Suppose that $A \in \mathbb{R}^{n \times n}$ is symmetric positive definite. In this case, solving $Ax = b$ with $x,b \in \mathbb{R}^{n}$ is equivalent to find \begin{align} \underset{x \in \mathbb{R}^{n}}{\arg\min}\ f(x) = \frac{1}{2} x^{\mathrm{T}}Ax - b^{\mathrm{T}}x. \end{align} So, I can use graident descent (GD) method to approximate it as: \begin{align} u_i = u_{i-1} - \alpha_{i-1}\nabla f(u_{i-1}) = u_{i-1} + \alpha_{i-1} (b - Au_{i-1}). \end{align} where $\alpha_{i-1} = 2 / (\lambda_{\max} + \lambda_{\min})$, $\lambda_{\max}$ and $\lambda_{\min}$ denote the maximum and minimum eigenvalues of $A$.
My question:
Now, consider the complex-valued case, i.e., $A \in \mathbb{C}^{n \times n}$ is Hermitian positive definite, and $x,b \in \mathbb{C}^{n}$.
I find that the same GD method can still be used to approximate $x = A^{-1}b$.
However, $f(x) = \frac{1}{2} x^{\mathrm{H}}Ax - b^{\mathrm{H}}x$ is not a function $\mathbb{C}^n \to \mathbb{R}$, as $b^{\mathrm{H}}x$ is not necessarily real-valued. So, how to explain this point?