If -z1-z2-z1-z2- then z1-argz2-A- 0B- -4C- -2D- -

The correct option is C ±π/2|z_1+z_2|^2=|z_1 z_2|^2 ⇒ z_1.z̅_2+z̅_1z_2=0 i.e z_1.z̅_2 +z̅_̅1̅.̅z̅_̅2̅=0 ⇒ Re(z_1.z̅_2)=0 Let z_1=r, e^i θ_1, z_2 =r_2e^iθ_2 ⇒ R ...

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Byju's Answer

Standard XII

Mathematics

Derivative of Standard Functions

If |z1+z2|=|z...

Question

If $| z_{1} + z_{2} | = | z_{1} - z_{2} |$ then $a r g z_{1} - a r g z_{2} =$

0

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\pm \frac{π}{4}

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\pm \frac{π}{2}

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π

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Solution

The correct option is C $\pm \frac{π}{2}$
$| z_{1} + z_{2} |^{2} = | z_{1} - z_{2} |^{2} \Rightarrow z_{1} . {¯ z}_{2} + {¯ z}_{1} z_{2} = 0$
i.e $z_{1} . {¯ z}_{2} + ¯ z_{1} . z_{2} = 0 \Rightarrow R e (z_{1} . {¯ z}_{2}) = 0$
Let $z_{1} = r, e^{i θ_{1}}, z_{2} = r_{2} e^{i θ_{2}} \Rightarrow R e z_{1} . {¯ z}_{2} = r_{1} r_{2} e^{i (θ_{1} - θ_{2})} = 0$
$\Rightarrow c o s (θ_{1} - θ_{2}) = 0 \Rightarrow θ_{1} - θ_{2} = \pm \frac{π}{2}$

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If |z1+z2|=|z1−z2| then argz1−argz2=

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If $| z_{1} + z_{2} | = | z_{1} - z_{2} |$ then $a r g z_{1} - a r g z_{2} =$