- #1
Kahsi
- 41
- 0
Hi
I just started to look at complex numbers.
Prove the ``Parallellogram law''
http://www.sosmath.com/complex/number/complexplane/img4.gif
This is how I solved it:
[tex]z=a+bi[/tex]
[tex]w=c+di[/tex]
[tex]|z+w|^2=\sqrt{(a+c)^2+(b+d)^2}=a^2+2ac+c^2+b^2+2bd+d^2[/tex]
then we have
[tex]|z-w|^2=\sqrt{(a-c)^2+(b-d)^2}=a^2-2ac+c^2+b^2-2bd+d^2[/tex]
[tex]2(|z|^2+|w|^2)=2((a^2+b^2)+(c^2+d^2)) = 2a^2+2b^2+2c^2+2d^2[/tex]
[tex]a^2+2ac+c^2+b^2+2bd+d^2+a^2-2ac+c^2+b^2-2bd+d^2=2a^2+2b^2+2c^2+2d^2[/tex]
My question is:
Was my calculation OK or was it a misscalculation (a lucky one which prooved the formula)
I just started to look at complex numbers.
Prove the ``Parallellogram law''
http://www.sosmath.com/complex/number/complexplane/img4.gif
This is how I solved it:
[tex]z=a+bi[/tex]
[tex]w=c+di[/tex]
[tex]|z+w|^2=\sqrt{(a+c)^2+(b+d)^2}=a^2+2ac+c^2+b^2+2bd+d^2[/tex]
then we have
[tex]|z-w|^2=\sqrt{(a-c)^2+(b-d)^2}=a^2-2ac+c^2+b^2-2bd+d^2[/tex]
[tex]2(|z|^2+|w|^2)=2((a^2+b^2)+(c^2+d^2)) = 2a^2+2b^2+2c^2+2d^2[/tex]
[tex]a^2+2ac+c^2+b^2+2bd+d^2+a^2-2ac+c^2+b^2-2bd+d^2=2a^2+2b^2+2c^2+2d^2[/tex]
My question is:
Was my calculation OK or was it a misscalculation (a lucky one which prooved the formula)
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