$m+ni+k\lambda,\,\Re(\lambda),\Im(\lambda)\notin \mathbb{Q}$ is dense in $\mathbb{C}$! 
As said in the comments below, it's needed to suppose $\{1,\Re(\lambda),\Im(\lambda)\}$ linearly independent over $\mathbb{Q}$, otherwise the result is false, according to Christian's example.

This is an exercise of the book Funções de uma Variável Complexa, by Alcides Lins Neto and, well, let's see it before I explain my doubt:

Consider $\lambda\in \mathbb{C}$ such that $\Re(\lambda)$ and $\Im(\lambda)$ are NOT rational numbers. Prove that the set
  $$\{m+ni+k\lambda; m,n,k\in \mathbb{Z}\}$$
  is a dense subset of $\mathbb{C}$.

This is exercise 19 of the very first section of exercises in the book. At exercise 18 is given a suggestion (which, I suppose, may also be used at the 19th) that is:

If $a\notin\mathbb{Q}$ then the set $\{m+na;m,n\in \mathbb{Z}\}$ is dense in $\mathbb{R}$.

Let's see what I've tried.
Let $z_0$ be any complex number and $\varepsilon>0$ be any "radius". In order to show that that set is dense in $\mathbb{C}$ is sufficient to show that there is an element of it in the square
$$R_\varepsilon(z_0)=\left\{u+iv\in \mathbb{C};|u-\Re(z_0)|<\varepsilon,\,\,\, |v-\Im(z_0)|<\varepsilon\right\}.$$
It is sufficient because the collection of "open squares" is a basis for topology on $\mathbb{C}\cong\mathbb{R}^2$.
Now, using the suggestion, we have that there are $m_1,n_1,m_2,n_2\in \mathbb{Z}$ such that
$$\begin{array}{c}\Re(z_0)-\varepsilon<m_1+n_1\Re(\lambda)<\Re(z_0)+\varepsilon\\
\Im(z_0)-\varepsilon<m_2+n_2\Im(\lambda)<\Im(z_0)+\varepsilon\end{array}.$$
So we have that the number $u+iv$ given by
$$\begin{array}{rcl}u&=&m_1+n_1\Re(\lambda),\\
v&=&m_2+n_2\Im(\lambda),\end{array}$$
is in the square $R_\varepsilon(z_0)$ "near" $z_0$.
But now we have the following:
$$u+iv=m_1+m_2i+n_1\Re(\lambda)+n_2\Im(\lambda)i,$$
and this IS NOT NECESSARILY a point of the form $m+ni+k\lambda$, since $n_1$ mights be different from $n_2$. After this I've tried to think something about least common multiple, tried to find a pair to which $n_1=n_2$, or tried to fix $n_1=n_2$ and then to search for suitable $m_1$ and $m_2$: all of them without any success... How to garantee $n_1=n_2$?? Or maybe there is a completely different way of proving it...  
Thank you guys!
 A: ${\bf Summary:}$
We first prove that we can replace $\lambda$ with a multiple $(j-j_1)\lambda$ of $\lambda$ such that $\|(j-j_1)\lambda\|\le\varepsilon$ in the torus $T=\Bbb{R}/\Bbb{Z}\times \Bbb{R}/\Bbb{Z}$, then we use that any line with irrational slope is dense in $T$.
Let $\varepsilon>0$ and $z\in \Bbb{C}$. We have to proof that 
$$
\exists m,n,k\in \Bbb{Z},\quad \text{such that}\quad \|m+ni+k\lambda-z\|<\varepsilon.
$$
Define $d_1(-,-):\Bbb{C}^2\to [0,\sqrt{2}]$ to be the distance 
of the reduced values, i.e., 
if $z=a+bi$ and $w=c+di$, then 
$$
d_1^2(z,w)=(Fr(Re(z))-Fr(Re(w))^2+(Fr(Im(z))-Fr(Im(w)))^2,
$$
Where $Fr$ denotes the fractional part of a real number: $x-Fr(x)\in\Bbb{Z}$ and
$0\le Fr(x)<1$ for all $x\in \Bbb{R}$.
We can assume that $z=a+bi$ with $a,b\in[0,1)$. We will prove that there exist
$n,m,k,r\in\Bbb{Z}$ and $c,d\in\Bbb{R}$ such that
$$
d_1(k\lambda,r(c+di))=0\quad\text{and}\quad \|r(c+di)-(n+a-(m-b)i)\|<\varepsilon.
$$
Let $t_1\ge\frac{\sqrt{2}}{\varepsilon}$. Then for $t\ge t_1^2$
there exist $0\le j,j_1\le t$, $j_1\ne j$,  such that 
$d_1(j\lambda,j_1\lambda)<\varepsilon$. In fact, all values of 
$Fr(Re(j\lambda))+iFr(Im(j\lambda))$ land in $[0,1]\times[0,1]$ and 
we can partition the square into $t_1^2$ little squares of side length
$1/t_1$. So necessarily two of the $t_1^2+1$ elements have to be in the 
same little square. The greatest distance between two points in a square is the length of its diagonal, in this case $\sqrt{2}/t_1\le \varepsilon$. 
Eventually interchanging $j$ and $j_1$ we can assume that $Re(j\lambda)>Re(j_1\lambda)$. 
Now consider $c=Fr(Re((j-j_1)\lambda))$ and define 
$$
d:=\begin{cases} Fr(Im((j-j_1)\lambda)),& \text{if 
$Im((j-j_1)\lambda)>0$,}\\ Fr(Im((j-j_1)\lambda))-1,& \text{otherwise}\end{cases}
$$
 Then $d_1(r(c+di),r(j-j_1)\lambda)=0$ for all $r\in\Bbb{Z}$ and $\|c+di\|\le\varepsilon$.
Moreover, since $1,Re(\lambda),Im(\lambda)$ are l.i. over $\Bbb{Q}$, we have
$s:=\frac{d}{c}\notin \Bbb{Q}$.
Define $R:=b-sa$. Then there exist $n,m\in \Bbb{Z}$ such that $|ns+m-R|<\varepsilon/2$.
But then 
$$
\| (n+a)(1+si)-(n+a-(m-b)i)\|=\|(ns+m+as-b)i\|=\|ns+m-R\|<\varepsilon/2.
$$
On the other hand, for each $u\in\Bbb{R}$ there exists $r\in\Bbb{Z}$ such that
$\|u(c+di)-r(d+ci)\|\le\varepsilon/2$, in particular, for $u=\frac{n+a}{c}(c+di)$ 
we find an $r$ such that
$$
\varepsilon/2\ge \|r(c+di)-u(c+di)\|=\|r(c+di)-(n+a)(1+si)\|,
$$
hence $\|r(c+di)-(n+a-(m-b)i)\|<\varepsilon.$
Since for $k=r(j-j_1)$ we have $d_1(k\lambda,r(c+di))=0$, this concludes the proof.
