Question

2) Find real values of for which
is purely real.
4+3i sin e
1-2i sin​

Answer 1

Answer:

Step-by-step explanation:

Given:  

⇒  

1−2isinθ

4+3isin(θ)

​  

    .......(i)

As the function is purely real, the imaginary part of eqn.(i) would be 0

rationalizing eqn.(i) we get,

⇒  

1−2isinθ

4+3isin(θ)

​  

×  

1+2isinθ

1+2isinθ

​  

 

⇒  

1  

2

−(2isinθ)  

2

 

(4+3isin(θ))×(1+2isin(θ))

​  

 

.......{since,(a+b)(a−b)=a  

2

−b  

2

 }

⇒  

1−4i  

2

sin  

2

θ

4+8isinθ+3isinθ+6i  

2

sin  

2

θ

​  

 

⇒  

1+4sin  

2

θ

4+11isinθ−6sin  

2

θ

​  

 

.......{since,i  

2

=−1 }

⇒  

1+4sinθ

1

​  

×[(4−6sin  

2

θ)+i(11sinθ)]

the imainary part is equal to zero for eqn. (i) to be purely real.

∴  

⇒  

1+4sin  

2

θ

11sinθ

​  

=0

⇒sinθ=0

the general solution for sinθ=0 is θ∈nπwhere,n∈Z

Hence, for θ∈nπwhere,n∈Z

⇒  

1−2isinθ

4+3isin(θ)

​  

 is purely real.