# Complex Analysis Question

## Homework Statement

(a) Use the polar form of the Cauchy-Riemann equations to show that:

g(z) = ln(r) + i(theta); r > 0 and 0 < (theta) < 2pi

is analytic in the given region and find its derivative.

(b) then show that the composite function G(z) = g(z^2 + 1) is analytic in the quadrant x > 0 and y > 0 and find its derivative.

## The Attempt at a Solution

Ive done part (a) and got the correct answer, but I am having some trouble with (b). The main question I have is, how do I write this composite function? I can write:
z^2 + 1 as r^2(cos(2(theta)) + 1) + ir^2(sin(2(theta))) but I dont know if that helps me.

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I think we should be a little more careful with our labels,

say we have a complex number $$z = re^{i\theta}$$, then $$z^2 = r^2e^{i2\theta}$$, but $$z^2 + 1 = \zeta$$ will be a new complex number we are examining with a different radius $$R$$ and phase $$\phi$$, we can begin with a form $$\zeta = u(x,y) + iv(x,y)$$, but the form of $$g(z)$$ is in terms of polar quantities, so it would probably be best to go back to a form $$\zeta = Re^{i\phi}$$ so you may directly insert those expressions for $$R$$, $$\phi$$ directly into $$g(z) \rightarrow g(\zeta ) = g(\R,\phi )$$ for $$r$$ and "$$\theta$$.

Note that the new parameters $$R = R(r)$$ and $$\phi = \phi (r,\theta )$$. This approach should work I think, where the proof may be furnished by recalling the results for part (a) which proves the analyticity of $$z$$ itself (i.e. $$r$$ and $$\theta$$ ). I have not thought it through that much, but it sounds ok to me so far.

What I ended up doing was using the result from part (a), in which I found the derivative of g(z) = 1/z. This implies that the derivative of g(z^2 + 1) = 2z/(z^2 +1)