$(3 - 2 i) \frac{1 + 2 i}{2 - i} \cdot (2 + 3 i) = 9 i^{2} + 12 i + 4$

Explanation

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$(3 - 2 i) \frac{1 + 2 i}{2 - i} \cdot (2 + 3 i) \overset{◯}{1} (3 - 2 i) \cdot 1 i \cdot (2 + 3 i) \overset{◯}{2} (3 i - 2 i^{2}) \cdot (2 + 3 i) \overset{◯}{3} (3 i + 2) \cdot (2 + 3 i) \overset{◯}{4} 6 i + 9 i^{2} + 4 + 6 i \overset{◯}{5} 9 i^{2} + 12 i + 4$
①	Divide $1 + 2 i$ by $2 - i$ to get $i$ . ( view steps )
②	$(3 - 2 i) \cdot i = 3 i - 2 i^{2}$
③	$- 2 i^{2} = - 2 \cdot (- 1) = 2$
④	Multiply each term of $(3 i + 2)$ by each term in $(2 + 3 i)$ . $(3 i + 2) \cdot (2 + 3 i) = 6 i + 9 i^{2} + 4 + 6 i$
⑤	Combine like terms: $6 i + 9 i^{2} + 4 + 6 i = 9 i^{2} + 12 i + 4$

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(3−2i)2−i1+2i​⋅(2+3i)=9i2+12i+4

$(3 - 2 i) \frac{1 + 2 i}{2 - i} \cdot (2 + 3 i) = 9 i^{2} + 12 i + 4$