Question

#Quality Question

@ Complex Numbers

Show that a real value of x will satisfy the equation :

$\frac{1 - i x}{1 + ix} = a + ib \\ if \: \: \: \: \: \: \: \: \bf {a}^{2} + {b}^{2} = 1$
​

Answer 1

$\large\underline{\sf{Given- }}$

$\rm :\longmapsto\:\dfrac{1 - i x}{1 + ix} = a + ib$

and

$\rm :\longmapsto\:{a}^{2} + {b}^{2} = 1$

$\large\underline{\sf{To\:Find- }}$

The real value of x satisfy the equation if

$\rm :\longmapsto\:{a}^{2} + {b}^{2} = 1$

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:\dfrac{1 - i x}{1 + ix} = a + ib - - - (1)$

Taking conjugate on both sides, we get

$\rm :\longmapsto\: \overline{ \bigg(\dfrac{1 - i x}{1 + ix} \bigg)}= \overline{a + ib}$

We know that,

$\: \: \: \: \: \: \: \: \: \: \: \: \underbrace{ \boxed{ \bf \: \overline{\bigg(\dfrac{a_1}{a_2} \bigg)} \: = \: \dfrac{\overline{a_1}}{\overline{a_2}}}}$

Using this, identity we get

$\rm :\longmapsto\: \dfrac{\overline{1 - i x}}{ \: \: \: \overline{1 + ix} \: \: \: } = \overline{a + ib}$

We know,

If

$\rm :\longmapsto\:z = a + ib$

then

$\rm :\longmapsto\:\overline{z} =\overline{ a + ib} = a - ib$

So, using this identity, we get

$\rm :\longmapsto\:\dfrac{1 + i x}{1 - ix} = a - ib - - - (2)$

On multiply equation (1) by (2), we get

$\rm :\longmapsto\: \bigg(\dfrac{1 + i x}{1 - ix} \bigg) \times \bigg(\dfrac{1 - i x}{1 + ix} \bigg) = (a - ib) \times (a + ib)$

$\rm :\longmapsto\:1 = {a}^{2} - {(ib)}^{2}$

$\rm :\longmapsto\:1 = {a}^{2} - {i}^{2} {b}^{2}$

As, we know that

$\underbrace{ \boxed{ \bf \: {i}^{2} = - 1}}$

So,

$\rm :\longmapsto\:1 = {a}^{2} + {b}^{2}$

Hence, justified.

Additional Information :-

If a and b are two complex numbers, then

$\rm :\longmapsto\:\overline{a + b} \: = \: \overline{a} + \overline{b}$

$\rm :\longmapsto\:\overline{a - b} \: = \: \overline{a} - \: \overline{b}$

$\rm :\longmapsto\:\overline{a b} \: = \: \overline{a} \: \: \: \overline{b}$

$\rm :\longmapsto\:a \: \overline{a} \: = \: { |a| }^{2}$

Answer 2

Answer:

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