87-88. Applications to $z^n$ and the geometrical series

87. The limit of $z^{n}$ as $n \to \infty$ , $z$ being any complex number.

Let us consider the important case in which $ϕ (n) = z^{n}$ . This problem has already been discussed for real values of $z$ in § 72.

If $z^{n} \to l$ then $z^{n + 1} \to l$ , by (1) of § 86. But, by (4) of § 86, $z^{n + 1} = z z^{n} \to z l,$ and therefore $l = z l$ , which is only possible if (a) $l = 0$ or (b) $z = 1$ . If $z = 1$ then $lim z^{n} = 1$ . Apart from this special case the limit, if it exists, can only be zero.

Now if $z = r (\cos θ + i \sin θ)$ , where $r$ is positive, then $z^{n} = r^{n} (\cos n θ + i \sin n θ),$ so that $| z^{n} | = r^{n}$ . Thus $| z^{n} |$ tends to zero if and only if $r < 1$ ; and it follows from (10) of § 86 that $lim z^{n} = 0$ if and only if $r < 1$ . In no other case does $z^{n}$ converge to a limit, except when $z = 1$ and $z^{n} \to 1$ .

88. The geometric series $1 + z + z^{2} + \dots$ when $z$ is complex.

Since $s_{n} = 1 + z + z^{2} + \dots + z^{n - 1} = (1 - z^{n}) / (1 - z),$ unless $z = 1$ , when the value of $s_{n}$ is $n$ , it follows that the series $1 + z + z^{2} + \dots$ is convergent if and only if $r = | z | < 1$ . And its sum when convergent is $1 / (1 - z)$ .

Thus if $z = r (\cos θ + i \sin θ) = r Cis θ$ , and $r < 1$ , we have $\begin{aligned} 1 + z + z^{2} + \dots & = 1 / (1 - r Cis θ), or 1 + r Cis θ + r^{2} Cis 2 θ + \dots & = 1 / (1 - r Cis θ) \\ = (1 - r \cos θ + i r \sin θ) / (1 - 2 r \cos θ + r^{2}) . \end{aligned}$ Separating the real and imaginary parts, we obtain $\begin{aligned} 1 + r \cos θ + r^{2} \cos 2 θ + \dots & = (1 - r \cos θ) / (1 - 2 r \cos θ + r^{2}), \\ r \sin θ + r^{2} \sin 2 θ + \dots & = r \sin θ / (1 - 2 r \cos θ + r^{2}), \end{aligned}$ provided $r < 1$ . If we change $θ$ into $θ + π$ , we see that these results hold also for negative values of $r$ numerically less than $1$ . Thus they hold when $- 1 < r < 1$ .

Example XXXIII

1. Prove directly that $ϕ (n) = r^{n} \cos n θ$ converges to $0$ when $r < 1$ and to $1$ when $r = 1$ and $θ$ is a multiple of $2 π$ . Prove further that if $r = 1$ and $θ$ is not a multiple of $2 π$ , then $ϕ (n)$ oscillates finitely; if $r > 1$ and $θ$ is a multiple of $2 π$ , then $ϕ (n) \to + \infty$ ; and if $r > 1$ and $θ$ is not a multiple of $2 π$ , then $ϕ (n)$ oscillates infinitely.

2. Establish a similar series of results for $ϕ (n) = r^{n} \sin n θ$ .

3. Prove that $\begin{matrix} z^{m} + z^{m + 1} + \dots = z^{m} / (1 - z), \\ z^{m} + 2 z^{m + 1} + 2 z^{m + 2} + \dots = z^{m} (1 + z) / (1 - z), \end{matrix}$ if and only if $| z | < 1$ . Which of the theorems of § 86 do you use?

4. Prove that if $- 1 < r < 1$ then $1 + 2 r \cos θ + 2 r^{2} \cos 2 θ + \dots = (1 - r^{2}) / (1 - 2 r \cos θ + r^{2}) .$

5. The series $1 + \frac{z}{1 + z} + {(\frac{z}{1 + z})}^{2} + \dots$ converges to the sum $1 / (1 - \frac{z}{1 + z}) = 1 + z$ if $| z / (1 + z) | < 1$ . Show that this condition is equivalent to the condition that $z$ has a real part greater than $- \frac{1}{2}$ .

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85-86. Limits of complex functions and series of complex terms

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MISCELLANEOUS EXAMPLES ON CHAPTER IV

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87. The limit of zn as n→∞, z being any complex number.

88. The geometric series 1+z+z2+… when z is complex.

87. The limit of $z^{n}$ as $n \to \infty$ , $z$ being any complex number.

88. The geometric series $1 + z + z^{2} + \dots$ when $z$ is complex.