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Question regarding complex conjugates.

0

$begingroup$

$$ overline{4z+4+4i} = z+13−14i$$

The conjugate becomes $4z+4-4i$, but whenever I then try to solve for $z$, I end up getting the wrong answer when checking the answer.

complex-numbers

share|cite|improve this question

edited Mar 26 at 0:06

Peter Phipps

2,15622034

asked Mar 25 at 23:48

JUJUJUJU

11

$endgroup$

• 
  
  
  

  
  6
  

  
  
  
  
  
  

  
  $begingroup$
  If $overline{z}$ is the conjugate of $z$, your hypothesis is $$overline{4z+4+4i} = z + 13 - 14i$$ right? The left hand side is $4 overline{z} + 4 - 4i$, not $4z+4-4i$.
  $endgroup$
  – D_S
  Mar 25 at 23:54
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @D_S Consider converting your comment into an answer so that this question might be removed form the unanswered tab.
  $endgroup$
  – Brian
  Mar 25 at 23:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @D_S Oh I see, but how would I then solve for z?
  $endgroup$
  – JUJU
  Mar 26 at 0:02
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You could write $z$ as $a+bi$ and solve for the real numbers $a$ and $b$
  $endgroup$
  – D_S
  Mar 26 at 0:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @D_S Alright I got the answer, thank you my friend :)
  $endgroup$
  – JUJU
  Mar 26 at 0:29
  

  
  
  
  

add a comment | 

0

$begingroup$

$$ overline{4z+4+4i} = z+13−14i$$

The conjugate becomes $4z+4-4i$, but whenever I then try to solve for $z$, I end up getting the wrong answer when checking the answer.

complex-numbers

share|cite|improve this question

edited Mar 26 at 0:06

Peter Phipps

2,15622034

asked Mar 25 at 23:48

JUJUJUJU

11

$endgroup$

• 
  
  
  

  
  6
  

  
  
  
  
  
  

  
  $begingroup$
  If $overline{z}$ is the conjugate of $z$, your hypothesis is $$overline{4z+4+4i} = z + 13 - 14i$$ right? The left hand side is $4 overline{z} + 4 - 4i$, not $4z+4-4i$.
  $endgroup$
  – D_S
  Mar 25 at 23:54
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @D_S Consider converting your comment into an answer so that this question might be removed form the unanswered tab.
  $endgroup$
  – Brian
  Mar 25 at 23:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @D_S Oh I see, but how would I then solve for z?
  $endgroup$
  – JUJU
  Mar 26 at 0:02
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You could write $z$ as $a+bi$ and solve for the real numbers $a$ and $b$
  $endgroup$
  – D_S
  Mar 26 at 0:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @D_S Alright I got the answer, thank you my friend :)
  $endgroup$
  – JUJU
  Mar 26 at 0:29
  

  
  
  
  

add a comment | 

$begingroup$

$$ overline{4z+4+4i} = z+13−14i$$

The conjugate becomes $4z+4-4i$, but whenever I then try to solve for $z$, I end up getting the wrong answer when checking the answer.

complex-numbers

share|cite|improve this question

edited Mar 26 at 0:06

Peter Phipps

2,15622034

asked Mar 25 at 23:48

JUJUJUJU

11

$endgroup$

share|cite|improve this question

edited Mar 26 at 0:06

Peter Phipps

2,15622034

asked Mar 25 at 23:48

JUJUJUJU

11

share|cite|improve this question

edited Mar 26 at 0:06

Peter Phipps

2,15622034

asked Mar 25 at 23:48

JUJUJUJU

11

edited Mar 26 at 0:06

Peter Phipps

2,15622034

edited Mar 26 at 0:06

Peter Phipps

2,15622034

asked Mar 25 at 23:48

JUJUJUJU

11

asked Mar 25 at 23:48

JUJUJUJU

11