If -z1-z2-1 and arg z1 z2-0- then

Let z_1=r_1( cosθ_1+isinθ_1) | z_1z_2|=1 ⇒ |z_1|= |z_2| ⇒ |z_1|= |z_2|=r_1 Now, (z_1z_2)=0 ⇒ nbsp; (z_1)+(z_2)=0 ⇒ nbsp; (z_2) = θ_1 Therefore, z_2=r_1 [c ...

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Byju's Answer

Standard XII

Mathematics

Polar Representation of a Complex Number

If |z1/z2|=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} + z_{2} | = 0

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Solution

The correct option is B $| z_{2} |^{2} = z_{1} z_{2}$
Let $z_{1} = r_{1} (cos θ_{1} + i sin θ_{1})$
$∣ ∣ ∣ \frac{z_{1}}{z_{2}} ∣ ∣ ∣ = 1$
$\Rightarrow | z_{1} | = | z_{2} |$
$\Rightarrow | z_{1} | = | z_{2} | = r_{1}$

Now, $arg (z_{1} z_{2}) = 0$
$\Rightarrow arg (z_{1}) + arg (z_{2}) = 0$
$\Rightarrow arg (z_{2}) = - θ_{1}$
Therefore, $z_{2} = r_{1} [cos (- θ_{1}) + i sin (- θ_{1})]$
$= r_{1} (cos θ_{1} - i sin θ_{1}) =_{1}$
$\Rightarrow_{1} = z_{2}$
$\Rightarrow z_{1}_{1} = z_{1} z_{2}$
$\Rightarrow | z_{1} |^{2} = z_{1} z_{2}$

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If ∣∣∣z1z2∣∣∣=1 and arg(z1z2)=0, then

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