# 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.

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 $$x$$.

• I am assuming that the $x$ in your answer refers to a complex variable and that in your 2nd to last line you intended ...for $|x|=1,\;x\neq -1.$ Given that, my intuition suggests that $\lim_{m\rightarrow \infty}\int_0^x \frac{t^m dt}{1+t}\;$ is in fact 0, but I don't know how to prove it; I request help here. For example, do I need to try to find the anti-derivative? Commented Oct 24, 2018 at 20:30
• my bad, I just got it. $\int_0^x \frac{t^m dt}{1+t} < \int_0^x t^m dt \;= \;\frac{x^{(m+1)}}{m+1},\;$ which does approach zero as $\;m\rightarrow\infty.$ Commented Oct 24, 2018 at 20:39

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

• In math.stackexchange.com/questions/2976199/…, zhw. and Felix Marin each provided an independent demonstration that the formula holds for any complex $z$ when $|z|=1$ and $z\neq -1.$ Thus, they each independently proved that the formula holds when $z=-i.$ On that basis, my answer above may be re-interpreted as an independent proof of James Gregory's formula, in the special case of $(-\pi/4).$ Commented Oct 29, 2018 at 22:23