is the complexification of a finitely strictly singular operator itself FSS? Let $X$ and $Y$ be real Banach spaces, and let $X_\mathbb{C}$ and $Y_\mathbb{C}$ denote their respective complexifications.  Suppose $T:X\to Y$ is a bounded linear operator which is finitely strictly singular.
(Q1)  Is it always true that the complexification $T_\mathbb{C}:X_\mathbb{C}\to Y_\mathbb{C}$ is finitely strictly singular?
This might reduce to the same question about strictly singular operators, via the following argument.  Let $\mathcal{U}$ be an arbitrary free ultrafilter on $\mathbb{N}$.  If $T$ is FSS then so is the ultrapower $T_\mathcal{U}$.  In particular, $T_\mathcal{U}$ is strictly singular.
(Q2)  Is the complexification $(T_\mathcal{U})_\mathbb{C}$ strictly singular?
(Q3)  Is it true that $(T_\mathcal{U})_\mathbb{C}=(T_\mathbb{C})_\mathcal{U}$?
If the answer is "yes" to (Q2) and (Q3), then it will follow that $T_\mathbb{C}$ is FSS, giving an affirmative answer to (Q1).  (This is due to the fact that an operator $T$ is FSS if and only if $T_\mathcal{U}$ is SS for every free ultrafilter $\mathcal{U}$ on $\mathbb{N}$.)
Otherwise, I could try to construct a counter-example.
 A: Nevermind, I answered it myself using an elementary argument.  Thanks anyway!
EDIT:  Here is the argument, in brief.  Consider the real Banach spaces $\widetilde{X}=X\oplus_{\ell_1}X$ and $\widetilde{Y}=Y\oplus_{\ell_1}Y$.  It is a simple exercise to show that $T\oplus 0$ and $0\oplus T$ both lie in $\mathcal{FSS}(\widetilde{X},\widetilde{Y})$, and hence so does $\widetilde{T}=T\oplus T$.  Now fix any $\epsilon>0$, and let $n\in\mathbb{Z}^+$ be such that if $(e_j)_{j=1}^n$ are linearly independent vectors in $\widetilde{X}$ then we can find $(\alpha_j)_{j=1}^n\in\mathbb{R}^n$ with $\|\widetilde{T}\sum_{j=1}^n\alpha_je_j\|_{\widetilde{Y}}<\frac{\epsilon}{2}\|\sum_{j=1}^n\alpha_je_j\|_{\widetilde{X}}$.   Now let $(f_j+ig_j)_{j=1}^n$ be linearly independent in $X_\mathbb{C}$.  Then $(f_j\oplus g_j)_{j=1}^n$ is linearly independent in $\widetilde{X}$, and hence we can find $(\beta_j)_{j=1}^n$ with $\|\widetilde{T}\sum_{j=1}^n\beta_j(f_j\oplus g_j)\|_{\widetilde{Y}}<\frac{\epsilon}{2}\|\sum_{j=1}^n\beta_j(f_j\oplus g_j)\|_{\widetilde{X}}$.  However, recall that $\frac{1}{2}(\|f\|_X+\|g\|_X)\leq\|f+ig\|_{X_\mathbb{C}}\leq\|f\|_X+\|g\|_X$ for all $f,g\in X$.  It follows that $\|T_\mathbb{C}\sum_{j=1}^n\beta_j(f_j+ig_j)\|_{Y_\mathbb{C}}<\epsilon\|\sum_{j=1}^n\beta_j(f_j+ig_j)\|_{X_\mathbb{C}}$.
