Request help verifying the Taylor series expansion of $\text{ln}(1 + e^{2ix})$

Question

To Prove: $\;\text{ln}(1 + e^{2ix})\; =\; \sum_{k=0}^{\infty}\dfrac{(-1)^k}{k+1}\;e^{2ix(k+1)}$

My Work:
I have worked through "Calculus Volume 1, 2nd Ed.", 1966, by Tom Apostol. I am familiar with Taylor series expansions applied to the real variable $x.$ In particular, I understand that when
$\;|x|<1, \;\dfrac{1}{x+1} = 1 - x + x^2 - x^3 + \cdots,\;$ with $\;x^n \rightarrow 0\;$ as $\;x \rightarrow \infty.$

From this I can say that when
$\;|x|<1,\; \text{ln}(1+x) \;= \;\int_0^x \dfrac{1}{1+t}dt \;= \int_0^x (1 - t + t^2 - t^3 + \cdots) dt$
$= x - \dfrac{x^2}{2} + \dfrac{x^3}{3} - \dfrac{x^4}{4} +\cdots \;= \sum_{k=0}^{\infty}\dfrac{(-1)^k}{k+1} \;x^{(k+1)}.$

I know little about complex analysis. My intuition suggests that when
complex $\;|z|<1,\;$ the same reasoning holds, so that
$\;\text{ln}(1+z) \;= \;\sum_{k=0}^{\infty}\dfrac{(-1)^k}{k+1}\;z^{(k+1)}.\;$

However, $|e^{2ix}| \;= \;|\text{cos}(2x) + i\times \text{sin}(2x)| \;= \;\sqrt{\text{cos}^2(2x) + \text{sin}^2(2x)} \;= \;1.\;$
Therefore, when complex $\;z = e^{2ix},\;$ my reasoning does not hold.

My Real Question:
Why does the formula hold even though $\;|e^{2ix}| \;= \;1?\;$ I recognize that for an in-depth understanding, I'll need to attack complex analysis from the ground up, via a book. However, all I'm looking for is an elementary (?over-simplified?) explanation that covers this particular issue.

My Research:
Searching on this forum for "complex taylor" showed many articles. I browsed a few of them but did not find any questions/comments/answers that dealt with my issue.

My Attempt To Answer The Question:
It just occured to me that the case of complex $\;|z| = 1\;$ needs special handling. In this case, it could be argued that if $\;z\neq -1,\;$ then it becomes irrelevant that $\;z^n\;$ does not approach zero, since (in the formula), $\;\frac{z^{k+1}}{k+1} \rightarrow 0\;$ as $\;k\rightarrow \infty.\;$ I am on very shaky ground here and would like a math pro to weigh in.

Answer 1

Look at the finite series $\sum_{k=0}^{m-1} x^k =\dfrac{1-x^m}{1-x} $ which is true whenever $x \ne 1$.

Putting $-x$ for $x$, $\sum_{k=0}^{m-1}(-1)^k x^k =\dfrac{1-(-1)^mx^m}{1+x} $ so $\dfrac{1}{1+x} =\sum_{k=0}^{m-1}(-1)^k x^k +\dfrac{(-1)^mx^m}{1+x} $.

Integrating this from $0$ to $x$,

$\begin{array}\\ \ln(1+x) &=\sum_{k=0}^{m-1}(-1)^k \int_0^xt^kdt +(-1)^m\int_0^x\dfrac{t^mdt}{1+t}\\ &=\sum_{k=0}^{m-1}(-1)^k\dfrac{x^{k+1}}{k+1} +(-1)^m\int_0^x\dfrac{t^mdt}{1+t}\\ \end{array} $

To show that $\lim_{m \to \infty}\sum_{k=0}^{m-1}(-1)^k\dfrac{x^{k+1}}{k+1} =\ln(1+x) $ for $|x = 1|, x\ne 1$, you need to show that $\lim_{m \to \infty}\int_0^x\dfrac{t^mdt}{1+t} =0 $ for these $x$.

Answer 2

Per my auxiliary query: Verify proof that $\lim_{m\rightarrow\infty} \int_0^z \frac{t^m}{1+t}dt = 0 : z\in\mathbb{C}, |z|=1.$, marty cohen's answer resolves all values of $z$ except for $z=-1,\;$ which is disallowed and $z = -i.$ The special case of $z = -i\;$ may be resolved manually.

$\ln(1-i) = \ln(\sqrt{2}e^{-i\pi/4}) = -i\pi/4 + \ln(\sqrt{2}).$

Let $A = \{(1/2) - (1/4) + (1/6) - (1/8) + ...\}$ and let $B = \{-(1) + (1/3) - (1/5) + (1/7) - ...\}.$

$\displaystyle \sum_{k=0}^{\infty}\frac{(-1)^k}{k+1}(-i)^{k+1} \;= \;A + iB.$

Per the "Other Work" section of this link: https://en.wikipedia.org/wiki/James_Gregory_(mathematician), $B = -(\pi/4).$

Since the special case of $z=1$ is resolved, $\displaystyle \ln(2) = \ln(1+1) = \sum_{k=0}^{\infty}\frac{(-1)^k}{k+1}(1)^{k+1} \;= \;1 - (1/2) + (1/3) - (1/4) + ... \;= 2A.$

Therefore, $A = (1/2)\ln(2) = \ln(\sqrt{2}).$