If |z1|=|z2| and arg (z1)+ arg (z2)=π / 2, then

Question

If |z1|=|z2| and arg (z1)+ arg (z2)=π / 2, then (A) z1 z2 is purely real (B) z1 z2 is purely imaginary (C) (z1+z2)2 is purely real. (D)  arg (z1-1)+ arg (z2-1)=(π/2)

Tardigrade Admin · Accepted Answer

Let |z1|=|z2|=r ⇒ z1=r( cos θ+i sin θ) and z2=r( cos ((π/2)-θ)+i sin ((π/2)-θ)) ⇒ z1 z2=r2 i, which is purely imaginary z1+z2=r[( cos θ+ sin θ)+i( cos θ+ sin θ)] ⇒ (z1+z2)2=2 r2 ⋅( cos θ+ sin θ)2 ⋅ i which is purely imaginary. Also arg (z1-1)+ arg (z2-1)=-(π/2).

Q. If $∣ z_{1} ∣ = ∣ z_{2} ∣$ and $ar g (z_{1}) + ar g (z_{2}) = π /2$ , then

$z_{1} z_{2}$ is purely real

$z_{1} z_{2}$ is purely imaginary

$(z_{1} + z_{2})^{2}$ is purely real.

$ar g (z_{1}^{- 1}) + ar g (z_{2}^{- 1}) = \frac{π}{2}$

Q. If ∣z1​∣=∣z2​∣ and arg(z1​)+arg(z2​)=π/2, then

z1​z2​ is purely real

z1​z2​ is purely imaginary

(z1​+z2​)2 is purely real.

arg(z1−1​)+arg(z2−1​)=2π​

Q. If $∣ z_{1} ∣ = ∣ z_{2} ∣$ and $ar g (z_{1}) + ar g (z_{2}) = π /2$ , then

$z_{1} z_{2}$ is purely real

$z_{1} z_{2}$ is purely imaginary

$(z_{1} + z_{2})^{2}$ is purely real.

$ar g (z_{1}^{- 1}) + ar g (z_{2}^{- 1}) = \frac{π}{2}$