Follow-up question to this one.
A complex lattice consists of a pair of $\mathbb{R}$-linearly independent vectors of the real-vector space $\mathbb{C}$. We call two lattices $(\lambda_1,\lambda_2)$ and $(\mu_1,\mu_2)$ equivalent if $\mathbb{Z}\lambda_1+\mathbb{Z}\lambda_2= \mathbb{Z}\mu_1+\mathbb{Z}\mu_2$. Is any complex lattice equivalent to a lattice of the form $(\lambda_1,\lambda_2)$ with ${\rm Im}(\lambda_1)\geq 0$ and ${\rm Re}(\lambda_1)\geq 0$? If so, can you additionally require that also ${\rm Im}(\lambda_2)\geq 0$ and ${\rm Re}(\lambda_2)\geq 0$?
Let us define $\lambda = \lambda_1+i\lambda_2$ and $\mu = \mu_1+i\mu_2$. Two lattices $(\lambda_1, \lambda_2)$ and $(\mu_1, \mu_2)$ are equivalent if and only if there exists an integer coefficients homography $\phi: z \mapsto {az+b \over cz+d}$, $a,b,c,d \in {\bf Z}$ such that $\phi(\lambda) = \mu$.
Lattices up to equivalence form the modular surface, which is the quotient of the Poincare half-plane by the modular group. Since this action has a fundamental domain included in the quarter plane, the answer to your question is positive.