Question
Simplify the following and express in the form $a + ib:$
$(1+3 i)^2(3+i)$
$(1+3 i)^2(3+i)$
✓
Answer
$(1+3 i)^2(3+i)$
$=\left(1+6 i+9 i^2\right)(3+i)$
$=(1+6 i-9)(3+i) \ldots \ldots . .\left[\because i^2=-1\right]$
$=(-8+6 i)(3+i)$
$=-24-8 i+18 i+6 i^2$
$=-24+10 i+6(-1)$
$=-24+10 i-6$
$=-30+10 i$
$=\left(1+6 i+9 i^2\right)(3+i)$
$=(1+6 i-9)(3+i) \ldots \ldots . .\left[\because i^2=-1\right]$
$=(-8+6 i)(3+i)$
$=-24-8 i+18 i+6 i^2$
$=-24+10 i+6(-1)$
$=-24+10 i-6$
$=-30+10 i$
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