$$u = (1+i, i), v = (1-i, 2i), w = (2,3+i)$$
I'm asked to find is there's $z$ such that: $$v = zu$$
So if I suppose $z = a+bi$ I have the system:
$$(1-i, 2i) = (a+bi)(1+i, i)\implies\\(1-i, 2i) = (a-b+(a+b)i, ai-b)$$
then I have the system:
$$1-i = a-b + ai + bi\\2i = ai-b$$
So by the second equation $b = ai-2i$, subtitute it in the first equation to get:
$$1-i = 2i+2$$
So I don't get any solution in terms of $a$ and $b$. Does it means that there is no such $z$ that satifies what was asked? (don't know how to verify it on wolfram alpha)
If, in your notation, you are doing coordinate-wise multiplication in ordered pairs of complex numbers, $v=zu$ leads to
$$z=\frac vu$$ $$=\frac{(1-i,2i)}{(1+i,i)}$$ $$=\left(\frac{1-i}{1-i},\frac{2i}i\right)$$ $$=\left(\frac{(1-i)(1-i)}{(1-i)(1+i)}, 2 \right)$$ $$=\left(\frac{1-2i+i^2}{1^2-i^2}, 2 \right)$$ $$=\left(\frac{1-2i-1}{1+1}, 2 \right)$$ $$=\left(\frac{-2i}2, 2 \right)$$ $$=\left(-i, 2 \right)$$
It is easy to check that this $z$ works. Your ordered pair $w$ is not used at all.