If a complex number z = frac{4 + 3i sin theta | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) For a complex number $z$ to be purely real,its imaginary part must be zero. Given $z = \frac{4 + 3i \sin 	heta}{1 - 2i \sin 	heta}$. Multiply th

Vedclass · Accepted Answer

$n \pi, n \in Z$

Vedclass · Answer

(D) For a complex number $z$ to be purely real,its imaginary part must be zero. Given $z = \frac{4 + 3i \sin 	heta}{1 - 2i \sin 	heta}$. Multiply the numerator and denominator by the conjugate of the denominator $(1 + 2i \sin 	heta)$: $z = \frac{(4 + 3i \sin 	heta)(1 + 2i \sin 	heta)}{(1 - 2i \sin 	heta)(1 + 2i \sin 	heta)}$ $z = \frac{4 + 8i \sin 	heta + 3i \sin 	heta + 6i^2 \sin^2 	heta}{1 + 4 \sin^2 	heta}$ Since $i^2 = -1$,we have: $z = \frac{4 - 6 \sin^2 	heta + i(11 \sin 	heta)}{1 + 4 \sin^2 	heta}$ For $z$ to be purely real,the imaginary part must be zero: $\frac{11 \sin 	heta}{1 + 4 \sin^2 	heta} = 0$ This implies $11 \sin 	heta = 0$,so $\sin 	heta = 0$. The general solution for $\sin 	heta = 0$ is $	heta = n \pi$,where $n \in Z$.

If a complex number $z = \frac{4 + 3i \sin \theta}{1 - 2i \sin \theta}$ (where $i = \sqrt{-1}$) is purely real,then the value of $\theta$ is

Similar Questions

The value of $\sum_{n=1}^{13}(i^{n}+i^{n+1})$, where $i=\sqrt{-1}$, is

$\sum\limits_{n = 1}^{50} {{i^{2n-1}}}$ is equal to (where $i = \sqrt{-1}$)

If $(3x+2)-(5y-3)i$ and $(6x+3)+(2y-4)i$ are conjugates of each other,then the value of $\frac{x-y}{x+y}$ is (where $i=\sqrt{-1}, x, y \in R$ ).

Solving $3 - 2yi = 9^x - 7i$,where $i^2 = -1$,for real values of $x$ and $y$,we get:

For any two complex numbers $z_{1}$ and $z_{2},$ prove that $\operatorname{Re}(z_{1} z_{2})=\operatorname{Re} z_{1} \operatorname{Re} z_{2}-\operatorname{Im} z_{1} \operatorname{Im} z_{2}.$

Vedclass Test Series

Exam Paper Generator

Online Exam Module