Find $z$ such that $(2+i)(1+i)=2+zi $

Question

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.

Answer 1

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 $z = i + 3 = 3+i$.

Answer 2

$$ (2+i)(1+i)=2+zi \\ 1+3i=2+zi \\ -1+3i=zi $$ Multiply both sides by $-i$. $$ i+3=z $$ So $z=3+i$.