find $z in mathbb C$ ($z$ equal to complex numbers) for this function $| z + 3i | = 3|z|$ [closed]

-1

$begingroup$

Can anybody explain how to solve this.

find $z in mathbb C$ for this function $$| z + 3i | = 3|z|$$

thanks in advance.

functions complex-numbers

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edited Jan 2 at 18:05

Andrei

11.6k21026

asked Jan 2 at 18:03

Donni Maulana SipaDonni Maulana Sipa

1

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closed as off-topic by José Carlos Santos, Davide Giraudo, A. Pongrácz, Holo, Chris Godsil Jan 3 at 2:11

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Davide Giraudo, A. Pongrácz, Holo, Chris Godsil

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Geometrically, the left hand side represent circles centred at $3i$ while the right hand side represent circles centred at $0.$ To "solve" the "equation" simply write $z = a + ib$ and expand; you will likely find that $a$ or $b$ is function of the other.
  $endgroup$
  – Will M.
  Jan 2 at 18:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @WillM. LHS represents a circle centered at $-3i$...
  $endgroup$
  – Doug M
  Jan 2 at 18:20
  

  
  
  
  

add a comment | 

-1

$begingroup$

Can anybody explain how to solve this.

find $z in mathbb C$ for this function $$| z + 3i | = 3|z|$$

thanks in advance.

functions complex-numbers

share|cite|improve this question

edited Jan 2 at 18:05

Andrei

11.6k21026

asked Jan 2 at 18:03

Donni Maulana SipaDonni Maulana Sipa

1

$endgroup$

closed as off-topic by José Carlos Santos, Davide Giraudo, A. Pongrácz, Holo, Chris Godsil Jan 3 at 2:11

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Davide Giraudo, A. Pongrácz, Holo, Chris Godsil

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Geometrically, the left hand side represent circles centred at $3i$ while the right hand side represent circles centred at $0.$ To "solve" the "equation" simply write $z = a + ib$ and expand; you will likely find that $a$ or $b$ is function of the other.
  $endgroup$
  – Will M.
  Jan 2 at 18:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @WillM. LHS represents a circle centered at $-3i$...
  $endgroup$
  – Doug M
  Jan 2 at 18:20
  

  
  
  
  

add a comment | 

-1

-1

-1

$begingroup$

Can anybody explain how to solve this.

find $z in mathbb C$ for this function $$| z + 3i | = 3|z|$$

thanks in advance.

functions complex-numbers

share|cite|improve this question

edited Jan 2 at 18:05

Andrei

11.6k21026

asked Jan 2 at 18:03

Donni Maulana SipaDonni Maulana Sipa

1

$endgroup$

Can anybody explain how to solve this.

find $z in mathbb C$ for this function $$| z + 3i | = 3|z|$$

thanks in advance.

functions complex-numbers

functions complex-numbers

share|cite|improve this question

edited Jan 2 at 18:05

Andrei

11.6k21026

asked Jan 2 at 18:03

Donni Maulana SipaDonni Maulana Sipa

1

share|cite|improve this question

edited Jan 2 at 18:05

Andrei

11.6k21026

asked Jan 2 at 18:03

Donni Maulana SipaDonni Maulana Sipa

1

share|cite|improve this question

share|cite|improve this question

edited Jan 2 at 18:05

Andrei

11.6k21026

edited Jan 2 at 18:05

Andrei

11.6k21026

edited Jan 2 at 18:05

Andrei

11.6k21026

11.6k21026

asked Jan 2 at 18:03

Donni Maulana SipaDonni Maulana Sipa

1

asked Jan 2 at 18:03

Donni Maulana SipaDonni Maulana Sipa

1

asked Jan 2 at 18:03

Donni Maulana SipaDonni Maulana Sipa

1

1

closed as off-topic by José Carlos Santos, Davide Giraudo, A. Pongrácz, Holo, Chris Godsil Jan 3 at 2:11

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Davide Giraudo, A. Pongrácz, Holo, Chris Godsil

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by José Carlos Santos, Davide Giraudo, A. Pongrácz, Holo, Chris Godsil Jan 3 at 2:11

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Davide Giraudo, A. Pongrácz, Holo, Chris Godsil

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Geometrically, the left hand side represent circles centred at $3i$ while the right hand side represent circles centred at $0.$ To "solve" the "equation" simply write $z = a + ib$ and expand; you will likely find that $a$ or $b$ is function of the other.
  $endgroup$
  – Will M.
  Jan 2 at 18:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @WillM. LHS represents a circle centered at $-3i$...
  $endgroup$
  – Doug M
  Jan 2 at 18:20
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Geometrically, the left hand side represent circles centred at $3i$ while the right hand side represent circles centred at $0.$ To "solve" the "equation" simply write $z = a + ib$ and expand; you will likely find that $a$ or $b$ is function of the other.
  $endgroup$
  – Will M.
  Jan 2 at 18:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @WillM. LHS represents a circle centered at $-3i$...
  $endgroup$
  – Doug M
  Jan 2 at 18:20
  

  
  
  
  

$begingroup$
Geometrically, the left hand side represent circles centred at $3i$ while the right hand side represent circles centred at $0.$ To "solve" the "equation" simply write $z = a + ib$ and expand; you will likely find that $a$ or $b$ is function of the other.
$endgroup$
– Will M.
Jan 2 at 18:06

$begingroup$
Geometrically, the left hand side represent circles centred at $3i$ while the right hand side represent circles centred at $0.$ To "solve" the "equation" simply write $z = a + ib$ and expand; you will likely find that $a$ or $b$ is function of the other.
$endgroup$
– Will M.
Jan 2 at 18:06

$begingroup$
@WillM. LHS represents a circle centered at $-3i$...
$endgroup$
– Doug M
Jan 2 at 18:20

$begingroup$
@WillM. LHS represents a circle centered at $-3i$...
$endgroup$
– Doug M
Jan 2 at 18:20

add a comment | 

2 Answers
2

active

oldest

votes

1

$begingroup$

Without breaking down to $x+iy$:

Square both sides and you have
$$(z+3i)(bar{z}-3i)=9zbar{z}$$
$$zbar{z}+3ibar{z}-3iz+9=9zbar{z}$$ Which, after algebraic rearrangement, is equivalent to: $$zbar{z}-frac38ibar{z}+frac38iz-frac98=0$$
Now "complete the square" on this: $$zbar{z}-frac38ibar{z}+frac38iz+frac9{64}-frac{9}{64}-frac98=0$$
$$zbar{z}-frac38ibar{z}+frac38iz+frac9{64}=frac{81}{64}$$
$$left(z-frac38iright)left(bar{z}+frac38iright)=frac{81}{64}$$
And square root:
$$leftlvert z-frac38irightrvert=frac98$$
So it is a circle of radius $frac98$ centered at $frac38i$.

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answered Jan 2 at 18:38

alex.jordanalex.jordan

38.9k560120

$endgroup$

add a comment | 

0

$begingroup$

Hint: With $$z=x+iy$$ we get
$$sqrt{x^2+(y+3)^2}=3sqrt{x^2+y^2}$$ Can you solve this?

share|cite|improve this answer

answered Jan 2 at 18:08

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

73.7k42864

$endgroup$

add a comment | 

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

1

$begingroup$

Without breaking down to $x+iy$:

Square both sides and you have
$$(z+3i)(bar{z}-3i)=9zbar{z}$$
$$zbar{z}+3ibar{z}-3iz+9=9zbar{z}$$ Which, after algebraic rearrangement, is equivalent to: $$zbar{z}-frac38ibar{z}+frac38iz-frac98=0$$
Now "complete the square" on this: $$zbar{z}-frac38ibar{z}+frac38iz+frac9{64}-frac{9}{64}-frac98=0$$
$$zbar{z}-frac38ibar{z}+frac38iz+frac9{64}=frac{81}{64}$$
$$left(z-frac38iright)left(bar{z}+frac38iright)=frac{81}{64}$$
And square root:
$$leftlvert z-frac38irightrvert=frac98$$
So it is a circle of radius $frac98$ centered at $frac38i$.

share|cite|improve this answer

answered Jan 2 at 18:38

alex.jordanalex.jordan

38.9k560120

$endgroup$

add a comment | 

1

$begingroup$

Without breaking down to $x+iy$:

Square both sides and you have
$$(z+3i)(bar{z}-3i)=9zbar{z}$$
$$zbar{z}+3ibar{z}-3iz+9=9zbar{z}$$ Which, after algebraic rearrangement, is equivalent to: $$zbar{z}-frac38ibar{z}+frac38iz-frac98=0$$
Now "complete the square" on this: $$zbar{z}-frac38ibar{z}+frac38iz+frac9{64}-frac{9}{64}-frac98=0$$
$$zbar{z}-frac38ibar{z}+frac38iz+frac9{64}=frac{81}{64}$$
$$left(z-frac38iright)left(bar{z}+frac38iright)=frac{81}{64}$$
And square root:
$$leftlvert z-frac38irightrvert=frac98$$
So it is a circle of radius $frac98$ centered at $frac38i$.

share|cite|improve this answer

answered Jan 2 at 18:38

alex.jordanalex.jordan

38.9k560120

$endgroup$

add a comment | 

1

1

1

$begingroup$

Without breaking down to $x+iy$:

Square both sides and you have
$$(z+3i)(bar{z}-3i)=9zbar{z}$$
$$zbar{z}+3ibar{z}-3iz+9=9zbar{z}$$ Which, after algebraic rearrangement, is equivalent to: $$zbar{z}-frac38ibar{z}+frac38iz-frac98=0$$
Now "complete the square" on this: $$zbar{z}-frac38ibar{z}+frac38iz+frac9{64}-frac{9}{64}-frac98=0$$
$$zbar{z}-frac38ibar{z}+frac38iz+frac9{64}=frac{81}{64}$$
$$left(z-frac38iright)left(bar{z}+frac38iright)=frac{81}{64}$$
And square root:
$$leftlvert z-frac38irightrvert=frac98$$
So it is a circle of radius $frac98$ centered at $frac38i$.

share|cite|improve this answer

answered Jan 2 at 18:38

alex.jordanalex.jordan

38.9k560120

$endgroup$

Without breaking down to $x+iy$:

Square both sides and you have
$$(z+3i)(bar{z}-3i)=9zbar{z}$$
$$zbar{z}+3ibar{z}-3iz+9=9zbar{z}$$ Which, after algebraic rearrangement, is equivalent to: $$zbar{z}-frac38ibar{z}+frac38iz-frac98=0$$
Now "complete the square" on this: $$zbar{z}-frac38ibar{z}+frac38iz+frac9{64}-frac{9}{64}-frac98=0$$
$$zbar{z}-frac38ibar{z}+frac38iz+frac9{64}=frac{81}{64}$$
$$left(z-frac38iright)left(bar{z}+frac38iright)=frac{81}{64}$$
And square root:
$$leftlvert z-frac38irightrvert=frac98$$
So it is a circle of radius $frac98$ centered at $frac38i$.

share|cite|improve this answer

answered Jan 2 at 18:38

alex.jordanalex.jordan

38.9k560120

share|cite|improve this answer

share|cite|improve this answer

answered Jan 2 at 18:38

alex.jordanalex.jordan

38.9k560120

answered Jan 2 at 18:38

alex.jordanalex.jordan

38.9k560120

answered Jan 2 at 18:38

alex.jordanalex.jordan

38.9k560120

38.9k560120

add a comment | 

add a comment | 

0

$begingroup$

Hint: With $$z=x+iy$$ we get
$$sqrt{x^2+(y+3)^2}=3sqrt{x^2+y^2}$$ Can you solve this?

share|cite|improve this answer

answered Jan 2 at 18:08

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

73.7k42864

$endgroup$

add a comment | 

0

$begingroup$

Hint: With $$z=x+iy$$ we get
$$sqrt{x^2+(y+3)^2}=3sqrt{x^2+y^2}$$ Can you solve this?

share|cite|improve this answer

answered Jan 2 at 18:08

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

73.7k42864

$endgroup$

add a comment | 

0

0

0

$begingroup$

Hint: With $$z=x+iy$$ we get
$$sqrt{x^2+(y+3)^2}=3sqrt{x^2+y^2}$$ Can you solve this?

share|cite|improve this answer

answered Jan 2 at 18:08

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

73.7k42864

$endgroup$

Hint: With $$z=x+iy$$ we get
$$sqrt{x^2+(y+3)^2}=3sqrt{x^2+y^2}$$ Can you solve this?

share|cite|improve this answer

answered Jan 2 at 18:08

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

73.7k42864

share|cite|improve this answer

share|cite|improve this answer

answered Jan 2 at 18:08

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

73.7k42864

answered Jan 2 at 18:08

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

73.7k42864

answered Jan 2 at 18:08

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

73.7k42864

73.7k42864

add a comment | 

add a comment | 

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