Question

find the multiplicative inverse of 4-3i

Answer 1

hope this is helpful.......

Answer 2

$Multiplicative \: \: inverse \\ \\ Let, \: \: a \: \: be \: \: any \: \: number. \\ Then, \: \: the \: \: multiplicative \: \: inverse \\ of \: \: a \: \: is \: \: \frac{1}{a} \\ Since, \: \: a \times \frac{1}{a} = 1. \\ \\ Now, the \: \: given \: \: complex \: \: number \: \: is \\ = 4 - 3i \\ \\ So, \: \: the \: \: multiplicative \: \: inverse \: \: be \\ \\ = \frac{1}{4 - 3i} \\ \\ = \frac{(4 + 3i)}{(4 - 3i)(4 + 3i)} \\ \\ = \frac{4 + 3i}{16 - 9 {i}^{2} } \: \: \\ (using \: \: (x + y)(x - y) = {x}^{2} - {y}^{2} ) \\ \\ = \frac{4 + 3i}{16 + 9} \: \: (since, \: \: {i}^{2} = - 1) \\ \\ = \frac{4 + 3i}{25} \\ \\ = \frac{1}{25} (4 + 3i) \\ \\ Thank \: \: you \: \: for \: \: your \: \: question.$