Request help verifying the Taylor series expansion of $\text{ln}(1 + e^{2ix})$ 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. 
 A: 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 $x$.
A: 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, $A = (1/2)\ln(2) = \ln(\sqrt{2}).$
