MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Properties of Modulus

Prove the ide...

Question

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

Open in App

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,

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

Suggest Corrections

thumbs-up

0

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

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}}{z_{2}} ∣ ∣ ∣ = 1

arg (z_{1} z_{2}) = 0

z_{1}

z_{2}

be two complex number such that

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

find the value of

k

z_{1}

z_{2}

are complex numbers and

u = \sqrt{z_{1} z_{2}}

, then prove that

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

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Properties of Modulus

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Properties of Modulus

Standard XII Mathematics

Join BYJU'S Learning Program

CrossIcon