If z 1 and z 2 are conjugate to each other- and - z 1 z 2- k - then k -

Z_1 and z_2 are conjugate to each other, i.e., z_2 = z_1. nbsp; ∴( z_1 z_2) =kπ ⇒ kπ= ( z_1 z_1) ⇒ kπ= ( |z_1|^2) ⇒ nbsp; kπ=(a negative real number) ⇒ kπ= ...

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Standard XII

Mathematics

Properties of Argument

If z 1 and z ...

Question

If $z_{1}$ and $z_{2}$ are conjugate to each other, and $arg (- z_{1} z_{2}) = k π$ , then $k =$

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Solution

$z_{1}$ and $z_{2}$ are conjugate to each other, i.e., $z_{2} =_{1}$ .
$∴ arg (- z_{1} z_{2}) = k π \Rightarrow k π = arg (- z_{1}_{1})$
$\Rightarrow k π = arg (- | z_{1} |^{2})$
$\Rightarrow k π = arg$ $($ a negative real number $)$
$\Rightarrow k π = π ∴ k = 1$

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Similar questions

Q. If

z_{1}

and

z_{2}

are conjugate to each other, then find

a r g (- z_{1} z_{2})

Q. If

z_{1}

and

z_{2}

are conjugate to each other, then find agr (-

z_{1}

z_{2}

Q. If

z_{1}

z_{2}

are conjugate complex numbers and arg

(z_{1} z_{2}) =

Q. Assertion :If

z_{1} + z_{2} = a

and

z_{1} z_{2} = b

where

a = ¯ ¯ ¯ a

and

b = ¯ ¯ b

, then

a r g (z_{1} z_{2}) = 0.

Reason: The sum and product of two complex numbers are real if and only if they are conjugate of each other.

Q. Statement 1: If

z_{1} + z_{2} = a

and

z_{1} z_{2} = b

, where

a = a

and

b = ¯ ¯ b

, then

a r g (z_{1} z_{2}) = 0

.
Statement 2: The sum and product of two complex numbers are real if and only if they are conjugate of each other.

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Properties of Argument

Standard XII Mathematics

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If z1 and z2 are conjugate to each other, and arg(−z1z2)=kπ, then k=

z1 and z2 are conjugate to each other, i.e., z2=¯¯¯¯¯z1. ∴arg(−z1z2)=kπ⇒kπ=arg(−z1¯¯¯¯¯z1) ⇒kπ=arg(−|z1|2) ⇒kπ=arg(a negative real number) ⇒kπ=π∴k=1

If $z_{1}$ and $z_{2}$ are conjugate to each other, and $arg (- z_{1} z_{2}) = k π$ , then $k =$

$z_{1}$ and $z_{2}$ are conjugate to each other, i.e., $z_{2} =_{1}$ .
$∴ arg (- z_{1} z_{2}) = k π \Rightarrow k π = arg (- z_{1}_{1})$
$\Rightarrow k π = arg (- | z_{1} |^{2})$
$\Rightarrow k π = arg$ $($ a negative real number $)$
$\Rightarrow k π = π ∴ k = 1$