Question

Show that the real value of x will satisfy the equation
$\frac{1-ix}{1+ix}$ = a - ib if $a^2+b^2$ = 1 (a, b real)

Answer 1

$\frac{1-ix}{1+ix}=\frac{a-ib}{1}$
By componendo and dividendo we have,
$\frac{(1+ix)-(1-ix)}{(1+ix)+(1-ix)}=\frac{1-(a-ib)}{1+(a-ib)}$
or $\frac{2ix}{2}=\frac{1-(a-ib)}{1+(a-ib)}$
or ix = $\frac{(1-a+ib)(1+a+ib)}{(1+a{)}^2-i^2{b}^2}=\frac{1-a^2-b^2+2ib}{(1+a{)}^2+b^2}$
If $a^2+b^2=1$, the equation (1) reduces to
ix = $-\frac{2ib}{(1+a{)}^2+b^2}$
or x = $\frac{2b}{(1+a{)}^2+b^2}$ which is real.
Alternative method :
Let us suppose that x is real, then
$\frac{1-ix}{1+ix}$ = a - ib (Given)............(1)
Or $\left(\frac{1-ix}{1+ix}\right)$ = a - ib
On taking conjugate of both sides,
or $\frac{1+ix}{1-ix}$ = a + ib ..... (2)
Multiplying (1) and (2) we get,
$\frac{1+x^2}{1+x^2}=a^2+b^{2}or{a}^2+b^2=1$ which is true by given condition.
Hence, our supposition that x is real is correct.