Question

x+ib/x-ib=solution?
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Answer 1

Answer:

$\frac{ {( x + ib)}^{2} }{ {x}^{2} + {b}^{2} } \\ = \frac{ {x}^{2} - {b}^{2} + 2ixb }{ {x}^{2} + {b}^{2} } \\ = \frac{( {x}^{2} - {b}^{2} ) }{( {x {}^{2} + {b}^{2} )} } + i \frac{2xb}{( {x}^{2} + {b}^{2} )}$