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Complex numbers and reflection

  • Thread starter erisedk
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  • #1
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Homework Statement


Reflection of the line ##\bar{a}z + a\bar{z} = 0## in the real axis is

Homework Equations




The Attempt at a Solution


I know that a line in the complex plane is represented as ##\bar{a}z + a\bar{z} + b= 0## and that its slope ##μ = \dfrac{-a}{\bar{a}}##. I'm not sure how to do this problem. I'm also not very good with complex geometry so please help.
 

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  • #2
haruspex
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Homework Statement


Reflection of the line ##\bar{a}z + a\bar{z} = 0## in the real axis is

Homework Equations




The Attempt at a Solution


I know that a line in the complex plane is represented as ##\bar{a}z + a\bar{z} + b= 0## and that its slope ##μ = \dfrac{-a}{\bar{a}}##. I'm not sure how to do this problem. I'm also not very good with complex geometry so please help.
What is the reflection of z in the real axis?
 
  • #3
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Its conjugate.
 
  • #4
haruspex
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Its conjugate.
Right. So if you have two points ##z## and ##\bar w##, how would you write their reflections notationally? What is the general rule you see here?
 
  • #5
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As ##\bar{z}## and ##w##? That I need to take the conjugate of the equation of the line? But that gives me back the original line. However, the funny thing is if I take the conjugate of only ##z##, I get the desired answer, i.e. ##\bar{a}\bar{z} + az = 0##. I can't really explain that though.
 
  • #6
ehild
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However, the funny thing is if I take the conjugate of only ##z##, I get the desired answer, i.e. ##\bar{a}\bar{z} + az = 0##. I can't really explain that though.
You need to conjugate those complex numbers z which are on that line instead of conjugating the equation.
You have a line in the x,y plane. What is the equation of that line?
What line do you get when you reflect the original line on the x axis?
How can you write the complex numbers z1 with their real and imaginary parts which are on the original line ? What are those complex numbers z2 which are on the reflected line?
 
  • #7
haruspex
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As ##\bar{z}## and ##w##? That I need to take the conjugate of the equation of the line? But that gives me back the original line. However, the funny thing is if I take the conjugate of only ##z##, I get the desired answer, i.e. ##\bar{a}\bar{z} + az = 0##. I can't really explain that though.
If you have an equation for z that specifies a point, z=a say, then how do you write the equation for the reflection of that point? You would write ##\bar z=a## or ##z=\bar a##, not ##\bar z=\bar a##.
 
  • #8
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Ok, got it thank you!
 

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