Complex Number probs

[Crash]

Prove the triangle inequaltiy as follows

a) [ math ]|z_{1} + z_{2}|^{2} = |z_{1}|^{2} + |z_{2}|^{2} + (z_{1}Z_{2} + z_{2}Z_{1})[ /math ]

(b) [ math ]z_{1}Z_{2} + z_{2}Z_{1} = 2Re (z_{1}Z_{2})[ /math ]

[ math ]<2|z_{1}Z_{2}|\[ /math ]

[ math ]=2 |z_{1}][z_{2}|\[ /math ]

 

Where capital Z's are the conjugate,

Oh and < means greater than or equal to

 

Cheers

[Dapthar]

Prove the triangle inequaltiy as follows

a) [math]|z_{1} + z_{2}|^{2} = |z_{1}|^{2} + |z_{2}|^{2} + (z_{1}Z_{2} + z_{2}Z_{1})[/math]

(b) [math]z_{1}Z_{2} + z_{2}Z_{1} = 2Re (z_{1}Z_{2})[/math]

[math]<2|z_{1}Z_{2}|\[/math]

[math]=2 |z_{1}][z_{2}|\[/math]

 

Where capital Z's are the conjugate' date='

Oh and < means greater than or equal to

 

Cheers[/quote']It helps if you delete the spaces between the math tags. Also, if you just write [math]z=x+i \cdot y[/math], and do some basic algebraic manipulation, the inequalities become obvious.

[Crash]

No they still dont become obvious, can some one please help me on this? even my calculus teacher cant do this............

[Guest drcyber]

Well, first we have the identity |z|^2=zZ. You have applied thath to obtain |z_1+z_2|^2=(z_1+z_2)(Z_1+Z_2), since the conjugate of z_1+z_2 is Z_1+Z+2. ( Remember? conjugate of an addition is the addition of conjugates?)

 

Then you can see that z_1Z_2 is the conjugate of Z_1z_2 and vice versa. since addition of conjugates leave twice the real part you get the other stuff.

 

Hope this helps...