Complex equation z+z*=0 giving a point or a line

[loonychune]

Just want a check of this please:

We have a complex equation of the form az+bz*+c=0
where a, b and z are complex #s, c is real...
If you take the real and imaginary parts of such an equation you obtain two linear equations in x and y, whose solutions of each gives rise to a line (L_1 and L_2 respectively)...
The set then, of solutions, is L_1 unison L_2

Now, the set of solutions of the complex equation is either empty, a point, or a line......the book gives these 3 examples as each case:
z + z* = i

z+2z* = 0

z+z* = 0


I don't understand how z+z*=0 is a line... for we in fact have
RE(Z+Z*)=2x=0
IM(Z+Z*)= 0y = 0
which then gives rise to a point solution does it not??

if it was z+z* = c say, then i could see that having a line of solutions but as it is, i reckon the book has made an error... is this the case?

(perhaps i have confused the issue and if that is the case maybe then you coudl point out how my thinking is wrong)

THANKYOU :)

 

[Vagrant]

x=0 and y=0 refer not to points but to the y- and x- axis respectively, which you can see are lines, it's like saying for every value of y, x=0 or any constant so it's not a point but infact a line. I hope this explanation was coherent.

 

[uart]

Just want a check of this please:
I don't understand how z+z*=0 is a line... for we in fact have
RE(Z+Z*)=2x=0
IM(Z+Z*)= 0y = 0
which then gives rise to a point solution does it not??
:)

It does not.

Yes 2x = 0 only has the solution that x=0. However 0y = 0 does not just have the solution that y=0, rather it has the solution y=anything. This makes the solutuion the entire imaginary axis (x=0), does that make sense.

 

[loonychune]

Yeah it makes sense, thanks a lot the both of you...
Physicsforums again proves a real gem..