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For all complex numbers `z_(1) and z_(2)`, prove that `|z_(1)+z_(2)|^(2)+|z_(1)-z_(2)|^(2)=2(|z_(1)|^(2)+|z_(2)|^(2))`. |
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Answer» We have `|z_(1) + z_(2)|^(2)+|z_(1)-z_(2)|^(2)=2(|z_(1)|^(2)+|z_(2)|^(2))`. `=(z_(1)+z_(2))(bar(z)_(1)+bar(z)_(2))" "[because bar((z_(1)+z_(2)))=bar(z)_(1)+bar(z)_(2)]` `=z_(1)bar(z)_(1)+z_(2)bar(z)_(2)+z_(1)bar(z)_(2)+z_(2)bar(z)_(1)` `=|z_(1)|^(2)+|z_(2)|^(2)+z_(1) bar(z)_(2) + z_(2) bar(z)_(1)." "...(i)` `|z_(1)-z_(2)|^(2)=(z_(1)-z_(2))(bar(z_(1)-z_(2)))=(z_(1)-z_(2))(bar(z)_(1)-bar(z)_(2))` `=z_(1) bar(z)_(1)+z_(2)bar(z)_(2)-z_(1)bar(z)_(2)-z_(2)bar(z)_(1)` `=|z_(1)|^(2)+|z_(2)|^(2)-z_(1)bar(z)_(2)-z_(2)bar(z)_(1)." "...(ii)` On adding the corresponding sides of (i) and (ii), we get `|z_(1)+z_(2)|^(2)+|z_(1)-z_(2)|^(2)=2(|z_(1)|^(2)+|z_(2)|^(2))`. |
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