## 234. The logarithmic series.

We found in § 213 that $\begin{equation*} \log(1 + z) = z – \tfrac{1}{2} z^{2} + \tfrac{1}{3} z^{3} – \dots \tag{1} \end{equation*}$ when $$z$$ is real and numerically less than unity. The series on the right-hand side is convergent, indeed absolutely convergent, when $$z$$ has any complex value whose modulus is less than unity. It is naturally suggested that the equation (1) remains true for such complex values of $$z$$. That this is true may be proved by a modification of the argument of § 213. We shall in fact prove rather more than this, viz. that (1) is true for all values of $$z$$ such that $$|z| \leq 1$$, with the exception of the value $$-1$$.

It will be remembered that $$\log(1 + z)$$ is the principal value of $$\log(1 + z)$$, and that $\log(1 + z) = \int_{C} \frac{du}{u},$ where $$C$$ is the straight line joining the points $$1$$ and $$1 + z$$ in the plane of the complex variable $$u$$. We may suppose that $$z$$ is not real, as the formula (1) has been proved already for real values of $$z$$.

If we put $z = r(\cos\theta + i\sin\theta) = \zeta r,$ so that $$|r| \leq 1$$, and $u = 1 + \zeta t,$ then $$u$$ will describe $$C$$ as $$t$$ increases from $$0$$ to $$r$$. And \begin{aligned} \int_{C} \frac{du}{u} &= \int_{0}^{r} \frac{\zeta\, dt}{1 + \zeta t} \\ &= \int_{0}^{r} \left\{\zeta – \zeta^{2} t + \zeta^{3} t^{2} – \dots + (-1)^{m-1} \zeta^{m} t^{m-1} + \frac{(-1)^{m} \zeta^{m+1} t^{m}}{1 + \zeta t}\right\} dt \\ &= \zeta r – \frac{(\zeta r)^{2}}{2} + \frac{(\zeta r)^{3}}{3} – \dots + (-1)^{m-1} \frac{(\zeta r)^{m}}{m} + R_{m} \\ &= z – \frac{z^{2}}{2} + \frac{z^{3}}{3} – \dots + (-1)^{m-1} \frac{z^{m}}{m} + R_{m}, \begin{equation*} \tag{2} \end{equation*}\end{aligned} where $\begin{equation*} R_{m} = (-1)^{m} \zeta^{m+1} \int_{0}^{r} \frac{t^{m}\, dt}{1 + \zeta t}. \tag{3} \end{equation*}$

It follows from (1) of § 164 that $\begin{equation*} |R_{m}| \leq \int_{0}^{r} \frac{t^{m}\, dt}{|1 + \zeta t|}. \tag{4} \end{equation*}$ Now $$|1 + \zeta t|$$ or $$|u|$$ is never less than $$\varpi$$, the perpendicular from $$O$$ on to the line $$C$$.1 Hence $|R_{m}| \leq \frac{1}{\varpi} \int_{0}^{r} t^{m}\, dt = \frac{r^{m+1}}{(m + 1) \varpi} \leq \frac{1}{(m + 1) \varpi},$ and so $$R_{m} \to 0$$ as $$m \to \infty$$. It follows from that $\begin{equation*} \log(1 + z) = z – \tfrac{1}{2} z^{2} + \tfrac{1}{3} z^{3} – \dots. \tag{5} \end{equation*}$

We have of course shown in the course of our proof that the series is convergent: this however has been proved already (Ex. LXXX. 4). The series is in fact absolutely convergent when $${|z|} < 1$$ and conditionally convergent when $$|z| = 1$$.

Changing $$z$$ into $$-z$$ we obtain $\begin{equation*} \log \left(\frac{1}{1 – z}\right) = -\log(1 – z) = z + \tfrac{1}{2} z^{2} + \tfrac{1}{3} z^{3} + \dots. \tag{6} \end{equation*}$

## 235.

Now \begin{aligned} \log(1 + z) &= \log\{(1 + r \cos\theta) + ir\sin\theta\} \\ &= \tfrac{1}{2} \log(1 + 2r\cos\theta + r^{2}) + i\arctan \left(\frac{r\sin\theta}{1 + r\cos\theta}\right).\end{aligned} That value of the inverse tangent must be taken which lies between $$-\frac{1}{2}\pi$$ and $$\frac{1}{2}\pi$$. For, since $$1 + z$$ is the vector represented by the line from $$-1$$ to $$z$$, the principal value of $$\operatorname{am}(1 + z)$$ always lies between these limits when $$z$$ lies within the circle $$|z| = 1$$.2

Since $$z^{m} = r^{m}(\cos m\theta + i\sin m\theta)$$, we obtain, on equating the real and imaginary parts in equation (5) of § 234, \begin{aligned} \tfrac{1}{2} \log(1 + 2r\cos\theta + r^{2}) &= r\cos\theta – \tfrac{1}{2}r^{2} \cos 2\theta + \tfrac{1}{3}r^{3} \cos 3\theta – \dots, \\ \arctan \left(\frac{r\sin\theta}{1 + r\cos\theta}\right) &= r\sin\theta – \tfrac{1}{2}r^{2} \sin 2\theta + \tfrac{1}{3}r^{3} \sin 3\theta – \dots.\end{aligned} These equations hold when $$0 \leq r \leq 1$$, and for all values of $$\theta$$, except that, when $$r = 1$$, $$\theta$$ must not be equal to an odd multiple of $$\pi$$. It is easy to see that they also hold when $$-1 \leq r \leq 0$$, except that, when $$r = -1$$, $$\theta$$ must not be equal to an even multiple of $$\pi$$.

A particularly interesting case is that in which $$r = 1$$. In this case we have \begin{aligned} \log(1 + z) = \log(1 + \operatorname{Cis}\theta) &= \tfrac{1}{2} \log(2 + 2\cos\theta) + i\arctan\left(\frac{\sin\theta}{1 + \cos\theta}\right) \\ &= \tfrac{1}{2} \log(4\cos^{2} \tfrac{1}{2}\theta) + \tfrac{1}{2}i\theta,\end{aligned} if $$-\pi < \theta < \pi$$, and so \begin{aligned} {4} \cos\theta &- \tfrac{1}{2} \cos 2\theta &&+ \tfrac{1}{3} \cos 3\theta &&- \dots &&= \tfrac{1}{2} \log(4\cos^{2} \tfrac{1}{2}\theta), \\ \sin\theta &- \tfrac{1}{2} \sin 2\theta &&+ \tfrac{1}{3} \sin 3\theta &&- \dots &&= \tfrac{1}{2} \theta.\end{aligned} The sums of the series, for other values of $$\theta$$, are easily found from the consideration that they are periodic functions of $$\theta$$ with the period $$2\pi$$. Thus the sum of the cosine series is $$\frac{1}{2} \log(4\cos^{2} \frac{1}{2}\theta)$$ for all values of $$\theta$$ save odd multiples of $$\pi$$ (for which values the series is divergent), while the sum of the sine series is $$\frac{1}{2} (\theta – 2k\pi)$$ if $$(2k – 1)\pi < \theta < (2k + 1)\pi$$, and zero if $$\theta$$ is an odd multiple of $$\pi$$. The graph of the function represented by the sine series is shown in Fig. 58. The function is discontinuous for $$\theta = (2k + 1)\pi$$. If we write $$iz$$ and $$-iz$$ for $$z$$ in , and subtract, we obtain $\frac{1}{2i} \log\left(\frac{1 + iz}{1 – iz}\right) = z – \tfrac{1}{3}z^{3} + \tfrac{1}{5}z^{5} – \dots.$ If $$z$$ is real and numerically less than unity, we are led, by the results of § 231, to the formula $\arctan z = z – \tfrac{1}{3}z^{3} + \tfrac{1}{5}z^{5} – \dots,$ already proved in a different manner in § 214.

Example XCVII
1. Prove that, in any triangle in which $$a > b$$, $\log c = \log a – \frac{b}{a} \cos C – \frac{b^{2}}{2a^{2}} \cos 2C – \dots.$

[Use the formula $$\log c = \frac{1}{2} \log(a^{2} + b^{2} – 2ab\cos C )$$.]

2. Prove that if $$-1 < r < 1$$ and $$-\frac{1}{2}\pi < \theta < \frac{1}{2}\pi$$ then $r\sin 2\theta – \tfrac{1}{2}r^{2} \sin 4\theta + \tfrac{1}{3}r^{3} \sin 6\theta – \dots = \theta – \arctan \left\{\left(\frac{1 – r}{1 + r}\right) \tan\theta\right\},$ the inverse tangent lying between $$-\frac{1}{2}\pi$$ and $$\frac{1}{2}\pi$$. Determine the sum of the series for all other values of $$\theta$$.

3. Prove, by considering the expansions of $$\log(1 + iz)$$ and $$\log(1 – iz)$$ in powers of $$z$$, that if $$-1 < r < 1$$ then \begin{gathered} \begin{alignedat}{4} r\sin\theta &+ \tfrac{1}{2}r^{2} \cos 2\theta &&- \tfrac{1}{3}r^{3} \sin 3\theta &&- \tfrac{1}{4}r^{4} \cos 4\theta + \dots &&= \tfrac{1}{2} \log(1 + 2r \sin\theta + r^{2}),\\ r\cos\theta &+ \tfrac{1}{2}r^{2} \sin 2\theta &&- \tfrac{1}{3}r^{3} \cos 3\theta &&- \tfrac{1}{4}r^{4} \sin 4\theta + \dots &&= \arctan \left(\frac{r\cos\theta}{1 – r\sin\theta}\right), \end{alignedat} \\ \begin{alignedat}{2} r\sin\theta &- \tfrac{1}{3}r^{3} \sin 3\theta + \dots &&= \tfrac{1}{4} \log\left(\frac{1 + 2r \sin\theta + r^{2}} {1 – 2r \sin\theta + r^{2}}\right),\\ r\cos\theta &- \tfrac{1}{3}r^{3} \cos 3\theta + \dots &&= \tfrac{1}{2} \arctan \left(\frac{2r\cos\theta}{1 – r^{2}}\right), \end{alignedat}\end{gathered} the inverse tangents lying between $$-\frac{1}{2}\pi$$ and $$\frac{1}{2}\pi$$.

4. Prove that \begin{aligned} {3} \cos\theta \cos\theta &- \tfrac{1}{2} \cos 2\theta \cos^{2}\theta &&+ \tfrac{1}{3} \cos 3\theta \cos^{3} \theta – \dots &&= \tfrac{1}{2} \log(1 + 3\cos^{2} \theta),\\ \sin\theta \sin\theta &- \tfrac{1}{2} \sin 2\theta \sin^{2}\theta &&+ \tfrac{1}{3} \sin 3\theta \sin^{3} \theta – \dots &&= \operatorname{arccot} (1 + \cot\theta + \cot^{2}\theta),\end{aligned} the inverse cotangent lying between $$-\frac{1}{2}\pi$$ and $$\frac{1}{2}\pi$$; and find similar expressions for the sums of the series $\cos\theta \sin\theta – \tfrac{1}{2} \cos 2\theta \sin^{2}\theta + \dots,\quad \sin\theta \cos\theta – \tfrac{1}{2} \sin 2\theta \cos^{2}\theta + \dots.$

1. Since $$z$$ is not real, $$C$$ cannot pass through $$O$$ when produced. The reader is recommended to draw a figure to illustrate the argument.↩︎
2. See the preceding footnote.↩︎