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Question

Answer

$\frac{3 ( a + bi ) ^{2} - 3}{( a + bi ) ^{2} + 4 ( a + bi ) + 1} = \frac{3 a ^{2} + 6 abi + 3 b ^{2} i ^{2} - 3}{b ^{2} i ^{2} + 2 abi + a ^{2} + 4 bi + 4 a + 1}$

Explanation

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$\frac{3 ( a + bi ) ^{2} - 3}{( a + bi ) ^{2} + 4 ( a + bi ) + 1} \overset{◯}{1} \overset{◯}{2} \overset{◯}{3} \frac{(( a + bi ) ^{2} - 1 ) \cdot 3}{a ^{2} + 2 abi + b ^{2} i ^{2} + 4 a + 4 bi + 1} \overset{◯}{4} \overset{◯}{5} \frac{( 1 a ^{2} + 2 abi + b ^{2} i ^{2} - 1 ) \cdot 3}{b ^{2} i ^{2} + 2 abi + a ^{2} + 4 bi + 4 a + 1} \overset{◯}{6} \frac{3 a ^{2} + 6 abi + 3 b ^{2} i ^{2} - 3}{b ^{2} i ^{2} + 2 abi + a ^{2} + 4 bi + 4 a + 1}$
①	Use the distributive property.
②	Find $(a + bi)^{2}$ using formula. $(A + B)^{2} = A^{2} + 2 \cdot A \cdot B + B^{2}$ where $A = a$ and $B = bi$ . $(a + bi)^{2} = a^{2} + 2 \cdot a \cdot bi + (bi)^{2} = a^{2} + 2 abi + b^{2} i^{2}$
③	Multiply $4$ by $(a + bi)$ $4 \cdot (a + bi) = 4 a + 4 bi$
④	Find $(a + bi)^{2}$ using formula. $(A + B)^{2} = A^{2} + 2 \cdot A \cdot B + B^{2}$ where $A = a$ and $B = bi$ . $(a + bi)^{2} = a^{2} + 2 \cdot a \cdot bi + (bi)^{2} = a^{2} + 2 abi + b^{2} i^{2}$
⑤	Combine like terms: $a^{2} + 2 abi + b^{2} i^{2} + 4 a + 4 bi = b^{2} i^{2} + 2 abi + a^{2} + 4 bi + 4 a$
⑥	$(a^{2} + 2 abi + b^{2} i^{2} - 1) \cdot 3 = 3 a^{2} + 6 abi + 3 b^{2} i^{2} - 3$

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