= Solution
{c}
Use the same <principal square root> $S(z)=\sqrt{(z-a)/(z-b)}$, which is <holomorphic> on $A$ because its argument avoids the nonpositive <real axis>. Then
$$
\boxed{g(z)=(z-b)\sqrt{\frac{z-a}{z-b}},\qquad z\in A,}
$$
is a <holomorphic function> satisfying
$$
g(z)^2=(z-b)^2\frac{z-a}{z-b}=(z-a)(z-b).
$$
Thus it is a <holomorphic branch> of the requested <square root>. This construction does not require a global <holomorphic logarithm> of $z-a$ or $z-b$ separately. Its sign is fixed by $S(z)\to1$ at infinity, so $g(z)/z\to1$. The negative of this <function> is the other <holomorphic branch>; any two branches have a <continuous> quotient taking values in $\{1,-1\}$, which is constant because $A$ is <connected>.
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