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Question

Answer

$\frac{( 1 + i ) ^{2} + ( 1 - i ) ^{2}}{( 1 + i ) ^{2} - ( 1 - i ) ^{2}} = 0$

Explanation

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$\frac{( 1 + i ) ^{2} + ( 1 - i ) ^{2}}{( 1 + i ) ^{2} - ( 1 - i ) ^{2}} \overset{◯}{1} \overset{◯}{2} \overset{◯}{3} \overset{◯}{4} \frac{1 + 2 i + i ^{2} + 1 - 2 i + i ^{2}}{1 + 2 i + i ^{2} - ( 1 - 2 i + i ^{2} )} \overset{◯}{5} \overset{◯}{6} \overset{◯}{7} \overset{◯}{8} \frac{1 + 2 i - 1 + 1 - 2 i - 1}{1 + 2 i - 1 - ( 1 - 2 i - 1 )} \overset{◯}{9} \overset{◯}{10} \overset{◯}{11} \overset{◯}{12} \frac{2 i - 2 i}{2 i - - 2 i} \overset{◯}{13} \frac{0}{2 i + 2 i} \overset{◯}{14} \frac{0}{4 i} 0$
①	Find $(1 + i)^{2}$ using formula. $(A + B)^{2} = A^{2} + 2 \cdot A \cdot B + B^{2}$ where $A = 1$ and $B = i$ . $(1 + i)^{2} = 1^{2} + 2 \cdot 1 \cdot i + i^{2} = 1 + 2 i + i^{2}$
②	Find $(1 - i)^{2}$ using formula. $(A - B)^{2} = A^{2} - 2 \cdot A \cdot B + B^{2}$ where $A = 1$ and $B = i$ . $(1 - i)^{2} = 1^{2} - 2 \cdot 1 \cdot i + i^{2} = 1 - 2 i + i^{2}$
③	Find $(1 + i)^{2}$ using formula. $(A + B)^{2} = A^{2} + 2 \cdot A \cdot B + B^{2}$ where $A = 1$ and $B = i$ . $(1 + i)^{2} = 1^{2} + 2 \cdot 1 \cdot i + i^{2} = 1 + 2 i + i^{2}$
④	Find $(1 - i)^{2}$ using formula. $(A - B)^{2} = A^{2} - 2 \cdot A \cdot B + B^{2}$ where $A = 1$ and $B = i$ . $(1 - i)^{2} = 1^{2} - 2 \cdot 1 \cdot i + i^{2} = 1 - 2 i + i^{2}$
⑤	$i^{2} = - 1$
⑥	$i^{2} = - 1$
⑦	$i^{2} = - 1$
⑧	$i^{2} = - 1$
⑨	Combine like terms: $1 + 2 i - 1 = 2 i$
⑩	Combine like terms: $1 - 2 i - 1 = - 2 i$
⑪	Combine like terms: $1 + 2 i - 1 = 2 i$
⑫	Combine like terms: $1 - 2 i - 1 = - 2 i$
⑬	Simplify numerator $2 i - 2 i = 0$
⑭	Simplify denominator $2 i + 2 i = 4 i$

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