Question

multiplicative inverse of (i+1)(i+2)/(i-1)(i-2)

Answer 1

(i+1)(i+2)/(i-1)(i-2)
=(i^2+2i+i+2)/(i^2-2i-i+2)
=(-1+3i+2)/(-1-3i+2)
=(1+3i)/(1-3i)
=(1+3√-1)/(1-3√-1)

Answer 2

Answer:

Required multiplicative inverse is $-\frac{5}{13} (3i + 2)$

Step-by-step explanation:

Given,

$\frac{(i + 1)(i + 2)}{(i - 1)(i - 2)}$

Here numerator is $(i + 1)(i + 2)$

and denominator is $(i - 1)(i - 2)$

First we are multiplying numerator and denominator by (i+1),

$\frac{(i + 1)(i + 2)(i + 1)}{(i - 1)(i - 2)(i + 1)} \\ = \frac{ {(i + 1)}^{2}(i + 2) }{(i - 2)( {i}^{2} - {1}^{2} )} \\ = \frac{( {i}^{2} + 2i + 1)(i + 2) }{(i - 2)( - 1 - 1)} \\ = {\frac{( - 1 + 2i + 1)(i + 2)}{ - 2(i - 2)} } \\ = \frac{2i(i + 2)}{ - 2(i - 2)} \\ = - i \frac{(i + 2)}{(i - 2)}$

We are multiplying denominator and numerator by (i+2),

$= - i \frac{(i + 2)(i + 2)}{(i + 2)(i - 2)} \\ = - i \frac{ {(i + 2)}^{2} }{ {i}^{2} - {2}^{2} } \\ = - i \frac{ {i}^{2} + 2i + {2}^{2} }{ - 1 - 4} \\ = - i \frac{ - 1 + 2i + 4}{ - 5} \\ = i \frac{3 + 2i}{5} \\ = \frac{3i + 2 {i}^{2} }{5}$

$= \frac{3i - 2}{5}$

Therefore

$\frac{(i + 1)(i + 2)}{(i - 1)(i - 2)} = \frac{3i - 2}{5}$

Multiplicative inverse of

$\frac{(i + 1)(i + 2)}{(i - 1)(i - 2)}$

= multiplicative inverse of $\frac{3i - 2}{5}$

is $\frac{5}{3i - 2}$

or,

$\frac{5 \times ( - 3i - 2)}{( 3i - 2)( - 3i - 2)} \\ = + \frac{5(3i + 2)}{ {( 3i)}^{2} - {2}^{2} } \\ = \frac{5(3i + 2)}{ - 9 - 4} \\ = -\frac{5}{13} (3i + 2)$

Here applied formulas are

${i}^{2} = - 1 \\ {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} \\ {x}^{2} - {y}^{2} = (x + y)(x - y)$