<s>[INST] Find $z$ such that $(2+i)(1+i)=2+zi $ I tried multiplying both of the complex numbers, but the answer is  $1+3i $.
Sorry for the poor English; I'm not a native speaker.
 [/INST] Let $z = a + bi$ where $a$ and $b$ are real.
$(2+i)(1+i) = 2 +zi$
$1 + 3i = 2+ (a+bi)i$
$-1 + 3i = ai -b = -b + ai$
So $a =3$ and $-b = -1$ so $b = 1$.
So $z = 3 + i$.
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You should get used to the fact that Complex numbers are just like real numbers.
If you $m=a + dz$ then $z = \frac {m-a}d$ and this is no different.
$(2+i)(1+i)  = 2 + zi$ means
$zi = (2+i)(1+i) - 2$
$z = \frac {(2+i)(1+i) - 2}i$.
You've multiplied out $(2+i)(1+i) = 1+3i$ so 
$z = \frac {(1+3i)-2}i = \frac {-1 + 3i}i = \frac {-1}i + 3$.
Now it's a matter of figuring out what $\frac {-1}i$ is.
$\frac {-1}i = \frac {-1}i\cdot \frac ii = \frac {-i}{-1} = i$.
So $z = i + 3 = 3+i$.
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