# If a and b are real numbers such that

Question:

If $a$ and $b$ are real numbers such that $a^{2}+b^{2}=1$ then show that a real value

of $x$ satisfies the equation, $\frac{1-i x}{1+i x}=(a-i b)$

Solution:

We have,

$\frac{1-i x}{1+i x}=(a-i b)=\frac{a-i b}{1}$

Applying componendo and dividendo, we get

$\frac{(1-i x)+(1+i x)}{(1-i x)-(1+i x)}=\frac{a-i b+1}{a-i b-1}$

$\Rightarrow \frac{1-i x+1+i x}{1-i x-1+i x}=\frac{a-i b+1}{a-i b-1}$

$\Rightarrow \frac{2}{-2 i x}=\frac{a-i b+1}{-(-a+i b+1)}$

$\Rightarrow i x=\frac{1-a+i b}{1+a-i b} \times \frac{1+a+i b}{1+a+i b}$

$=\frac{1+a+i b-a-a^{2}-a i b+i b+a i b+i^{2} b^{2}}{(1+a)^{2}-i^{2} b^{2}}$

$\Rightarrow i x=\frac{1-a^{2}-b^{2}+2 i b}{(1+a)^{2}-i^{2} b^{2}}=\frac{1-a^{2}-b^{2}+2 i b}{(1+a)^{2}+b^{2}}=\frac{1-\left(a^{2}+b^{2}\right)+2 i b}{1+a^{2}+2 a+b^{2}}$

$\Rightarrow i x=\frac{1-\left(a^{2}+b^{2}\right)+2 i b}{1+2 a+\left(a^{2}+b^{2}\right)}$

$\Rightarrow i x=\frac{1-1+2 i b}{1+2 a+1}\left[\because a^{2}+b^{2}=1\right]$

$\Rightarrow i x=\frac{2 i b}{2+2 a}$

$\Rightarrow x=\frac{2 b}{2+2 a}=$ Real value