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Complex Numbers

Prove that fo...

Question

Prove that for two complex numbers $z_{1}, z_{2}, {(z_{1} - z_{2})}^{2} = z_{1}^{2} - {2 z}_{1} z_{2} + z_{2}^{2}$

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Solution

Let $z_{1} = x_{1} + i y_{1}$

And $z_{2} = x_{2} + i y_{2}$

Consider LHS

$(z_{1} - z_{2})^{2} = ((x_{1} + i y_{1}) - (x_{2} + i y_{2}))^{2}$

$= (x_{1} + i y_{1})^{2} + (x_{2} + i y_{2})^{2} - 2 (x_{1} + i y_{1}) (x_{2} + i y_{2})$

$(z_{1} - z_{2})^{2} = z_{1}^{2} + z_{2}^{2} - 2 z_{1} z_{2}$

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z_{1}, z_{2}

are two complex numbers

(z_{1} \neq z_{2})

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

z_{1}

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are any two complex numbers, then

∣ ∣ ∣ z_{1} + \sqrt{z_{1}^{2} - z_{2}^{2}} ∣ ∣ ∣ + ∣ ∣ ∣ z_{1} - \sqrt{z_{1}^{2} - z_{2}^{2}} ∣ ∣ ∣

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∣ ∣ ∣ z_{1^{+}} \sqrt{z_{1}^{2} - z_{2}^{2}} ∣ ∣ ∣ + ∣ ∣ ∣ z_{1} - \sqrt{z_{1}^{2} - z_{2}^{2}} ∣ ∣ ∣

z_{1}, z_{2}

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

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