Gsyfuy

Difference between i++ and (i)++ in C

What are "industrial chops"?

Why has the mole been redefined for 2019?

Looking for access to original paper for Category O

How to say "Brexit" in Latin?

Making him into a bully (how to show mild violence)

Moving from one room/cave to another within the same dungeon

My cat mixes up the floors in my building. How can I help him?

How did Ancient Greek 'πυρ' become English 'fire?'

Why is working on the same position for more than 15 years not a red flag?

Why would Pakistan closing its air space cancel flights not headed to Pakistan itself?

What is 6÷2×(1+2) =?

Traveling through the asteriod belt?

Early credit roll before the end of the film

What is the postion of Lord Shiva as per Srimad-Bhagavatam

Can an insurance company drop you after receiving a bill and refusing to pay?

Why did the villain in the first Men in Black movie care about Earth's Cockroaches?

Graph with overlapping labels

Publishing research using outdated methods

Can we use the stored gravitational potential energy of a building to produce power?

How to limit sight distance to 1 KM

Is boss over stepping boundary/micromanaging?

Dilemma of explaining to interviewer that he is the reason for declining second interview

One Half of Ten; A Riddle

finding the number of complex number

Simplified way of finding a complex number raised to another complex numberNumber of Infinities in complex numbersComplex number polar form equationFinding numbers $a$ and $b$ for a complex numberFinding a Trigonometric Form of Complex NumberFinding the root of a complex numberFind the polar form of the complex numberInequation complex number with fractionsThe Geometry of Complex Number and Its Multiplicative InverseFinding a complex number

2

$begingroup$

How many complex number $z$ are there such that $|z+1|= |z+i|$ and $|z|=5$?

My attempt :

I got $2$, that is $ z=-2, z= +2$ , $|z| = {sqrt{ 2^2+1}}$, $|z| = {sqrt{(-2^2) +1}}$

Is it true ?

geometry proof-verification complex-numbers euclidean-geometry analytic-geometry

share|cite|improve this question

edited 28 mins ago

Anirban Niloy

599118

asked 2 hours ago

jasminejasmine

1,806418

$endgroup$

add a comment | 

2

$begingroup$

How many complex number $z$ are there such that $|z+1|= |z+i|$ and $|z|=5$?

My attempt :

I got $2$, that is $ z=-2, z= +2$ , $|z| = {sqrt{ 2^2+1}}$, $|z| = {sqrt{(-2^2) +1}}$

Is it true ?

geometry proof-verification complex-numbers euclidean-geometry analytic-geometry

share|cite|improve this question

edited 28 mins ago

Anirban Niloy

599118

asked 2 hours ago

jasminejasmine

1,806418

$endgroup$

add a comment | 

$begingroup$

How many complex number $z$ are there such that $|z+1|= |z+i|$ and $|z|=5$?

My attempt :

I got $2$, that is $ z=-2, z= +2$ , $|z| = {sqrt{ 2^2+1}}$, $|z| = {sqrt{(-2^2) +1}}$

Is it true ?

geometry proof-verification complex-numbers euclidean-geometry analytic-geometry

share|cite|improve this question

edited 28 mins ago

Anirban Niloy

599118

asked 2 hours ago

jasminejasmine

1,806418

$endgroup$

share|cite|improve this question

edited 28 mins ago

Anirban Niloy

599118

asked 2 hours ago

jasminejasmine

1,806418

share|cite|improve this question

edited 28 mins ago

Anirban Niloy

599118

asked 2 hours ago

jasminejasmine

1,806418

edited 28 mins ago

Anirban Niloy

599118

edited 28 mins ago

Anirban Niloy

599118

asked 2 hours ago

jasminejasmine

1,806418

asked 2 hours ago

jasminejasmine

1,806418

2 Answers
2

active

oldest

votes

7

$begingroup$

From $|z+1|= |z+i|$ we get $z$ at equaly distance from $-1$ and $-i$ so $z$ is on perpendicular bisetor for segment between $-1$ and $-i$, that is line $y=x$ so $$z=x+xi$$ for some real $x$. Then $|x|sqrt{1+1} = 5$ so $x=pm{5sqrt{2}over 2}$

share|cite|improve this answer

edited 2 hours ago

answered 2 hours ago

greedoidgreedoid

45k1156111

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  (+1) for explaining it geometrically
  $endgroup$
  – José Carlos Santos
  2 hours ago
  

  
  
  
  

add a comment | 

3

$begingroup$

Let $A(-1,0)$ and $B(0,-1)$.

Thus, we need to find a number of intersect points of the perpendicular bisector of $AB$

with the circle $x^2+y^2=25.$

Now we see that they are indeed two points because $A$ and $B$ are placed inside the circle.

share|cite|improve this answer

answered 2 hours ago

Michael RozenbergMichael Rozenberg

105k1893198

$endgroup$

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
});


}
});

draft saved

draft discarded

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

Name

Email

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3130333%2ffinding-the-number-of-complex-number%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

7

$begingroup$

From $|z+1|= |z+i|$ we get $z$ at equaly distance from $-1$ and $-i$ so $z$ is on perpendicular bisetor for segment between $-1$ and $-i$, that is line $y=x$ so $$z=x+xi$$ for some real $x$. Then $|x|sqrt{1+1} = 5$ so $x=pm{5sqrt{2}over 2}$

share|cite|improve this answer

edited 2 hours ago

answered 2 hours ago

greedoidgreedoid

45k1156111

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  (+1) for explaining it geometrically
  $endgroup$
  – José Carlos Santos
  2 hours ago
  

  
  
  
  

add a comment | 

7

$begingroup$

From $|z+1|= |z+i|$ we get $z$ at equaly distance from $-1$ and $-i$ so $z$ is on perpendicular bisetor for segment between $-1$ and $-i$, that is line $y=x$ so $$z=x+xi$$ for some real $x$. Then $|x|sqrt{1+1} = 5$ so $x=pm{5sqrt{2}over 2}$

share|cite|improve this answer

edited 2 hours ago

answered 2 hours ago

greedoidgreedoid

45k1156111

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  (+1) for explaining it geometrically
  $endgroup$
  – José Carlos Santos
  2 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

From $|z+1|= |z+i|$ we get $z$ at equaly distance from $-1$ and $-i$ so $z$ is on perpendicular bisetor for segment between $-1$ and $-i$, that is line $y=x$ so $$z=x+xi$$ for some real $x$. Then $|x|sqrt{1+1} = 5$ so $x=pm{5sqrt{2}over 2}$

share|cite|improve this answer

edited 2 hours ago

answered 2 hours ago

greedoidgreedoid

45k1156111

$endgroup$

share|cite|improve this answer

edited 2 hours ago

answered 2 hours ago

greedoidgreedoid

45k1156111

answered 2 hours ago

greedoidgreedoid

45k1156111

answered 2 hours ago

greedoidgreedoid

45k1156111

3

$begingroup$

Let $A(-1,0)$ and $B(0,-1)$.

Thus, we need to find a number of intersect points of the perpendicular bisector of $AB$

with the circle $x^2+y^2=25.$

Now we see that they are indeed two points because $A$ and $B$ are placed inside the circle.

share|cite|improve this answer

answered 2 hours ago

Michael RozenbergMichael Rozenberg

105k1893198

$endgroup$

add a comment | 

3

$begingroup$

Let $A(-1,0)$ and $B(0,-1)$.

Thus, we need to find a number of intersect points of the perpendicular bisector of $AB$

with the circle $x^2+y^2=25.$

Now we see that they are indeed two points because $A$ and $B$ are placed inside the circle.

share|cite|improve this answer

answered 2 hours ago

Michael RozenbergMichael Rozenberg

105k1893198

$endgroup$

add a comment | 

$begingroup$

Let $A(-1,0)$ and $B(0,-1)$.

Thus, we need to find a number of intersect points of the perpendicular bisector of $AB$

with the circle $x^2+y^2=25.$

Now we see that they are indeed two points because $A$ and $B$ are placed inside the circle.

share|cite|improve this answer

answered 2 hours ago

Michael RozenbergMichael Rozenberg

105k1893198

$endgroup$

share|cite|improve this answer

answered 2 hours ago

Michael RozenbergMichael Rozenberg

105k1893198

answered 2 hours ago

Michael RozenbergMichael Rozenberg

105k1893198

answered 2 hours ago

Michael RozenbergMichael Rozenberg

105k1893198