iff proof of complex perpendicular lines *EDITED*

[Idealistic]

Let a, b, c, and d be complex numbers. Show that the two lines Re(az + b) and Re(cz + d) are perpendicular if and only if Re(ac*) = 0.

I'm pretty sure I should relate this back to the dot prduct formula with:

cos(theta) = Re(ac*)/mag(a)mag(c)

theta = 0 if Re(ac*) = 0

but to prove it do I use the Law of cosines? if so, do I break the lines up into a = a1 + ia2 and c = c1 + ic2 or just treat them as vectors as a starting point for the proof?

Re(az + b) = Re(a1 + ia2(x + iy) + b) = a1x - a2y + b
Re(cz + d) = Re(c1 + ic2)(s + it) + d) = c1s - c2t + d

EDIT. Sorry, its just supposed to be:

Let a, b, c, and d be complex numbers. Show that the two lines Re(az + b) and Re(cz + d) are perpendicular if and only if Re(ac*) = 0.

c* is just the conjugate of c.

 

[galactus]

Re: iff proof of complex perpendicular lines

What does ac* and mag mean?.

Is the problem supposed to read?:

"Prove the lines \(\displaystyle Re(az+b)=0\) and \(\displaystyle Re(cz+d)=0\) are perpendicular iff \(\displaystyle Re(a\overline{c})=0\)".