Question

#Quality Question

@Complex Numbers

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€R​

Answer 1

$\blue{\large\qquad \qquad \underline{ \pmb{{ \mathbb{ \maltese  \:  \:  \: GIVEN \:  \:  \:  \maltese }}}}}$

$\frac{1 - ix}{1 + ix} = a - ib\\ \\ \bold{ \mathfrak{ \large{a}^{2} + {b}^{2} = 1}}$

$\green{\large\qquad \qquad \underline{ \pmb{{ \mathbb{ \maltese  \:  \:  \: TO \: \: PROVE \:  \:  \:  \maltese }}}}}$

$\large\bf{x \: is \: real}$

$\red{\large\qquad \qquad \underline{ \pmb{{ \mathbb{ \maltese  \:  \:  \: SOLUTION \:  \:  \:  \maltese }}}}}$

$\frac{1 - ix}{1 + ix} = a - ib \\ \\ \implies \frac{1 - ix}{1 + ix} = \frac{a - ib}{1}$

$\huge \mathfrak{ \text{A}ppl\text{y}ing}\ \ \\ \large \green{\mathfrak{ \bigstar\: Componendo \: \: \text{A}nd \: \: Dividendo }}$

$\dfrac{(1 + ix) - (1 - ix)}{(1 + ix) + (1 - ix)} = \dfrac{1 - (a - ib)}{1 +( a - ib)} \\ \\ \implies \frac{2ix}{2} = \frac{(1 - a) + ib}{(1 + a) - ib} \\ \\ \implies ix = \frac{(1 - a) + ib}{(1 + a) - ib} \times \frac{(1 + a) + ib}{(1 + a) + ib} \\ \\ \implies ix = \frac{ \{(1 + ib) - a \} \{(1 + ib) + a \}}{( {1 + a)}^{2} - {(ib)}^{2} } \\ \\ \implies ix = \frac{(1 + ib) {}^{2} - {a}^{2} }{( {1 + a) {}^{2} + {b}^{2} } } \\ \\ \implies ix = \frac{1 - {a}^{2} - {b}^{2} + 2ib }{ {(1 + a)}^{2} + {b}^{2} }$

$\\ \\ \implies ix = \frac{2ib}{ {(1 + a)}^{2} + {b}^{2} } \\ \\ \implies \green{\boxed{ \boxed{ x = \frac{2b}{( {1 + a)}^{2} + {b}^{2} } }}}$

$\orange{ \huge\mathfrak{ \text{S}o \: \: \bold {x} \: \: is \: \: Real}}$

Answer 2

▬▬▬▬▬▬▬▬▬▬▬▬

$\underline{\underline{\sf{\red{Given\::-}}}}$

• $\tt \dfrac{1 - ix}{1 + ix} = a - ib$

• $\tt a^2 + b^2 = 1$

$\underline{\underline{\sf{\red{To\:Prove\::-}}}}$

• Value of x is is real !

$\underline{\underline{\sf{\red{Proof\::-}}}}$

$\longrightarrow\quad\tt \dfrac{1 - ix}{1 + ix} = a - ib$

★ By componendo and dividendo :-

$\longrightarrow\quad\tt \dfrac{(1 + ix) - (1 - ix)}{(1 + ix) + (1 - ix)} = \dfrac{1 - (a - ib)}{1 + (a + ib)}$

$\longrightarrow\quad\tt \dfrac{2ix}{2} = \dfrac{1 - (a - ib)}{1 + (a + ib)}$

$\longrightarrow\quad\tt \dfrac{\cancel{2}ix}{\cancel{2}} = \dfrac{1 - (a - ib)}{1 + (a + ib)}$

$\longrightarrow\quad\tt ix = \dfrac{1 - (a - ib)}{1 + (a + ib)}$

$\longrightarrow\quad\tt ix = \dfrac{(1 - a) + ib}{(1 + a) - ib}$

$\longrightarrow\quad\tt ix = \dfrac{(1 - a) + ib}{(1 + a) - ib}\:\times\:\dfrac{(1 + a) + ib}{(1 + a) + ib}$

$\longrightarrow\quad\tt ix = \dfrac{[(1 + ib) - a]\:[(1 + ib) + a]}{(1 + a)^2 - (ib)^2}$

$\longrightarrow\quad\tt ix = \dfrac{(1 + ib)^2 - a^2}{(1 + a)^2 - (ib)^2}$

$\longrightarrow\quad\tt ix = \dfrac{(1 + ib)^2 - a^2}{(1 + a)^2 + b^2}$

$\longrightarrow\quad\tt ix = \dfrac{1 - a^2 - b^2 + 2ib}{(1 + a)^2 + b^2}\:\:\bigg\lgroup eq^{n}\:(1) \bigg\rgroup$

• If a² + b² = 1 , then [eqⁿ (1)] reduces to :-

$\longrightarrow\quad\tt ix = -\dfrac{2ib}{1 + a^2 + b^2}$

$\longrightarrow\quad\bf{ x = \pink{\dfrac{2ib}{1 + a^2 + b^2}}}$

$\therefore\:{\underline{\sf{Hence,\:value\:of\:x\:is\:real.}}}$

$\qquad\qquad\quad\:\:\underline{\underline{\sf{\red{Hence,\:Proved!}}}}$

★ Alternative Method :-

$\underline{\underline{\sf{\red{Proof\::-}}}}$

• Let us assume that x is real.

$\longrightarrow\quad\tt \dfrac{1 - ix}{1 + ix} = a - ib \qquad\qquad\bigg\lgroup Given \bigg \rgroup$

$\longrightarrow\quad\tt \bigg(\dfrac{1 - ix}{1 + ix}\bigg) = a + ib$

★ On taking conjugate of both sides :-

$\longrightarrow\quad\tt \dfrac{1 + ix}{1 - ix} = a + ib\qquad\quad\bigg\lgroup eq^{n}\:(2) \bigg\rgroup$

★ By multiplying [eqⁿ (1)] and [eqⁿ (2)] we get :-

$\longrightarrow\quad\tt \dfrac{1 + x^2}{1 + x^2} = a^2 + b^2$

$\longrightarrow\quad\tt \dfrac{\cancel{1 + x^2}}{\cancel{1 + x^2}} = a^2 + b^2$

$\longrightarrow\quad{\bf{\purple{a^2 + b^2 = 1}}}$

ㅤ★ This is true by given condition ★

Hence,

• Our assumption that x is real is correct !!

$\qquad\qquad\quad\:\:\underline{\underline{\sf{\red{Hence,\:Proved!}}}}$

▬▬▬▬▬▬▬▬▬▬▬▬