- #1
rajeshmarndi
- 319
- 0
- Homework Statement
- w+w* = (s+t*)z + (t+s*)z* + r+r* = 0, is a straight line. Then couldn't find how does
w-w* = (s-t*)z + (t-s*)z* + r-r* = 0, is also a straight line.
- Relevant Equations
- w = sz+tz*+r=0
sz+tz*+r=0=say w
so w* = s*z* + t*z + r*=0
Now ,
w+w* = (s+t*)z + (t+s*)z* + r+r* = 0
= p*z + pz* + k = 0...eq(1) ( k is a constant or twice real part of w)
which is in complex straight line equation form i.e ab* + a*b + c = 0 ( a,b are complex number and c a real number.
Now, again,
w-w* = (s-t*)z + (t-s*)z* + r-r* = 0
I couldn't understand, in the solution, how this is also termed as a complex straight line like eq(1).
Since when this is worked out, it comes to be as,
q*z - qz* + id = 0 ( since r-r* will give imaginary number)
This is not in the form of a complex straight line equation.
Thanks.
so w* = s*z* + t*z + r*=0
Now ,
w+w* = (s+t*)z + (t+s*)z* + r+r* = 0
= p*z + pz* + k = 0...eq(1) ( k is a constant or twice real part of w)
which is in complex straight line equation form i.e ab* + a*b + c = 0 ( a,b are complex number and c a real number.
Now, again,
w-w* = (s-t*)z + (t-s*)z* + r-r* = 0
I couldn't understand, in the solution, how this is also termed as a complex straight line like eq(1).
Since when this is worked out, it comes to be as,
q*z - qz* + id = 0 ( since r-r* will give imaginary number)
This is not in the form of a complex straight line equation.
Thanks.