## Studentc14 Group Title 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)? one year ago one year ago

1. asnaseer Group Title

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

2. Studentc14 Group Title

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)

3. asnaseer Group Title

$(a+ib)^2=(a+ib)(a+ib)=a^2+2abi+i^2b^2=a^2+2abi-b^2=a^2-b^2+2abi$

4. asnaseer Group Title

That last term is:$=a^2-b^2+2abi$

5. Studentc14 Group Title

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)

6. asnaseer Group Title

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

7. Studentc14 Group Title

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:

8. asnaseer Group Title

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

9. Studentc14 Group Title

yes

10. asnaseer Group Title

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

11. Studentc14 Group Title

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

12. asnaseer Group Title

yes

13. Studentc14 Group Title

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

14. asnaseer Group Title

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

15. Studentc14 Group Title

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?

16. asnaseer Group Title

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

17. hartnn Group Title

question is correct, they are not same

18. Studentc14 Group Title

hartnn- how are they not the same?

19. asnaseer Group Title

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

20. hartnn Group Title

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

21. hartnn Group Title

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

22. asnaseer Group Title

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

23. hartnn Group Title

oh dear, sorry, i worked out right side incorrectly

24. asnaseer Group Title

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

25. Studentc14 Group Title

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

26. asnaseer Group Title

yw :)