If z 1- Z 2 are complex numbers- then the maximum value of z 1 z 2- z 1 z 2- z 1 z 2- z 1 z 2- z 1 z 2- is equal to

Z_1=r_1e^iθ_1, z_2=r_2e^iθ_2 ⇒z_1z̅_̅2̅+z̅_̅1̅z_2+z_1z_2+z̅_̅1̅z̅_̅2̅/|z_1z_2|=2 cos(θ_1 θ_2)+2cos(θ_1+θ_2)≤4 nbsp; ...

You visited us 6 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Euler's Representation

If z 1, Z 2 a...

Question

If $z_{1}, z_{2}$ are complex numbers, then the maximum value of $\frac{z_{1}_{2} +_{1} z_{2} + z_{1} z_{2} +_{1}_{2}}{| z_{1} z_{2} |}$ is equal to

Open in App

Solution

$z_{1} = r_{1} e^{i θ_{1}}, z_{2} = r_{2} e^{i θ_{2}}$

$\Rightarrow \frac{z_{1}_{2} +_{1} z_{2} + z_{1} z_{2} +_{1}_{2}}{| z_{1} z_{2} |} = 2 cos (θ_{1} - θ_{2}) + 2 cos (θ_{1} + θ_{2}) \leq 4$

Suggest Corrections

Similar questions

Q. If

∣ ∣ ∣ \frac{z_{1}}{z_{2}} ∣ ∣ ∣ = 1

and

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

, then

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) $a r g (\frac{z_{1}}{z_{2}}) = a r g (z_{1}) - a r g (z_{2})$