Power series expansion for Log z

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SUMMARY

The power series expansion of Log z about the point z = i-2 is derived as Log z = (z - (i-2)) - (1/2)(z - (i-2))^2 + (1/3)(z - (i-2))^3 - ... The radius of convergence for this series is confirmed to be R = √5, which is calculated based on the distance from the center point i-2 to the origin. The analysis indicates that Log z cannot be extended analytically in any neighborhood containing the origin, reinforcing the conclusion about the radius of convergence.

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Homework Statement


Find the power series expansion of Log z about the point z = i-2. Show that the radius of convergence of the series is R = [tex]\sqrt{5}[/tex].


Homework Equations


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The Attempt at a Solution


I know that Log z = (z-1) - (1/2)(z-1)^2 + (1/3)(z-1)^3 -...
So wouldn't this become (z-3-i) - (1/2)(z-3-i)^2 + (1/3)(z-3-i)^3 -... about the point z = i-2? What would a_k be for the series? Also, since values of Log z are unbounded in any neighborhood of the origin, I would think that this function cannot be extended analytically on any open disk containing the origin. But wouldn't it be analytic on the disk of radius |i-2| = |-2+i| = Sqrt((-2)^2 + 1^2) = Sqrt(5)? Then I would say that the radius of convergence would be Sqrt(5) for the disk centered at i-2, right? I would appreciate any suggestions, thanks in advance! :-)
 
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"about the point z = i-2" means set z=ω-(i-2) and find log(z) as a power series in ω.
 

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