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Complex Analysis Question

  • Thread starter tylerc1991
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  • #1
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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.
Thank you for your help!
 

Answers and Replies

  • #2
I think we should be a little more careful with our labels,

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

Note that the new parameters [tex]R = R(r)[/tex] and [tex]\phi = \phi (r,\theta )[/tex]. This approach should work I think, where the proof may be furnished by recalling the results for part (a) which proves the analyticity of [tex]z[/tex] itself (i.e. [tex]r[/tex] and [tex]\theta[/tex] ). I have not thought it through that much, but it sounds ok to me so far.
 
  • #3
166
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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)
 

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