We have SDE $$dX_t = X_t(1-X_t)dW_t$$ where $W$ is standard Brownian motion and $X_0 = x_0 \in (0,1)$. Assume that holds $P(X_t \in (0,1))=1.$
For any $u$, we can define $f(x) = (\frac{x}{1-x})^u \sqrt{x(1-x)}$.
So first i have to prove that if we take $\lambda = u^2 - \frac{1}{4}$, for any $u$ the process $$Y_t = e^{\frac{-\lambda t}{2}}f(X_t)$$ is local martingale.
That part i did it. But now i have to take complex $u$. If i take $ui$ with real part $u$, then the process $Y_t$ is bounded and martingale. How i prove this?
If $y>0$ is a positive number and $u \in \mathbb{C}$ complex, then
$$y^u = \exp(u \log(y)).$$
Now if $u = i v$ for $v \in \mathbb{R}$, then this implies that
$$|y^u| = |\exp(i v \log(y))| \leq 1 \quad \text{for all $y>0$} \tag{1}$$
as $|e^{ix}|=1$ for any real $x$. Since the stochastic process satisfies, by assumption, $X_t \in (0,1)$ almost surely, we can use $(1)$ for $y=X_t/(1-X_t)$ to obtain that $|f(X_t)| \leq 1$ almost surely. In particular,
$$|Y_t| \leq |f(X_t)| \leq 1,$$
and this shows that $(Y_t)_{t \geq 0}$ is bounded. Since any bounded local martingale is a "true" martingale you are done if you can show that $(Y_t)_{t \geq 0}$ is a local martingale. To prove this, you can use exactly the same reasoning as you did in the real-valued case (i.e. apply Itô's formula and check that the process can be written as a stochastic integral with respect to Brownian motion).