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\[(a+ib)^2=(a+ib)(a+ib)=a^2+2abi+i^2b^2=a^2+2abi-b^2=a^2-b^2+2abi\]

That last term is:\[=a^2-b^2+2abi\]

yes, so now find \(\bar{z}^2\) using the same technique, i.e.:\[\bar{z}^2=(a-ib)^2=(a-ib)(a-ib)=?\]

i did that and i got a^2-b^2-2abi... so this is the first part of the problem that i have to compare to z^2(with a line over the entire thing including the square... look at the attatched picture, im not sure if I'm typing it right, but this is the question at the top there:

So is this the question - prove that:\[(\bar{z})^2\ne\bar{(z^2)}\]

yes

for step 3, would the complex conjugate of z^2 be a^2-b^2-2abi?

yes

yes - so do I. Are you 100% sure you have the correct question?

yes - that is what I believe the question would have stated

question is correct, they are not same

hartnn- how are they not the same?

{z(bar) }^2 = left side = (a-bi)(a-bi)

right side =conjugate of (a+bi)(a+bi)

correct - and they both work out to be the same value

oh dear, sorry, i worked out right side incorrectly

so I believe the question should have been:
Prove that \((\bar{z})^2=\bar{(z^2)}\)

yes i beleive that was the question. okay, thanks for all your help!

yw :)