Is every complex Banach space with Schur's property hereditarily $l^1$? An infinite-dimensional Banach space $X$ is hereditarily $l^1$ if every infinite-dimensional subspace of $X$ contains a subspace isomorphic to $l^1$.  And $X$ has Schur's property if every weakly convergent sequence in $X$ is norm convergence.  Now this 1974 journal paper shows that every real Banach space with Schur's property is hereditarily $l^1$.
My question is, is the same true for complex Banach spaces?  If not, does anyone know of a counterexample?

Yes, an infinite dimensional Banach space (real or complex) with Schur's property contains a copy of $\ell_1$.
This follows from the Rosenthal-Dor Theorem: in order that each bounded sequence in the Banach space $X$ have a weakly Cauchy subsequence, it is both necessary and sufficient that $X$ contain no isomorphic copy of $\ell_1$. 
For real Banach spaces, Rosenthal proved this in your linked paper. I believe Dor's proof for the complex case is in: 1975, On sequences spanning a complex $\ell_1$ space, Proceedings of the American Mathematical Society, 57, 515-516.
Your result is an immediate consequence of the Rosenthal-Dor Theorem and the following observation:
Let $(x_n)$ be weakly Cauchy in the Banach space $X$.  Then for any two strictly increasing sequences $(n_k)$ and $(m_k)$ of positive integers, $(x_{n_k}-x_{m_k})$ is weakly null. So then if $X$ has Schur's property, $(x_n)$ would be norm-Cauchy and thus norm convergent. 
