If Z=cos⁡ϕ+isin⁡ϕ ∀ϕ∈(π)/3,π, then the value of argZ^2-Z is equal to (where, arg(Z) represents the argument of the complex number Z lying in the…

Question

If Z=cos⁡ϕ+isin⁡ϕ ∀ϕ∈(π)/3,π, then the value of argZ^2-Z is equal to (where, arg(Z) represents the argument of the complex number Z lying in the interval -π,π and i^2=-1)

MARKS Content Team · Accepted Answer

Correct answer: (C) 3/2ϕ-π. Z^2-Z=ZZ-1 =cos⁡ϕ+isin⁡ϕcos⁡ϕ-1+isin⁡ϕ =cos⁡ϕ+isin⁡ϕ-2sin^2(ϕ)/2+i2sin(ϕ)/2cos(ϕ)/2 =2isin(ϕ)/2cos(3ϕ)/2+isin(3ϕ)/2 =2sin(ϕ)/2cos(3ϕ+π)/2+isin(3ϕ+π)/2 Now, 3ϕ∈π,3π⇒(3ϕ+π)/2∈π,2π But, the argument lies in -π,π, hence argZ^2-Z=(3ϕ+π)/2-2π=3/2ϕ-π

If $Z = \cos ϕ + i \sin ϕ (\forall ϕ \in (\frac{π}{3}, π)),$ then the value of $\arg (Z^{2} - Z)$ is equal to (where, $\arg (Z)$ represents the argument of the complex number $Z$ lying in the interval $(- π, π]$ and $i^{2} = - 1$ )

Solution

More Complex Number Questions

If Z=cos⁡ϕ+isin⁡ϕ ∀ϕ∈π3,π, then the value of argZ2-Z is equal to (where, arg(Z) represents the argument of the complex number Z lying in the interval -π,π and i2=-1)

Solution

More Complex Number Questions

If $Z = \cos ϕ + i \sin ϕ (\forall ϕ \in (\frac{π}{3}, π)),$ then the value of $\arg (Z^{2} - Z)$ is equal to (where, $\arg (Z)$ represents the argument of the complex number $Z$ lying in the interval $(- π, π]$ and $i^{2} = - 1$ )