$(y + 4 i)^{2} = 8 i y + y^{2} - 16$

Explanation

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$(y + 4 i)^{2} \overset{◯}{1} y^{2} + 8 i y + 16 i^{2} \overset{◯}{2} y^{2} + 8 i y + (- 16) \overset{◯}{3} 8 i y + y^{2} - 16$
①	Find $(y + 4 i)^{2}$ using formula. $(A + B)^{2} = A^{2} + 2 \cdot A \cdot B + B^{2}$ where $A = y$ and $B = 4 i$ . $(y + 4 i)^{2} = y^{2} + 2 \cdot y \cdot 4 i + (4 i)^{2} = y^{2} + 8 i y + 16 i^{2}$
②	$16 i^{2} = 16 \cdot (- 1) = - 16$
③	Combine like terms: $8 i y + y^{2} - 16 = 8 i y + y^{2} - 16$

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(y+4i)2=8iy+y2−16

$(y + 4 i)^{2} = 8 i y + y^{2} - 16$