Subjective Type

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

Solution

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

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