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 $g(z)$.

Can anyone please help me to mention what I am getting wrong?

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.

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