Question

If z1 and z2 are complex numbers, prove that:

(z1 + (conjugate z1)(z2+ (conjugate z2) = z1z2 + (conjugate z1z2) + z1(conjugate z2) + (conjugate(z1(conjugate z2)

I'm not sure if I should be writing this in a different format. I don't even think you'll be able to help me with this one.

Answer 1

\[(z _{1}+(conjugatez _{1})) (z _{2}+(conjugatez _{2}) = z _{1}z _{2}+(conjugate z _{1}z _{2}) + z _{1}(conjugatez _{2}+(conjugatez _{1}(conjugatez _{2}))\]

Answer 2

Just foil it out..
\[\large (z_1 + \widetilde{z_1})(z_2 + \widetilde{z_2}) \]\[\large = z_1z_2 + \widetilde{z_1}z_2 + z_1\widetilde{z_2} + \widetilde{z_1}\widetilde{z_2}\]

Answer 3

All goes to 0?

Answer 4

So once it has been foiled out - can we let:

z1 = a+bi
z1(conjugate)= a-bi
z2 = c+di
z2 (conjugate) = c-di?

then do some algebra and manipulation of the two to make them equal?

Answer 5

yep

Answer 6

Yep, that'll work.

Answer 7

good