- z 1- z 2-1 and z 1 z 2-0- thenA- z 1- z 2B- z 1-2 z 2C- z 1 z 2-1D- - z 2-2- z 1 z 2

Name: JEE- Grade 11- Mathematics- Complex Numbers- Session 05- W09
Uploaded: 2022-07-04T10:36:42+05:30
Description: Given: | z_1z_2 |=1, ( z_1z_2)=0 | z_1z_2 |=1 ⇒ | z_1 |= | z_2 | ...(i) nbsp; So, let z_1z_2=k, (k gt;0) ...(ii) ⇒ |z_1z_2|=k ⇒ |z_1||z_2|=k ⇒ |z ...

Given: | z_1z_2 |=1, ( z_1z_2)=0 | z_1z_2 |=1 ⇒ | z_1 |= | z_2 | ...(i) nbsp; So, let z_1z_2=k, (k gt;0) ...(ii) ⇒ |z_1z_2|=k ⇒ |z_1||z_2|=k ⇒ |z ...

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Byju's Answer

Standard XII

Mathematics

Geometrical Representation of Argument and Modulus

| z 1/ z 2|=1...

Question

If $∣ ∣ ∣ \frac{z_{1}}{z_{2}} ∣ ∣ ∣ = 1$ and $arg (z_{1} z_{2}) = 0$ , then

z_{1} = z_{2}

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

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z_{1} z_{2} = 1

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z_{1} = 2 z_{2}

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Solution

The correct option is B $| z_{2} |^{2} = z_{1} z_{2}$
Given that $arg (z_{1} z_{2}) = 0$
So, let $z_{1} z_{2} = k, (k > 0)$
$\Rightarrow | z_{1} z_{2} | = k$
$\Rightarrow | z_{1} | | z_{2} | = k$
$∵ ∣ ∣ ∣ \frac{z_{1}}{z_{2}} ∣ ∣ ∣ = 1 \Rightarrow | z_{1} | = | z_{2} |$
$\Rightarrow | z_{2} | | z_{2} | = k$
$∵ k = z_{1} z_{2}$
$∴ | z_{2} |^{2} = z_{1} z_{2}$

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Similar questions

For any complex numbers $z, z_{1}$ and $z_{2}$ prove that:
(i) $a r g (¯ z) = - a r g (z)$
(ii) $a r g (z_{1} z_{2}) = a r g (z_{1}) + a r g (z_{2})$
(iii) $a r g (z_{1}_{2}) = a r g (z_{1}) - a r g (z_{2})$
(iv) $a r g (\frac{z_{1}}{z_{2}}) = a r g (z_{1}) - a r g (z_{2})$