Show $\forall_{z \in \mathbb{C}, n \in \mathbb{N}} : \exists_{u,v \in \mathbb{Q}} |z - (u+ vi)| < \frac{1}{n}$ 
Show $\forall_{z \in \mathbb{C}, n \in \mathbb{N}} : \exists_{u,v \in \mathbb{Q}} |z - (u+ vi)| < \frac{1}{n}$
In other words, show that you can make an infinitely small complex number (> 0) by using rational coefficients (uses density of $\mathbb{Q}$ in $\mathbb{R)}$.

What I've done is simplifiy the equation so that we can work with it:
$|z - (u+ vi)| < \frac{1}{n} \Leftrightarrow $ (substituting $(a+bi)$ for $z$, and $x$ for $\frac{1}{n}$)
$|(a+bi) - (u + vi)| < x \Leftrightarrow$
$|(a-u) + (b - v)i| < x \Leftrightarrow$
Now when I try to solve for $u$ or $v$ im getting stuck. I've also tried to make cases like
($a = 0, b = 0$), ($a \ne 0, b = 0$), ($a = 0, b \ne 0$), ($a \ne 0, b \ne 0$)
but it didn't work out either.

Hint: choose $u\in\mathbb Q$ such that $\left|a-u\right|\leqslant 1/(2n)$ and $v\in\mathbb Q$ such that $\left|b-v\right|\leqslant 1/(2n)$.
