234–235. The logarithmic series

234. The logarithmic series.

We found in § 213 that $$\begin{array}{}\text{(1)}& \log (1+z)=z–\frac{1}{2}{z}^2+\frac{1}{3}{z}^3–\dots \end{array}$$ 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 $|z|\le 1$, 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 $|r|\le 1$, and $$u=1+\zeta t,$$ then $u$ will describe $C$ as $t$ increases from $0$ to $r$. And $$\begin{array}{rl}{\int }_C\frac{du}{u}& ={\int }_0^r\frac{\zeta dt}{1+\zeta t}\\ & ={\int }_0^r\{\zeta –{\zeta }^{2}t+{\zeta }^3{t}^2–\cdots +(-1{)}^{m-1}{\zeta }^m{t}^{m-1}+\frac{(-1{)}^m{\zeta }^{m+1}{t}^m}{1+\zeta t}\}dt\\ & =\zeta r–\frac{(\zeta r{)}^2}{2}+\frac{(\zeta r{)}^3}{3}–\cdots +(-1{)}^{m-1}\frac{(\zeta r{)}^m}{m}+R_m\\ & =z–\frac{z^2}{2}+\frac{z^3}{3}–\cdots +(-1{)}^{m-1}\frac{z^m}{m}+R_m,\begin{array}{}\text{(2)}& \end{array}\end{array}$$ where $$\begin{array}{}\text{(3)}& R_m=(-1{)}^m{\zeta }^{m+1}{\int }_0^r\frac{t^{m}dt}{1+\zeta t}.\end{array}$$

It follows from (1) of § 164 that $$\begin{array}{}\text{(4)}& |R_m|\le {\int }_0^r\frac{t^{m}dt}{|1+\zeta t|}.\end{array}$$ Now $|1+\zeta t|$ or $|u|$ is never less than $\varpi$, the perpendicular from $O$ on to the line $C$.¹ Hence $$|R_m|\le \frac{1}{\varpi }{\int }_0^r{t}^{m}dt=\frac{r^{m+1}}{(m+1)\varpi }\le \frac{1}{(m+1)\varpi },$$ and so $R_m\to 0$ as $m\to \mathrm{\infty }$. It follows from that $$\begin{array}{}\text{(5)}& \log (1+z)=z–\frac{1}{2}{z}^2+\frac{1}{3}{z}^3–\dots .\end{array}$$

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 $$\begin{array}{}\text{(6)}& \log \left(\frac{1}{1–z}\right)=-\log (1–z)=z+\frac{1}{2}{z}^2+\frac{1}{3}{z}^3+\dots .\end{array}$$

235.

Now $$\begin{array}{rl}\log (1+z)& =\log \{(1+r\cos \theta )+ir\sin \theta \}\\ & =\frac{1}{2}\log (1+2r\cos \theta +r^2)+i\arctan \left(\frac{r\sin \theta }{1+r\cos \theta }\right).\end{array}$$ 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 $\mathrm{am}(1+z)$ 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, $$\begin{array}{rl}\frac{1}{2}\log (1+2r\cos \theta +r^2)& =r\cos \theta –\frac{1}{2}{r}^2\cos 2\theta +\frac{1}{3}{r}^3\cos 3\theta –\dots ,\\ \arctan \left(\frac{r\sin \theta }{1+r\cos \theta }\right)& =r\sin \theta –\frac{1}{2}{r}^2\sin 2\theta +\frac{1}{3}{r}^3\sin 3\theta –\dots .\end{array}$$ These equations hold when $0\le r\le 1$, 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 $-1\le r\le 0$, 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 $$\begin{array}{rl}\log (1+z)=\log (1+\mathrm{Cis}\theta )& =\frac{1}{2}\log (2+2\cos \theta )+i\arctan \left(\frac{\sin \theta }{1+\cos \theta }\right)\\ & =\frac{1}{2}\log (4{\cos }^2\frac{1}{2}\theta )+\frac{1}{2}i\theta ,\end{array}$$ if $-\pi <\theta <\pi$, and so $$\begin{array}{rlrlrlrl}4\cos \theta & -\frac{1}{2}\cos 2\theta & & +\frac{1}{3}\cos 3\theta & & -\dots & & =\frac{1}{2}\log (4{\cos }^2\frac{1}{2}\theta ),\\ \sin \theta & -\frac{1}{2}\sin 2\theta & & +\frac{1}{3}\sin 3\theta & & -\dots & & =\frac{1}{2}\theta .\end{array}$$ 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 $$\frac{1}{2i}\log \left(\frac{1+iz}{1–iz}\right)=z–\frac{1}{3}{z}^3+\frac{1}{5}{z}^5–\dots .$$ If $z$ is real and numerically less than unity, we are led, by the results of § 231, to the formula $$\arctan z=z–\frac{1}{3}{z}^3+\frac{1}{5}{z}^5–\dots ,$$ already proved in a different manner in § 214.