Question

if(a+in)=1+i/1-i,then prove that (a^2+b^2)=1

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Answer 1

$\large\underline\blue{\bold{Given \:  Question :-  }}$

$\bf \:  If \: a \: + \: ib = \dfrac{1 + i}{1 - i} \: then \: prove \: that \: {a}^{2} + {b}^{2} = 1$

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Concept and theory:-

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Conjugate of a Complex Number

Conjugate of a complex number z = x + iy is denoted by  It is the reflection of the complex number about the real axis on Arg and’s plane or the image of the complex number about the real axis on Arg and’s plane. If we replace the ‘i’ with ‘- i’, we get conjugate of the complex number.

$\bf \:  ⟼ If \: z \: = x + iy \: then \: \bar z= x \: – \: iy.$

☆Properties of the Conjugate of a Complex Number

$\bf \:  ⟼ \: (1). \:  \overline{z_1 \pm z_2} = \bar z_1 \pm \bar z_2$

$\bf \:  ⟼ \: (2). \:  \overline{z_1 z_2} = \bar z_1 \bar z_2$

$\bf \:  ⟼ (3) . \: \overline{ \bigg(\dfrac{z_1}{ z_2} \bigg)} \: = \dfrac{\bar z_1}{\bar z_2}$

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$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \large\underline\purple{\bold{Solution :- }}$

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$\bf \:  \red{ \underline{According \: to \: statement}}$

$\bf \:  ⟼ \: a \: + \: ib = \dfrac{1 + i}{1 - i} \: \sf \:  ⟼ \: (1)$

☆ Take Conjugates on both sides, we get

$\sf \:  ⟼ \overline{\: a \: + \: ib} = \: \overline{ \bigg(\dfrac{1 + i}{1 - i} \bigg)}$

$\sf \:  ⟼ \: a \: - \: ib \: = \: \dfrac{1 - i}{1 + i} \: \sf \:  ⟼ \: (2)$

☆ Now, multiply (1) with (2), we get

$\sf \:  ⟼ \: (a + ib)(a - ib) = \cancel{\dfrac{1 + i}{1 - i} } \times \cancel{\dfrac{1 - i}{1 + i} }$

$\sf \:  ⟼ \: {a}^{2} - {(ib)}^{2} = 1$

$\sf \:  ⟼ {a}^{2} - {i}^{2} {b}^{2} = 1$

$\sf \:  ⟼ {a}^{2} + {b}^{2} = 1 \: \: \: \: \: \: \: \: \: ( \because \: {i}^{2} = - 1)$

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$\large{\boxed{\boxed{\bf{Hence, Proved}}}}$

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