- anonymous

How to find prove that the complex conjugate of Z(with a line over it)^2 is not the same as Z^2(with a line over it)?

- chestercat

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- asnaseer

If you write the complex number z as:\[z=a+ib\]Then do you know what \(\bar{z}\) equals in terms a and b?

- anonymous

i know that z=a+bi, and z(line over)=a-bi, but when i squared the first one i got (a^2-b^2)-2abi and i got the same thing when i solve (a+bi)^2(all with a line over it)

- asnaseer

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- asnaseer

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

- anonymous

yeah i got that part... that's the z^2 part right? because after that the problem wants z^2(with a line over z^2)

- asnaseer

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

- anonymous

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:

##### 1 Attachment

- asnaseer

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

- anonymous

yes

- asnaseer

ok, then steps are:
1. set z = a + bi
2. find \(z^2\) using this
3. take the complex conjugate of this expression
4. set \(\bar{z}\)=a=bi
5. find \(\bar{z}^2\) using this
6. compare the two results

- anonymous

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

- asnaseer

yes

- anonymous

then isn't it the same as the other result? i got a^2-b^2-2abi as the complex conjugate squared [z^2(with a line over it)] as well

- asnaseer

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

- anonymous

not 100%... it was from a test that we started today and are finishing tomorrow. I remember the z terms correctly but im not exactly sure what it was asking... i guess maybe the question was to prove that they ARe the same?

- asnaseer

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

- hartnn

question is correct, they are not same

- anonymous

hartnn- how are they not the same?

- asnaseer

@hartnn - the /revised/ question is:
\[(\bar{z})^2\ne\bar{(z^2)}\]

- hartnn

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

- hartnn

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

- asnaseer

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

- hartnn

oh dear, sorry, i worked out right side incorrectly

- asnaseer

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

- anonymous

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

- asnaseer

yw :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.