anonymous
  • anonymous
How do I prove this parallelogram equality? |z1+z2|^2 +|z1-z2|^2 = 2(|z1|^2+|z2|^2)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
Is z a complex number ?
anonymous
  • anonymous
Yes
anonymous
  • anonymous
I just don't know what to do with the absolute if I regard it as positive only I get:

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anonymous
  • anonymous
\[z_1^2+2z_1z_2+z_2^2+z_1^2-2z_1z_2+z_2^2 \]
anonymous
  • anonymous
\[2(z_1^2+z_2^2)\]
anonymous
  • anonymous
but then I don't have the absolute signs
anonymous
  • anonymous
First, if z is a complex number we have : \[|z|^2=z\bar z\] So : \[RHS=|z_1+z_2|^2+|z_1-z_2|^2\\ ~~~~~~=(z_1+z_2)(\overline{z_1+z_2})+(z_1-z_2)(\overline{z_1-z_2})\\ ~~~~~~=z_1\overline {z_1}+z_1\overline {z_2}+z_2\overline {z_1}+z_2\overline {z_2}+z_1\overline {z_1}-z_1\overline {z_2}-z_2\overline {z_1}+z_2\overline {z_2}\\ ~~~~~=2z_1\overline {z_1}+2z_2\overline {z_2}\\ ~~~~~~=2(|z_1|^2+|z_2|^2)\\ ~~~~=LHS \]
anonymous
  • anonymous
Oh right I forgot about that! Thank you I get it now

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