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Properties of Modulus

Prove the ide...

Question

Prove the identity, $| 1 + z_{1}_{2} |^{2} + | z_{1} - z_{2} |^{2} = (1 + | z_{1} |^{2}) (1 + | z_{2} |^{2})$

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Solution

Given,

$| 1 - z_{1} {¯ z}_{2} |^{2} - | z_{1} - z_{2} |^{2}$

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

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

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

$(1 - | z_{1} |^{2}) (1 - | z_{2} |^{2})$

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

Comparing (1) and (2), it's clear that,

$| 1 - z_{1} {¯ z}_{2} |^{2} - | z_{1} - z_{2} |^{2} = (1 - | z_{1} |^{2}) (1 - | z_{2} |^{2})$

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Similar questions

z_{1}, z_{2}

be complex numbers with

| z_{1} | = | z_{2} | = 1

| z_{1} + 1 | + | z_{2} + 1 | + | z_{1} z_{2} + 1 | \geq 2

z_{1}

z_{2}

are two complex numbers, then prove that

| 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}} ∣ ∣ ∣

Q. Find the value of

k

, if for the complex numbers

z_{1}

z_{2}

{1 - ∣ ∣_{1} z_{2} ∣ ∣}^{2} - {| z_{1} + z_{2} |}^{2} = k (1 - {| z_{1} |}^{2}) (1 - {| z_{2} |}^{2})

| \frac{z_{1} - 2 z_{2}}{2 - z_{1}_{2}} | = 1

| z_{2} | \neq 1,

z_{1} a n d z_{2}

are complex numbers. Find the value of

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Z_{1}

Z_{2}

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c > 0,

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{| z_{1} + z_{2} |}_{1}^{2} \leq (1 + c) {| z_{1} |}_{1}^{2} + (1 + c^{- 1}) {| z_{2} |}_{2}^{2}

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