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Question Number 19313 by Tinkutara last updated on 09/Aug/17

Prove that ∣ z_{1} \pm z_{2} ∣^{2} = ∣ z_{2} ∣^{2} + ∣ z_{1} ∣^{2} \pm

2 Re (z_{1} {\bar{z}}_{2}) = ∣ z_{1} ∣^{2} + ∣ z_{2} ∣^{2} \pm 2 Re ({\bar{z}}_{1} . z_{2})

Answered by ajfour last updated on 09/Aug/17

let z_{1} = x_{1} + {iy}_{1}

z_{2} = x_{2} + {iy}_{2}

∣ z_{1} \pm z_{2} ∣^{2} = {(x_{1} \pm x_{2})}^{2} + {(y_{1} \pm y_{2})}^{2}

= x_{1}^{2} + y_{1}^{2} + x_{2}^{2} + y_{2}^{2} \pm 2 (x_{1} x_{2} + y_{1} y_{2})

and z_{1} {\bar{z}}_{2} = (x_{1} + {iy}_{1}) (x_{2} - {iy}_{2})

Re (z_{1} {\bar{z}}_{2}) = x_{1} x_{2} + y_{1} y_{2}

So, ∣ z_{1} \pm z_{2} ∣^{2} =∣ z_{1} ∣^{2} + ∣ z_{2} ∣^{2} \pm 2 Re (z_{1} {\bar{z}}_{2}) .

Commented by Tinkutara last updated on 09/Aug/17

Thank you very much Sir!