If $A$ is positive and invertible, and $B$ is Hermitian, then $A+iB$ is invertible. Exercise 14(b) from SEC. 82 of Finite-Dimensional Vector Spaces - 2nd Ed, from Paul R. Halmos.
Prove or disprove: if (operator) $A$ is positive and invertible and if (operator) $B$ is Hermitian, then $A+iB$ is invertible.
(The dimension for the underlying unitary space is not specified as finite or infinite.)
My solution requirement: if the assertion is to be established in finite dimensional spaces alone, then I request a proof in finite dimensions together with a counterexample in infinite dimensions. Else, if the assertion is to be proved in both types of spaces, then I request a proof in infinite dimensions. Else, if the assertion is to be shown as invalid in both types of spaces, then I would appreciate a counterexample in finite dimensions.

First edit:
My original (and scatterbrained) attempts in the finite dimensions were around showing that $A+iB$ is normal, and that it does not have $0$ as an Eigenvalue, thereby invertible. Once successful, the next hope was to extend the argument to infinite dimensions.
(PS: the trouble with my original approach was pointed out on this network after me posting the problem. Thereafter, I posted the second edit as follows.)

Second edit:
I am able to establish the finite-dimensional case in a relatively straightforward manner. Proof: in finite-dimensions, towards showing that $A+iB$ is invertible for the given $A$ and $B$, it is sufficient to show that $(A+iB)x = 0 \implies x = 0$. If $(A+iB)x = 0$ for some $x$, then $Ax = -iBx$. It follows that
$$
\begin{align}
(Ax, x) &= (-iBx, x) = (x, iB^*x) = (x, iBx) = (x, -Ax) = -(A^*x, x) \\
&= -(Ax, x)\\ 
&\implies (Ax, x) = 0 \implies (\sqrt A \sqrt Ax, x) = 0 \\
&\implies (\sqrt Ax, {\sqrt A}^*x) = 0 \implies (\sqrt Ax, \sqrt Ax) = 0 \implies \Vert \sqrt Ax \Vert^2 = 0 \\
&\implies \sqrt Ax = 0 \implies \sqrt A \sqrt Ax = 0 \implies Ax = 0 \\
&\implies x = 0
\end{align}
$$ since $A$ is invertible. In summary, $(A+iB)x = 0 \implies x = 0$. Thus, $A+iB$ is invertible. This proof exploits the finite-dimensional assumption on two occasions: firstly the invertibility criteria for $A+iB$, and secondly the theory of positive square-roots for positive transformations. Both exploitations aren't readily extendable to the infinite-dimensional case in my understanding. Would appreciate some pointers.
(Credits: the above argument was developed after I was rightly pointed out on this network that my earlier approach was misguided. Specifically, I was advised that $A+iB$ was not normal, and that the normality wasn't needed anyway.)
 A: Hint: Although it is not explicitly specified, I suspect (given the title of the book) that we are meant to assume that the space is finite dimensional.
One approach is as follows. First, note that $A$ has a positive square root, and $A + iB$ is invertible iff the matrix
$$
A^{-1/2}(A + iB)A^{-1/2} = I + i(A^{-1/2}BA^{-1/2})
$$
is invertible.  We note that $M = A^{-1/2}BA^{-1/2}$ is necessarily Hermitian.
Now, suppose for the purpose of contradiction that the equation $(I + iM)x = 0$ has a solution $x \neq 0$ and proceed.

Alternatively, we could note that
$$
\langle (I + iM)x, (I + iM )x\rangle = \\
\langle x,x \rangle + \langle Mx, Mx \rangle + 2 \operatorname{Re}(\langle x, iMx \rangle) =\\
\|x\|^2 + \|Mx\|^2.
$$
A: @Omnomnomnom' answer can be adapted to work on general Hilbert spaces.
$A$ is positive and invertible and hence has an invertible square root $A^{1/2}$. Therefore
$$A+iB \,\text{ is invertible} \iff A^{-1/2}(A + iB)A^{-1/2} = I + i(A^{-1/2}BA^{-1/2})\,\text{ is invertible}.$$
Moreover, $A^{-1/2}BA^{-1/2}$ is hermitian so its spectrum is contained in $\Bbb{R}$. Therefore
$$\sigma(I + i(A^{-1/2}BA^{-1/2})) = 1 + i\sigma(A^{-1/2}BA^{-1/2}) \subseteq 1+i\Bbb{R}$$
so certainly $0 \notin \sigma(I + i(A^{-1/2}BA^{-1/2}))$ and hence $A+iB$ is invertible.
