:00




$begingroup$
I agree with you. That's definitely not an ellipse, although it looks really close to one.
$endgroup$
– Jairo Peña
Aug 24 '17 at 6:00










2 Answers
2






active

oldest

votes


















2












$begingroup$

The locus of the points $P$ which have the product of their distances from two given points (foci) constant is known as a the Cassini oval. Thus I also think that there should be a + sign rather than multiplication in your expression.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I completely agree with you . However, I'm surprised by the fact that plotting this equation does in fact show an ellipse. Check it out: wolframalpha.com/input/… Btw, I'm new to this site, so forgive me if my etiquette is not up to standards.
    $endgroup$
    – Jairo Peña
    Aug 24 '17 at 5:52






  • 2




    $begingroup$
    @JairoPeña It looks like an ellipse, but it isn't. This is a special case of the Cassini oval in which the parameter $e$ from the Wikipedia page is equal to $sqrt{2}$.
    $endgroup$
    – user1337
    Aug 24 '17 at 5:54












  • $begingroup$
    I completely understand now. Just out of curiosity, what do you think was the mistake in the second exercise, where I'm asked to find a straight line from the equation. (I plotted that equation and it's definitely not a straight line lol)
    $endgroup$
    – Jairo Peña
    Aug 24 '17 at 5:58










  • $begingroup$
    @JairoPeña Not sure about it, but it could be $|z-1| div |z+i|=1$.
    $endgroup$
    – user1337
    Aug 24 '17 at 7:49



















4












$begingroup$

This expression doesn't define an ellipse. It's the set of those points of the plane such that the products of its distances to $i$ and to $-i$ is equal to $2$. If we were talking about the sum of the distances, then, yes, it would be an ellipse.



In order to prove that it is not an ellipse, note that four of the points of this set are $pm1$ and $pmsqrt3,i$. Furthermore, the set is symmetric with respect to both axis. There is one and only one ellipse fulfilling these conditions:$$left{x+yiinmathbb{C},middle|,x^2+frac{y^2}3=1right}.$$One of the points of this ellipse is $z_0=frac12+frac{3i}2$. But $|z_0-i|.|z_0+i|=frac{sqrt{13}}2$. Therefore, your set is not an ellipse.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    @JairoPeña Please read the new paragraph.
    $endgroup$
    – José Carlos Santos
    Aug 24 '17 at 6:00










  • $begingroup$
    Thank you for your thorough response. It is definitely not an ellipse. Just to tickle your brain, what do you think was the mistake in the second exercise, where they want us to find a straight line.
    $endgroup$
    – Jairo Peña
    Aug 24 '17 at 6:04










  • $begingroup$
    @JairoPeña I have no idea,
    $endgroup$
    – José Carlos Santos
    Aug 24 '17 at 6:06










  • $begingroup$
    Considering that this whole question was based on a mistake, I'm going to delete it from the site.
    $endgroup$
    – Jairo Peña
    Aug 24 '17 at 6:08






  • 1




    $begingroup$
    @JairoPeña Don't delete! The problem and solution of noticing it's not an ellipse is instructive. Someone with similar problems might find this question by searching. You might change the title though.
    $endgroup$
    – JiK
    Aug 24 '17 at 15:11











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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2404258%2fstrange-way-to-define-an-ellipse-in-the-complex-plane%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









2












$begingroup$

The locus of the points $P$ which have the product of their distances from two given points (foci) constant is known as a the Cassini oval. Thus I also think that there should be a + sign rather than multiplication in your expression.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I completely agree with you . However, I'm surprised by the fact that plotting this equation does in fact show an ellipse. Check it out: wolframalpha.com/input/… Btw, I'm new to this site, so forgive me if my etiquette is not up to standards.
    $endgroup$
    – Jairo Peña
    Aug 24 '17 at 5:52






  • 2




    $begingroup$
    @JairoPeña It looks like an ellipse, but it isn't. This is a special case of the Cassini oval in which the parameter $e$ from the Wikipedia page is equal to $sqrt{2}$.
    $endgroup$
    – user1337
    Aug 24 '17 at 5:54












  • $begingroup$
    I completely understand now. Just out of curiosity, what do you think was the mistake in the second exercise, where I'm asked to find a straight line from the equation. (I plotted that equation and it's definitely not a straight line lol)
    $endgroup$
    – Jairo Peña
    Aug 24 '17 at 5:58










  • $begingroup$
    @JairoPeña Not sure about it, but it could be $|z-1| div |z+i|=1$.
    $endgroup$
    – user1337
    Aug 24 '17 at 7:49
















2












$begingroup$

The locus of the points $P$ which have the product of their distances from two given points (foci) constant is known as a the Cassini oval. Thus I also think that there should be a + sign rather than multiplication in your expression.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I completely agree with you . However, I'm surprised by the fact that plotting this equation does in fact show an ellipse. Check it out: wolframalpha.com/input/… Btw, I'm new to this site, so forgive me if my etiquette is not up to standards.
    $endgroup$
    – Jairo Peña
    Aug 24 '17 at 5:52






  • 2




    $begingroup$
    @JairoPeña It looks like an ellipse, but it isn't. This is a special case of the Cassini oval in which the parameter $e$ from the Wikipedia page is equal to $sqrt{2}$.
    $endgroup$
    – user1337
    Aug 24 '17 at 5:54












  • $begingroup$
    I completely understand now. Just out of curiosity, what do you think was the mistake in the second exercise, where I'm asked to find a straight line from the equation. (I plotted that equation and it's definitely not a straight line lol)
    $endgroup$
    – Jairo Peña
    Aug 24 '17 at 5:58










  • $begingroup$
    @JairoPeña Not sure about it, but it could be $|z-1| div |z+i|=1$.
    $endgroup$
    – user1337
    Aug 24 '17 at 7:49














2












2








2





$begingroup$

The locus of the points $P$ which have the product of their distances from two given points (foci) constant is known as a the Cassini oval. Thus I also think that there should be a + sign rather than multiplication in your expression.






share|cite|improve this answer









$endgroup$



The locus of the points $P$ which have the product of their distances from two given points (foci) constant is known as a the Cassini oval. Thus I also think that there should be a + sign rather than multiplication in your expression.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Aug 24 '17 at 5:44









user1337user1337

16.6k43391




16.6k43391












  • $begingroup$
    I completely agree with you . However, I'm surprised by the fact that plotting this equation does in fact show an ellipse. Check it out: wolframalpha.com/input/… Btw, I'm new to this site, so forgive me if my etiquette is not up to standards.
    $endgroup$
    – Jairo Peña
    Aug 24 '17 at 5:52






  • 2




    $begingroup$
    @JairoPeña It looks like an ellipse, but it isn't. This is a special case of the Cassini oval in which the parameter $e$ from the Wikipedia page is equal to $sqrt{2}$.
    $endgroup$
    – user1337
    Aug 24 '17 at 5:54












  • $begingroup$
    I completely understand now. Just out of curiosity, what do you think was the mistake in the second exercise, where I'm asked to find a straight line from the equation. (I plotted that equation and it's definitely not a straight line lol)
    $endgroup$
    – Jairo Peña
    Aug 24 '17 at 5:58










  • $begingroup$
    @JairoPeña Not sure about it, but it could be $|z-1| div |z+i|=1$.
    $endgroup$
    – user1337
    Aug 24 '17 at 7:49


















  • $begingroup$
    I completely agree with you . However, I'm surprised by the fact that plotting this equation does in fact show an ellipse. Check it out: wolframalpha.com/input/… Btw, I'm new to this site, so forgive me if my etiquette is not up to standards.
    $endgroup$
    – Jairo Peña
    Aug 24 '17 at 5:52






  • 2




    $begingroup$
    @JairoPeña It looks like an ellipse, but it isn't. This is a special case of the Cassini oval in which the parameter $e$ from the Wikipedia page is equal to $sqrt{2}$.
    $endgroup$
    – user1337
    Aug 24 '17 at 5:54












  • $begingroup$
    I completely understand now. Just out of curiosity, what do you think was the mistake in the second exercise, where I'm asked to find a straight line from the equation. (I plotted that equation and it's definitely not a straight line lol)
    $endgroup$
    – Jairo Peña
    Aug 24 '17 at 5:58










  • $begingroup$
    @JairoPeña Not sure about it, but it could be $|z-1| div |z+i|=1$.
    $endgroup$
    – user1337
    Aug 24 '17 at 7:49
















$begingroup$
I completely agree with you . However, I'm surprised by the fact that plotting this equation does in fact show an ellipse. Check it out: wolframalpha.com/input/… Btw, I'm new to this site, so forgive me if my etiquette is not up to standards.
$endgroup$
– Jairo Peña
Aug 24 '17 at 5:52




$begingroup$
I completely agree with you . However, I'm surprised by the fact that plotting this equation does in fact show an ellipse. Check it out: wolframalpha.com/input/… Btw, I'm new to this site, so forgive me if my etiquette is not up to standards.
$endgroup$
– Jairo Peña
Aug 24 '17 at 5:52




2




2




$begingroup$
@JairoPeña It looks like an ellipse, but it isn't. This is a special case of the Cassini oval in which the parameter $e$ from the Wikipedia page is equal to $sqrt{2}$.
$endgroup$
– user1337
Aug 24 '17 at 5:54






$begingroup$
@JairoPeña It looks like an ellipse, but it isn't. This is a special case of the Cassini oval in which the parameter $e$ from the Wikipedia page is equal to $sqrt{2}$.
$endgroup$
– user1337
Aug 24 '17 at 5:54














$begingroup$
I completely understand now. Just out of curiosity, what do you think was the mistake in the second exercise, where I'm asked to find a straight line from the equation. (I plotted that equation and it's definitely not a straight line lol)
$endgroup$
– Jairo Peña
Aug 24 '17 at 5:58




$begingroup$
I completely understand now. Just out of curiosity, what do you think was the mistake in the second exercise, where I'm asked to find a straight line from the equation. (I plotted that equation and it's definitely not a straight line lol)
$endgroup$
– Jairo Peña
Aug 24 '17 at 5:58












$begingroup$
@JairoPeña Not sure about it, but it could be $|z-1| div |z+i|=1$.
$endgroup$
– user1337
Aug 24 '17 at 7:49




$begingroup$
@JairoPeña Not sure about it, but it could be $|z-1| div |z+i|=1$.
$endgroup$
– user1337
Aug 24 '17 at 7:49











4












$begingroup$

This expression doesn't define an ellipse. It's the set of those points of the plane such that the products of its distances to $i$ and to $-i$ is equal to $2$. If we were talking about the sum of the distances, then, yes, it would be an ellipse.



In order to prove that it is not an ellipse, note that four of the points of this set are $pm1$ and $pmsqrt3,i$. Furthermore, the set is symmetric with respect to both axis. There is one and only one ellipse fulfilling these conditions:$$left{x+yiinmathbb{C},middle|,x^2+frac{y^2}3=1right}.$$One of the points of this ellipse is $z_0=frac12+frac{3i}2$. But $|z_0-i|.|z_0+i|=frac{sqrt{13}}2$. Therefore, your set is not an ellipse.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    @JairoPeña Please read the new paragraph.
    $endgroup$
    – José Carlos Santos
    Aug 24 '17 at 6:00










  • $begingroup$
    Thank you for your thorough response. It is definitely not an ellipse. Just to tickle your brain, what do you think was the mistake in the second exercise, where they want us to find a straight line.
    $endgroup$
    – Jairo Peña
    Aug 24 '17 at 6:04










  • $begingroup$
    @JairoPeña I have no idea,
    $endgroup$
    – José Carlos Santos
    Aug 24 '17 at 6:06










  • $begingroup$
    Considering that this whole question was based on a mistake, I'm going to delete it from the site.
    $endgroup$
    – Jairo Peña
    Aug 24 '17 at 6:08






  • 1




    $begingroup$
    @JairoPeña Don't delete! The problem and solution of noticing it's not an ellipse is instructive. Someone with similar problems might find this question by searching. You might change the title though.
    $endgroup$
    – JiK
    Aug 24 '17 at 15:11
















4












$begingroup$

This expression doesn't define an ellipse. It's the set of those points of the plane such that the products of its distances to $i$ and to $-i$ is equal to $2$. If we were talking about the sum of the distances, then, yes, it would be an ellipse.



In order to prove that it is not an ellipse, note that four of the points of this set are $pm1$ and $pmsqrt3,i$. Furthermore, the set is symmetric with respect to both axis. There is one and only one ellipse fulfilling these conditions:$$left{x+yiinmathbb{C},middle|,x^2+frac{y^2}3=1right}.$$One of the points of this ellipse is $z_0=frac12+frac{3i}2$. But $|z_0-i|.|z_0+i|=frac{sqrt{13}}2$. Therefore, your set is not an ellipse.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    @JairoPeña Please read the new paragraph.
    $endgroup$
    – José Carlos Santos
    Aug 24 '17 at 6:00










  • $begingroup$
    Thank you for your thorough response. It is definitely not an ellipse. Just to tickle your brain, what do you think was the mistake in the second exercise, where they want us to find a straight line.
    $endgroup$
    – Jairo Peña
    Aug 24 '17 at 6:04










  • $begingroup$
    @JairoPeña I have no idea,
    $endgroup$
    – José Carlos Santos
    Aug 24 '17 at 6:06










  • $begingroup$
    Considering that this whole question was based on a mistake, I'm going to delete it from the site.
    $endgroup$
    – Jairo Peña
    Aug 24 '17 at 6:08






  • 1




    $begingroup$
    @JairoPeña Don't delete! The problem and solution of noticing it's not an ellipse is instructive. Someone with similar problems might find this question by searching. You might change the title though.
    $endgroup$
    – JiK
    Aug 24 '17 at 15:11














4












4








4





$begingroup$

This expression doesn't define an ellipse. It's the set of those points of the plane such that the products of its distances to $i$ and to $-i$ is equal to $2$. If we were talking about the sum of the distances, then, yes, it would be an ellipse.



In order to prove that it is not an ellipse, note that four of the points of this set are $pm1$ and $pmsqrt3,i$. Furthermore, the set is symmetric with respect to both axis. There is one and only one ellipse fulfilling these conditions:$$left{x+yiinmathbb{C},middle|,x^2+frac{y^2}3=1right}.$$One of the points of this ellipse is $z_0=frac12+frac{3i}2$. But $|z_0-i|.|z_0+i|=frac{sqrt{13}}2$. Therefore, your set is not an ellipse.






share|cite|improve this answer











$endgroup$



This expression doesn't define an ellipse. It's the set of those points of the plane such that the products of its distances to $i$ and to $-i$ is equal to $2$. If we were talking about the sum of the distances, then, yes, it would be an ellipse.



In order to prove that it is not an ellipse, note that four of the points of this set are $pm1$ and $pmsqrt3,i$. Furthermore, the set is symmetric with respect to both axis. There is one and only one ellipse fulfilling these conditions:$$left{x+yiinmathbb{C},middle|,x^2+frac{y^2}3=1right}.$$One of the points of this ellipse is $z_0=frac12+frac{3i}2$. But $|z_0-i|.|z_0+i|=frac{sqrt{13}}2$. Therefore, your set is not an ellipse.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Aug 24 '17 at 5:59

























answered Aug 24 '17 at 5:40









José Carlos SantosJosé Carlos Santos

155k22124227




155k22124227












  • $begingroup$
    @JairoPeña Please read the new paragraph.
    $endgroup$
    – José Carlos Santos
    Aug 24 '17 at 6:00










  • $begingroup$
    Thank you for your thorough response. It is definitely not an ellipse. Just to tickle your brain, what do you think was the mistake in the second exercise, where they want us to find a straight line.
    $endgroup$
    – Jairo Peña
    Aug 24 '17 at 6:04










  • $begingroup$
    @JairoPeña I have no idea,
    $endgroup$
    – José Carlos Santos
    Aug 24 '17 at 6:06










  • $begingroup$
    Considering that this whole question was based on a mistake, I'm going to delete it from the site.
    $endgroup$
    – Jairo Peña
    Aug 24 '17 at 6:08






  • 1




    $begingroup$
    @JairoPeña Don't delete! The problem and solution of noticing it's not an ellipse is instructive. Someone with similar problems might find this question by searching. You might change the title though.
    $endgroup$
    – JiK
    Aug 24 '17 at 15:11


















  • $begingroup$
    @JairoPeña Please read the new paragraph.
    $endgroup$
    – José Carlos Santos
    Aug 24 '17 at 6:00










  • $begingroup$
    Thank you for your thorough response. It is definitely not an ellipse. Just to tickle your brain, what do you think was the mistake in the second exercise, where they want us to find a straight line.
    $endgroup$
    – Jairo Peña
    Aug 24 '17 at 6:04










  • $begingroup$
    @JairoPeña I have no idea,
    $endgroup$
    – José Carlos Santos
    Aug 24 '17 at 6:06










  • $begingroup$
    Considering that this whole question was based on a mistake, I'm going to delete it from the site.
    $endgroup$
    – Jairo Peña
    Aug 24 '17 at 6:08






  • 1




    $begingroup$
    @JairoPeña Don't delete! The problem and solution of noticing it's not an ellipse is instructive. Someone with similar problems might find this question by searching. You might change the title though.
    $endgroup$
    – JiK
    Aug 24 '17 at 15:11
















$begingroup$
@JairoPeña Please read the new paragraph.
$endgroup$
– José Carlos Santos
Aug 24 '17 at 6:00




$begingroup$
@JairoPeña Please read the new paragraph.
$endgroup$
– José Carlos Santos
Aug 24 '17 at 6:00












$begingroup$
Thank you for your thorough response. It is definitely not an ellipse. Just to tickle your brain, what do you think was the mistake in the second exercise, where they want us to find a straight line.
$endgroup$
– Jairo Peña
Aug 24 '17 at 6:04




$begingroup$
Thank you for your thorough response. It is definitely not an ellipse. Just to tickle your brain, what do you think was the mistake in the second exercise, where they want us to find a straight line.
$endgroup$
– Jairo Peña
Aug 24 '17 at 6:04












$begingroup$
@JairoPeña I have no idea,
$endgroup$
– José Carlos Santos
Aug 24 '17 at 6:06




$begingroup$
@JairoPeña I have no idea,
$endgroup$
– José Carlos Santos
Aug 24 '17 at 6:06












$begingroup$
Considering that this whole question was based on a mistake, I'm going to delete it from the site.
$endgroup$
– Jairo Peña
Aug 24 '17 at 6:08




$begingroup$
Considering that this whole question was based on a mistake, I'm going to delete it from the site.
$endgroup$
– Jairo Peña
Aug 24 '17 at 6:08




1




1




$begingroup$
@JairoPeña Don't delete! The problem and solution of noticing it's not an ellipse is instructive. Someone with similar problems might find this question by searching. You might change the title though.
$endgroup$
– JiK
Aug 24 '17 at 15:11




$begingroup$
@JairoPeña Don't delete! The problem and solution of noticing it's not an ellipse is instructive. Someone with similar problems might find this question by searching. You might change the title though.
$endgroup$
– JiK
Aug 24 '17 at 15:11


















draft saved

draft discarded




















































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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2404258%2fstrange-way-to-define-an-ellipse-in-the-complex-plane%23new-answer', 'question_page');
}
);

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







-

Popular posts from this blog

Mario Kart Wii

Image
Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro...
Read more

The Binding of Isaac: Rebirth/Afterbirth

Image
Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! The Binding of Isaac: Rebirth/Afterbirth From The Cutting Room Floor < The Binding of Isaac: Rebirth Jump to: navigation, search This is a sub-page of The Binding of Isaac: Rebirth . Afterbirth is the first expansion pack for Rebirth and features a variety of improvements. Contents 1 Unused Graphics 1.1 Old Man Keeper 1.2 Skinless Hush 1.3 Miscellaneous Unused Hush Sprites 1.4 Generosity achievement 1.5 Afterbirth+ Room Editor 1.6 Bearded Ultra Greed 1.7 Early Ultra Greed 1.8 Unused Brownie Expression 1.9 Frail's Spinning Head 1.10 Mom's Dead Hand Alt 2 Unused Audio 2.1 Unused Dead Womb Layer 2.2 Unused Hush Battle Outro 2.3 The Stain Grunt Unused Graphics Old Man Keeper Image file: gfxuimain me...
Read more

What does “Dominus providebit” mean?

Image
5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1 ...
Read more