If - z1- - z2- and arg z1-z2- then z1-z2 is equal to

The correct option is A 0 z_1=re^iθ z_2=re^iϕ since |z_1|=|z_2| Then z_1z_2=e^i(θ ϕ) arg(z_1z_2)=θ ϕ =π Hence θ=π+ϕ Thus ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Geometrical Representation of Argument and Modulus

If | z1| =|...

Question

If $∣ z_{1} ∣=∣ z_{2} ∣$ and arg $(z_{1} / z_{2}) = π$ , then $z_{1} + z_{2}$ is equal to

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Purely imaginary

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Purely real

No worries! We‘ve got your back. Try BYJU‘S free classes today!

None of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

Suggest Corrections

Similar questions

If $z_{1} \neq 0, z_{2} \neq 0, | z_{1} |$ = $| z_{2} |$ and arg $z_{1} +$ arg $z_{2}$ = $π$ , then:

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

arg (\frac{z_{1}}{z_{2}}) = π

z_{1} + z_{2}

If $| z_{1} |$ = $| z_{2} |$ and arg $(\frac{z_{1}}{z_{2}})$ = $π$ , then $z_{1} + z_{2}$ is equal to

(\frac{z_{1}}{z_{2}}) = π,

[\frac{z_{1}}{z_{2}}]

=

\frac{π}{2},

∣ ∣ \frac{z_{1} + z_{2}}{z_{1} - z_{2}} ∣ ∣ .

Join BYJU'S Learning Program