Question

the modulus of the complex number a+ib/a-ib is​

Answer 1

Step-by-step explanation:

We have,

$z = \dfrac{a + ib}{a - ib}$

$\implies \: z = \dfrac{ \left(a + ib \right)\left(a + ib \right)}{ \left(a - ib \right)\left(a + ib \right)}$

$\implies \: z = \dfrac{ \left(a + ib \right)^{2} }{ {a}^{2} + {b}^{2} }$

$\implies \: z = \dfrac{ \left(a \right) ^{2} + \left( ib \right)^{2} + 2 \cdot \: a \cdot \: ib}{ {a}^{2} + {b}^{2} }$

$\implies \: z = \dfrac{ {a}^{2} - {b }^{2} + 2ab \: i}{ {a}^{2} + {b}^{2} }$

$\implies \: z = \dfrac{ {a}^{2} - {b }^{2} }{ {a}^{2} + {b}^{2} } + \dfrac{ 2ab }{ {a}^{2} + {b}^{2} }i$

$\implies \: | z | = \left| \dfrac{ {a}^{2} - {b }^{2} }{ {a}^{2} + {b}^{2} } + \dfrac{ 2ab }{ {a}^{2} + {b}^{2} } i\right|$

$\implies \: | z | = \left(\dfrac{ {a}^{2} - {b }^{2} }{ {a}^{2} + {b}^{2} } \right)^{2} + \left( \dfrac{ 2ab }{ {a}^{2} + {b}^{2} } \right)^{2}$

$\implies \: | z | = \dfrac{ \left({a}^{2} - {b }^{2} \right)^{2} }{ \left( {a}^{2} + {b}^{2} \right)^{2} } + \dfrac{ 4 {a}^{2} {b}^{2} }{ \left({a}^{2} + {b}^{2} \right)^{2} }$

$\implies \: | z | = \dfrac{ \left({a}^{2} - {b }^{2} \right)^{2} + 4 {a}^{2} {b}^{2} }{ \left( {a}^{2} + {b}^{2} \right)^{2} }$

$\implies \: | z | = \dfrac{ {a}^{4} + {b }^{4} - 2 {a}^{2} {b}^{2} + 4 {a}^{2} {b}^{2} }{ \left( {a}^{2} + {b}^{2} \right)^{2} }$

$\implies \: | z | = \dfrac{ {a}^{4} + {b }^{4} + 2 {a}^{2} {b}^{2} }{ \left( {a}^{2} + {b}^{2} \right)^{2} }$

$\implies \: | z | = \dfrac{ \left( {a}^{2} + {b}^{2} \right)^{2} }{ \left( {a}^{2} + {b}^{2} \right)^{2} }$

$\implies \: | z | =1$