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I'm reading Conway's complex analysis book and I'm trying to solve the exercise 5 from page 74.

In this exercise the author asks for the radius of convergence and power series expansion of $\log z$ about $i$.

Can I use the proposition 2.5 on page 35?

When I apply this theorem I get $a_0=\frac{\pi}{2}i, a_1=\frac{1}{i}, a_2=\frac{1}{2}, a_3=\frac{-1}{3i}$, etc.

Am I right? how can I use the results of the section 2 where this exercise come from?

Similar question: Taylor series expansion of $\log[z]$ about $z=1$ (different branches)

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user42912
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$$\frac1z =\frac1{i+(z-i)}=\frac1i\frac1{1-(i-z)/i}=\frac1i\sum_{n=0}^\infty\left(\frac{i-z}{i}\right)^n$$ The radius of convergence of the series is 1 (why?) and you can integrate it term to term.

EDIT: using the Weierstrass M-test, is easy to prove that for any $r\in(0,1)$ the series is uniformly convergent in the closed disk $\{z\in\Bbb C:|z-i|\le r\}$ and it is possible to exchange integral and summation.

Martín-Blas Pérez Pinilla
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