I found this question on my Algebra book but i couldn't answer it
Can you please explain step by step
Prove the following identities ..explain its geometric meaning
$|1+z_1\bar z_2|^2 + |z_1-z_2|^2 = (1+|z_1|^2)(1+|z_2|^2)$
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Ahmed emad
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4You're missing an exponent $2$, $\lvert 1 + z_1\overline{z}_2\rvert^2 + \lvert z_1 - z_2\rvert^2 = (1 + \lvert z_1\rvert^2)(1 + \lvert z_2\rvert^2)$. Then it is a simple expansion of $\lvert w\rvert^2 = w\overline{w}$. – Daniel Fischer Sep 05 '13 at 09:42
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@danielfisher but I am still confused about the left side never faced this before can you please simplify the left side step by step please – Sep 05 '13 at 10:37
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CHECK MY ANSWER – Shobhit Sep 05 '13 at 10:40
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Thank you very much but I can't even give you reputation I am too new but I am really glad for your help – Sep 05 '13 at 10:52
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@AhmedEmad: It appears that you have created two accounts. If you wish to have them merged, please follow the instructions listed here. – user642796 Sep 05 '13 at 10:58
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@AhmedEmad but you can accept the answer if you like by clicking the tick below the voteup option. – Shobhit Sep 05 '13 at 14:52
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Using $|z|^2=z\bar z$
we have L.H.S $$|1+z_1\bar z_2|^2+|z_1-z_2|^2=(1+z_1\bar z_2)(1+\bar z_1z_2)+(z_1-z_2)(\bar z_1-\bar z_2)$$
or $$1+z_1\bar z_2+\bar z_1z_2+|z_1|^2|z_2|^2+|z_1|^2-z_1\bar z_2-\bar z_1z_2+|z_2|^2$$
or $$1+|z_1|^2|z_2|^2+|z_1|^2+|z_2|^2$$
which is equal to R.H.S $$(1+|z_1|^2)(1+|z_2|^2)$$
Shobhit
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1@BabakS. the photo in the discription section of your profile what is it. – Shobhit Sep 11 '13 at 06:36