Indicate the point of the complex plane $z$ which satisfy the following equation.
Prove that $| z_{1} + z_{2} |^{2} + | z_{1} - z_{2} |^{2} = 2 (| z_{1} |^{2} + | z_{2} |^{2})$ for any two complex numbers $z_{1}$ and $z_{2} .$

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Solution

$| z_{1} + z_{2} |^{2} = (z_{1} + z_{2}) (_{1} +_{2})$
$⟹ | z_{1} |^{2} + | z_{2} |^{2} + z_{2}_{1} + z_{1}_{2}$
$⟹ | z_{1} |^{2} + | z_{2} |^{2} + 2 R e (z_{1} z_{2})$
$| z_{1} - z_{2} |^{2} = (z_{1} - z_{2}) (_{1} -_{2})$
$⟹ | z_{1} |^{2} + | z_{2} |^{2} - z_{2}_{1} - z_{1}_{2}$
$⟹ | z_{1} |^{2} + | z_{2} |^{2} - 2 R e (z_{1} z_{2})$
$| z_{1} z + z_{2} |^{2}$ $+ | z_{1} z - z_{2} |^{2}$ $= 2 (| z_{1}^{2} | + | z_{2} |^{2} + 2 R e (z_{1} z_{2}) - 2 R e (z_{1} z_{2}))$
$⟹ | z_{1} z + z_{2} |^{2}$ $+ | z_{1} z - z_{2} |^{2}$ $= 2 (| z_{1}^{2} | + | z_{2} |^{2})$

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z_{1}, z_{2}

(| z_{1} | + | z_{2} |) ∣ ∣ ∣ \frac{z_{1}}{| z_{1} |} + \frac{z_{2}}{| z_{2} |} ∣ ∣ ∣ \leq 2 (| z_{1} | + | z_{2} |)

Z_{1} = 2 + 3 i

z_{2} = 3 + 4 i

| z - z_{1} |^{2} + | z - z_{2} |^{2} = | z_{1} - z_{2} |^{2}

z_{1}, z_{2}

{| z_{1} + z_{2} |}_{1}^{2} = {| z_{1} |}_{1}^{2} + {| z_{2} |}_{2}^{2}

z_{1}

z_{2}

| z_{1} | + | z_{2} | = ∣ ∣ ∣ \frac{z_{1} + z_{2}}{2} + \sqrt{z_{1} z_{2}} ∣ ∣ ∣ + ∣ ∣ ∣ \frac{z_{1} + z_{2}}{2} - \sqrt{z_{1} z_{2}} ∣ ∣ ∣

z_{1} = 2 + 3 i

z_{2} = 3 + 4 i

z

{| z - z_{1} |}^{2} + {| z - z_{2} |}^{2} = {| z_{1} - z_{2} |}_{1}^{2}

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