Question

find the real and imaginary parts of the complex number a+ib/a-ib​

Answer 1

Answer:

a and b is the real parts of complex number.

i is the imaginary parts of the complex number.

Answer 2

Answer:-

• $\textbf{Real Part }\bf\longrightarrow\:\dfrac{a^2-b^2}{a^2+b^2}.$

• $\textbf{Imaginary Part}\bf\longrightarrow\:\dfrac{2ab}{a^2+b^2}.$

Explanation:-

Given Complex Number:-

• $\tt\leadsto \dfrac{a+ib}{a-ib}.$

To Find:-

• The real and imaginary parts of the number.

So,

Let us simplify the given complex number first.

$\\ \tt\mapsto \frac{a + ib}{a - ib}$

$\\ \tt = \dfrac{a + ib}{a - ib} \times \dfrac{a + ib}{a + ib} .$

$\\ \tt = \frac{(a + ib)(a + ib)}{(a + ib)(a - ib)} .$

$\\ \tt = \frac{ \bigg[(a + ib) \times a \bigg] \bigg[(a + ib) \times ib \bigg]}{ (a) {}^{2} - (ib) {}^{2} }. \\ \\ \rm [usi ng \: \: id \: \: (a + b)( a - b) = {a}^{2} - {b}^{2} ].$

$\\ \tt = \frac{ {a}^{2} + abi + abi + (ib) {}^{2} }{ {a}^{2} - (ib) {}^{2} } .$

$\\ \tt = \frac{ {a}^{2} + 2abi + {i}^{2} {b}^{2} }{ {a}^{2} - {i}^{2} {b}^{2} } .$

$\\ \tt = \frac{ {a}^{2} + 2abi + ( - 1)(b) }{a {}^{2} - ( - 1)(b) } . \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \rm as \: \: we \: \: know \: \: that \: \: value \: \: of \: \: {i}^{2} \: \: is \: \: always \: \: - 1.$

$\\ \tt = \frac{ {a}^{2} - {b}^{2} + 2abi}{ {a}^{2} + {b}^{2} } .$

$\\ \large\boxed{\bf= \frac{ {a}^{2} - {b}^{2} }{ {a}^{2} + {b}^{2} } + \frac{2abi}{ {a}^{2} + {b}^{2} } .}$

Therefore,

• $\textbf{Real Part }\bf =\:\dfrac{a^2-b^2}{a^2+b^2}.$

• $\textbf{Imaginary Part}\bf = \:\dfrac{2ab}{a^2+b^2}.$