$(4 z + 5 i)^{2} = 40 i z + 16 z^{2} - 25$

Explanation

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$(4 z + 5 i)^{2} \overset{◯}{1} 16 z^{2} + 40 i z + 25 i^{2} \overset{◯}{2} 16 z^{2} + 40 i z + (- 25) \overset{◯}{3} 40 i z + 16 z^{2} - 25$
①	Find $(4 z + 5 i)^{2}$ using formula. $(A + B)^{2} = A^{2} + 2 \cdot A \cdot B + B^{2}$ where $A = 4 z$ and $B = 5 i$ . $(4 z + 5 i)^{2} = (4 z)^{2} + 2 \cdot 4 z \cdot 5 i + (5 i)^{2} = 16 z^{2} + 40 i z + 25 i^{2}$
②	$25 i^{2} = 25 \cdot (- 1) = - 25$
③	Combine like terms: $40 i z + 16 z^{2} - 25 = 40 i z + 16 z^{2} - 25$

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(4z+5i)2=40iz+16z2−25

$(4 z + 5 i)^{2} = 40 i z + 16 z^{2} - 25$