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

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.

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

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...