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finding the number of complex number

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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 26 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 26 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 26 mins ago

Anirban Niloy

599118

asked 2 hours ago

jasminejasmine

1,806418

$endgroup$

share|cite|improve this question

edited 26 mins ago

Anirban Niloy

599118

asked 2 hours ago

jasminejasmine

1,806418

share|cite|improve this question

edited 26 mins ago

Anirban Niloy

599118

asked 2 hours ago

jasminejasmine

1,806418

edited 26 mins ago

Anirban Niloy

599118

edited 26 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 | 

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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