# Complex numbers + proof

1. Jan 28, 2005

### Kahsi

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:

$$z=a+bi$$
$$w=c+di$$

$$|z+w|^2=\sqrt{(a+c)^2+(b+d)^2}=a^2+2ac+c^2+b^2+2bd+d^2$$

then we have

$$|z-w|^2=\sqrt{(a-c)^2+(b-d)^2}=a^2-2ac+c^2+b^2-2bd+d^2$$

$$2(|z|^2+|w|^2)=2((a^2+b^2)+(c^2+d^2)) = 2a^2+2b^2+2c^2+2d^2$$

$$a^2+2ac+c^2+b^2+2bd+d^2+a^2-2ac+c^2+b^2-2bd+d^2=2a^2+2b^2+2c^2+2d^2$$

My question is:

Was my calculation OK or was it a misscalculation (a lucky one which prooved the formula)

Last edited: Jan 28, 2005
2. Jan 28, 2005

### dextercioby

Those 2 radicals should be squared.Or should be absent altogether.

The rest is okay.

Daniel.

3. Jan 29, 2005

### Kahsi

Oh....That was a typo

Thank you dextercioby

4. Jan 29, 2005

### mathwonk

vector methods are nice too. i.e. |z+w|^2 = (z+w).(z+w), (dot product),

and |z-w|^2 = (z-w).(z-w).

Expanding and adding, the cross terms cancel, leaving

z.z + w.w + z.z + w.w = 2(|z|^2 + |w|^2).

you can do this with complex numbers notation too, no vectors. i.e. let zbar be the conjugate of the complex number z. Then |z|^2 = z(zbar).

Hence |z+w|^2 = (z+w)([z+w]bar). But bar commutes with sums and products, so this equals

(z+w)(zbar + wbar) = z(zbar) + w(wbar) + wzbar + zwbar.

Similarly |z-w|^2 = z(zbar) + w(wbar) - wzbar - zwbar.

so the sum is 2 (z[zbar] + w[wbar]) = 2 (|z|^2 + |w|^2).

Last edited: Jan 29, 2005

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