Lattice generators

I hope this question isn't trivial. Let $L$ be a lattice in $\mathbb{C}$ generated by two complex numbers $w_1,w_2$ which are linearly independent over $\mathbb{R}$. Let $\gamma\in\mathbb{C}$ be a root of unity such that $\gamma\notin\mathbb{R}$ and $\gamma L=L$, and let $l\in L\backslash\{0\}$ be an element of shortest length. Is it true that the set $\{l,\gamma l\}$ generates $L$? It seems almost trivial, but I have been unable to prove it. Thanks!

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Since $\gamma L=L$, we have $\gamma w_1=aw_1+bw_2$, $\gamma w_2=cw_1+d w_2$ with integer $a,b,c,d$, which immediately implies that $\gamma$ is a root of a polynomial of degree $2$ with integer coefficients, so the root of unity in question can be of degree $3,4,6$ only (otherwise the cyclotomic polynomial is irreducible over $\mathbb{Z}$ and is of degree greater [or smaller] than $2$). In each of these cases the lattice generated by $l$ and $\gamma l$ satisfies the following property: for each point $A$ inside the fundamental parallelogram of that lattice there is a vertex $X$ of that parallelogram such that the distance between $A$ and $X$ is less than the length of $l$. Now, if our lattice is not $L$, take a point not in $L$ and bring it into the fundamental parallelogram by subtracting an integral combination of $l$ and $\gamma l$, thus finding a shorter element of $L$.
I don't see why that should be... What if $l=1$ and $\gamma=i$. Then $x=0.8+0.8i$ (for example) is in the fundamental parallelogram, but $|xi|=|x|=1.28$, $|x\cdot1|=1.28$ and $|x(1+i)|=2.56$; none of these are greater than 1, which is the length of $l$. – rfauffar May 17 '11 at 14:36