Is the Function g(z) = ln(r) + i(theta) Analytic and What Are Its Derivatives?

[tylerc1991]

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 don't know if that helps me.
Thank you for your help!

 

[PsychoSnowman]

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.

 

[tylerc1991]

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)