Question

find the real value of theta for which (3+2i sin theta/ 1-2i sin theta) is purely real

Answer 1

$\textbf{Given:}$

$z=\dfrac{3+2i\,\sin\theta}{1-2i\,\sin\theta}\;(say)$

$\textbf{To find:}$

$\text{Real value of \theta for which z is purely real}$

$\textbf{Solution:}$

$\text{Since z is purely real,we have}$

$\bf\,z=\overline{z}$

$\implies\dfrac{3+2i\,\sin\theta}{1-2i\,\sin\theta}=\overline{(\dfrac{3+2i\,\sin\theta}{1-2i\,\sin\theta})}$

$\implies\dfrac{3+2i\,\sin\theta}{1-2i\,\sin\theta}=\dfrac{3-2i\,\sin\theta}{1+2i\,\sin\theta}$

$\implies(3+2i\,\sin\theta)(1+2i\,\sin\theta)=(3-2i\,\sin\theta)(1-2i\,\sin\theta)$

$\implies\;3+6i\,\sin\theta+2i\,\sin\theta+4i^2\sin^2\theta=3-6i\,\sin\theta-2i\,\sin\theta+4i^2\sin^2\theta$

$\implies\;8i\,\sin\theta=-8i\,\sin\theta$

$\implies\;16i\,\sin\theta=0$

$\implies\;\sin\theta=0$

$\therefore\bf\theta=n\pi,\;\;n{\in}Z$

$\textbf{Answer:}$

$\bf\;\theta=n\,\pi,\;\;n{\in}Z$

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Answer 2

Given that,

The purely real function is

$\dfrac{3+2i\sin\theta}{1-2i\sin\theta}$

We need to solve it

Using given function

$\dfrac{3+2i\sin\theta}{1-2i\sin\theta}$

$\dfrac{3+2i\sin\theta}{1-2i\sin\theta}\times\dfrac{1+2i\sin\theta}{1+2i\sin\theta}$

$\dfrac{(3+2i\sin\theta)(1+2i\sin\theta)}{(1-2i\sin\theta)(1+2i\sin\theta)}$

$\dfrac{3+6i\sin\theta+2i\sin\theta+4i^2\sin^2\theta}{1^2-(2i\sin\theta)^2}$

$\dfrac{3+8i\sin\theta+4i^2\sin^2\theta}{1-4i^2\sin^2\theta}$

Put the value of i² = -1

$\dfrac{3+8i\sin\theta-4\sin^2\theta}{1+4\sin^2\theta}$

$\dfrac{3-4\sin^2\theta+8i\sin\theta}{1+4\sin^2\theta}$

Now divided in real and imaginary parts,

$\dfrac{3-4\sin^2\theta}{1+4\sin^2\theta}+i(\dfrac{8\sin\theta}{1+4\sin^2\theta})$

If $\dfrac{3+2i\sin\theta}{1-2i\sin\theta}$  is purely real.

So, the imaginary part will be equal to zero.

So, $\dfrac{8\sin\theta}{1+4\sin^2\theta}=0$

$8\sin\theta=0\times(1+4\sin^2\theta)$

$8\sin\theta=0$

$\sin\theta=0$

$\theta=0$

Hence, The real value of θ is zero.