Question

If, a and b are real numbers such that a² + b² =1, then value of x in equation \(\rm \frac{1+ix}{1-ix}=(a-ib)\)
1. Purly imaginary 
2. Purely real
3. Complex number
4. None of the above

Answer 1

Correct Answer - Option 2 : Purely real

Calculation:

Here, \(\rm \frac{1+ix}{1-ix}=\frac{(a-ib)}{1}\)

Now, using componendo and dividendo property

\(\Rightarrow \rm \frac{1+ix+1-ix}{1+ix-1+ix}=\frac{(a-ib+1)}{a-ib-1}\)

\(\Rightarrow \rm \frac{2}{2ix}=\frac{a-ib+1}{a-ib-1}\)

\(\Rightarrow \rm ix=\frac{(a-1)-ib}{(a+1)-ib}\)

\(\Rightarrow \rm ix=\frac{(a-1)-ib}{(a+1)-ib}\times \frac{(a-1)+ib}{(a+1)+ib}\)

\(\Rightarrow \rm ix=\frac{(a^2-1+b^2)+[(a-1)b-(a+1)b]i}{(a+1)^2+b^2}\)

\(\Rightarrow \rm ix=\frac{0-2bi}{(a+1)^2+b^2}=\frac{-2bi}{(a+1)^2+b^2}\)                                ...(∵a2 + b2 =1)

\(\Rightarrow \rm x=\frac{-2b}{(a+1)^2+b^2}\)

So, the value of x in the given equation is purely real.

Hence, option (2) is correct.