Let $\mathfrak{U}$ be a $C^\star$-algebra. Then let $$\mathfrak{U}_{\mathbb{R}}:=\{A \in \mathfrak{U}\,\,:\,\,A=A^*\}$$
Let's create an $\mathfrak{U}\ni \tilde A$ such that, I define
$$\tilde A:=A_1+iA_2$$
where $A_1,A_2$ are both self-adjoint.
My question is, it is possible that $\tilde A$ satisfyies that $\tilde A\in \mathfrak{U}^+$ where
$$\mathfrak{U}^+ := \{A\geq 0 \,\,\, : A \in \mathfrak{U}_{\mathbb{R}}\}$$
Where $A \geq 0 \iff \sigma(A) \subseteq [0,\infty)$
To me the question is just whether it is possible to be self-adjoint, because then we will get that $\tilde A^* \tilde A = \tilde A^2 \in \mathfrak{U}^+$ because we can prove that $\mathfrak{U}^+ = \{A^2\,\,:\,\, A=A^*\,\,\,, \,\,\, A \in \mathfrak{U}\}$. But it seams to me that it is not possible that it is self-adjoint because
$$\tilde A^* = (A_1+iA_2)^* = A_1^* - iA_2^* = A_1-iA_2 \neq \tilde A$$
Is this correct line of thought?
For any $A$, you have $$ A=\text{Re}\,A+i\,\text{Im}\,A,$$ where $\text{Re}\,A$ and $\text{Im}\,A$ are the selfadjoint operators $$ \text{Re}\,A=\frac{A+A^*}2,\ \ \ \ \text{Im}\,A=\frac{A-A^*}{2i}. $$ So any $A$ is of the form $A=A_1+iA_2$ with $A_1,A_2$ selfadjoint. This decomposition is unique: if $A_1+iA_2=B_1+iB_2$ with $A_1,A_2,B_1,B_2$ selfadjoint, then $$ A_1-B_1=i(B_2-A_2).$$ The left-hand-side has real spectrum, while the right-hand-side has imaginary spectrum. So $\sigma(A_1-B_1)=\{0\}$; being selfadjoint, $A_1=B_1$. And thus $A_2=B_2$. This computation, with $B_1=A_1+iA_2$ and $B_2=0$ shows that if $A_1+iA_2$ is selfadjoint, then $A_2=0$.