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Question

Answer

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

Explanation

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$\frac{2 i}{( 1 - i ) \cdot ( 2 - i ) \cdot ( 3 - i )} \overset{◯}{1} \frac{2 i}{( 2 - i - 2 i + i ^{2} ) \cdot ( 3 - i )} \overset{◯}{2} \frac{2 i}{( 1 i ^{2} - 3 i + 2 ) \cdot ( 3 - i )} \overset{◯}{3} \frac{2 i}{( - 1 - 3 i + 2 ) \cdot ( 3 - i )} \overset{◯}{4} \frac{2 i}{( - 3 i + 1 ) \cdot ( 3 - i )} \overset{◯}{5} \frac{2 i}{- 9 i + 3 i ^{2} + 3 - i} \overset{◯}{6} \frac{2 i}{- 9 i - 3 + 3 - i} \overset{◯}{7} \frac{2 i}{- 10 i} \overset{◯}{8} - \frac{1}{5}$
①	Multiply each term of $(1 - i)$ by each term in $(2 - i)$ . $(1 - i) \cdot (2 - i) = 2 - i - 2 i + i^{2}$
②	Combine like terms: $2 - i - 2 i + i^{2} = i^{2} - 3 i + 2$
③	$i^{2} = - 1$
④	Combine like terms: $- 1 - 3 i + 2 = - 3 i + 1$
⑤	Multiply each term of $(- 3 i + 1)$ by each term in $(3 - i)$ . $(- 3 i + 1) \cdot (3 - i) = - 9 i + 3 i^{2} + 3 - i$
⑥	$3 i^{2} = 3 \cdot (- 1) = - 3$
⑦	Simplify denominator $- 9 i - 3 + 3 - i = - 10 i$
⑧	Divide $2 i$ by $- 10 i$ to get $\frac{- 1}{5}$ . ( view steps )

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