The conjugate of \(\rm \frac{(3-i)(1+2i)}{(2+i)(1-3i)}\)  is . 

  1. - i 
  2.  i
  3.  -2i
  4.  i/2

Answer (Detailed Solution Below)

Option 1 : - i 
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Detailed Solution

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Concept:

Let z = x + iy be a complex number,

Where x is called the real part of the complex number or Re (z) and y is called the Imaginary part of the complex number or Im (z)

Conjugate of z = z̅ = x - iy  

 

Calculation:

Let z = \(\rm \frac{(3-i)(1+2i)}{(2+i)(1-3i)}\)

⇒ z = \(\rm \frac{3+6i-i-2i^{2}}{2-6i+i-3i^{2}}\)  = \(\rm \frac{3+6i-i+2}{2-6i+i+3}\)

⇒ z = \(\rm \frac{5+5i}{5-5i}\)  = \(\rm\frac{1+i}{1-i}\) 

⇒ z = \(\rm\frac{1+i}{1-i}\) × \(\rm\frac{1+i}{1+i}\) = \(\rm \frac{1+i^{2}+2i}{1-i^{2}}\) = i 

⇒ z =  

We know that, Conjugate of z = z̅ = x - iy   

⇒ \(\rm\overline{z}\) = - i  . 

The correct option is 1

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