Question

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)?

Answer 1

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

Answer 2

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)

Answer 3

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

Answer 4

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

Answer 5

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)

Answer 6

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

Answer 7

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:

Answer 8

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

Answer 9

yes

Answer 10

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

Answer 11

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

Answer 12

yes

Answer 13

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

Answer 14

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

Answer 15

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?

Answer 16

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

Answer 17

question is correct, they are not same

Answer 18

hartnn- how are they not the same?

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

oh dear, sorry, i worked out right side incorrectly

Answer 24

so I believe the question should have been:

Prove that \((\bar{z})^2=\bar{(z^2)}\)

Answer 25

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

Answer 26

yw :)