use z=a+bi and w=c+di to show that the 
conjugate of z*w =
the conjugate of z * the conjugate of w

First let's find z*w

z*w = (a+bi)(c+di) = ac + adi + bci + bdi² = ac + (ad+bd)i + bd(-1) =

ac + (ad+bd)i - bd = ac - bd + (ad+bd)i = (ac-bd) + (ad+bd)i

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Now let's find the conjugate of z*w:

"Find the conjugate" means "leave the real part but change the 
sign of the imaginary part".

So the conjugate of z*w is (ac-bd) - (ad+bd)i

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Now let's find the conjugates of z and w.

Again, "find the conjugate" means "leave the real part but change the 
sign of the imaginary part".

So the conjugate of z, which equals a+bi, is a-bi.
And the conjugate of w, which equals c+di, is c-di. 

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Now let's find (the conjugate of z)*(the conjugate of w)

(a-bi)(c-di) = ac - adi - bci + bdi² = ac - (ad+bd)i + bd(-1) =

ac - (ad+bd)i - bd = ac - bd - (ad+bd)i = (ac-bd) - (ad+bd)i

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So they are both equal since both the red expressions above are the 
same, namely (ac-bd) - (ad+bd)i. 


Edwin