# the composite function $G(z) = g(2z−2+i)$ is analytic in the half plane $x>1$

The following problem is from Brown Churchil's Complex Analysis book.

The function $$g(z)=√re^{iθ/2} (r > 0,−π < θ < π)$$ is analytic in its domain of deﬁnition, with derivative $$g(z) = \frac{1}{ 2g(z)}$$ . Show that the composite function $$G(z) = g(2z−2+i)$$ is analytic in the half plane $$x>1$$, with derivative $$G'(z) = \frac{1}{ g(2z−2+i)}$$ .

Suggestion: Observe that $$Re(2z−2+i) > 0$$ when $$x>1$$.

I can not understand the suggestion. I think the composite function $$G(z) = g(2z−2+i)$$ is analytic in $$x> 1$$ and $$y= -1/2$$. Because the negative axis has been excluded from the domain of $$g(z)$$.

Well first, the function is not analytic on $$y = -\frac{1}{2}$$. You solved for Im$$(2z-2+i) = 0$$, which tells you where the original function was not defined, thus you should conclude that $$f$$ is analytic on $$x > 1$$ and $$y \not = -\frac{1}{2}$$.
However, for a point $$z$$ to be on the negative real axis for the composite function, it needs to satisfy both requirements, Re$$(2z-2+i) \leq 0$$ and Im$$(2z-2+i) = 0$$. Thus after restricting to the right half plane, there are no points which hit the imaginary real axis under the map $$z\rightarrow (2z-2+i)$$.
So in general, you need to "intersect" your two constraints Re$$(z) > 0$$ and Im($$z) \not = 0$$, to find the domain of analyticity. In this case, we are intersecting the horizontal line $$y = -\frac{1}{2}$$ and the region $$x > 1$$, which is of course, just the region $$x > 1$$. Then the fact that the derivative is what they claim it is follows from a straightforward application of the chain rule.