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for complex numbers z1 and z2

prove that [tex] |z_{2}| - |z_{1}| \leq |z_{2} - z_{1}| [/tex]

the left hand side becomes

[tex] \sqrt{x_{2}^2 + y_{2}^2} - \sqrt{x_{1}^2 + y_{1}^2} [/tex]

the right hand side becomes

[tex] \sqrt{(x_{2} - x_{1})^2 + (y_{2} - y_{1})^2} [/tex]

now i tried squaring both sides and i get

left hand side

[tex] x_{2}^2 + y_{2}^2 + x_{1}^2 + y_{1}^2 - 2 \sqrt{(x_{2}x_{1})^2 + (x_{2}y_{1})^2 + (x_{1}y_{2})^2 + (y_{1}y_{2})^2} [/tex]

right hand side

[tex] (x_{2} - x_{1})^2 + (y_{2} - y_{1})^2 [/tex]

i put the two of them not equal to eah other and reduce and i ended up with

[tex] x_{2}^2 y_{1}^2 + x_{1}^2 y_{2}^2 \neq 2x_{1} x_{2} y_{1} y_{2} [/tex]

im stuck now...

please help

is there a simpler... less tedious way of doing this... by the way?

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# Homework Help: Triangle inequality

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