The discussion revolves around the reflection of a line represented in the complex plane, specifically the equation ##\bar{a}z + a\bar{z} = 0##, and how to determine its reflection in the real axis. Participants are exploring concepts related to complex geometry and the properties of complex conjugates.
Some participants have offered insights regarding the reflection process and the nature of complex conjugates, while others express confusion about the relationship between the original line and its reflection. The conversation is ongoing, with various interpretations being explored.
There is mention of the participants' varying levels of familiarity with complex geometry, which may influence their understanding of the problem. Additionally, the discussion includes questions about the general rules for reflecting points and lines in the complex plane.
What is the reflection of z in the real axis?erisedk said: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.
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?erisedk said:Its conjugate.
You need to conjugate those complex numbers z which are on that line instead of conjugating the equation.erisedk said: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##.erisedk said: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.