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

Let us consider the important case in which $$\phi(n) = z^{n}$$. 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, $z^{n+1} = zz^{n} \to zl,$ and therefore $$l = zl$$, 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 $$z = r(\cos\theta + i\sin\theta)$$, where $$r$$ is positive, then $z^{n} = r^{n} (\cos n\theta + i\sin n\theta),$ 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 $$z = r(\cos\theta + i\sin\theta) = r\operatorname{Cis}\theta$$, and $$r < 1$$, we have \begin{aligned} 1 + z + z^{2} + \dots &= 1/(1 – r\operatorname{Cis}\theta), \text{or} 1 + r \operatorname{Cis}\theta + r^{2} \operatorname{Cis} 2\theta + \dots &= 1/(1 – r\operatorname{Cis}\theta)\\ &= (1 – r\cos\theta + ir\sin\theta)/(1 – 2r\cos\theta + r^{2}).\end{aligned} Separating the real and imaginary parts, we obtain \begin{aligned} 1 + r\cos\theta + r^{2}\cos 2\theta + \dots &= (1 – r\cos\theta)/(1 – 2r\cos\theta + r^{2}),\\ r\sin\theta + r^{2}\sin 2\theta + \dots &= r\sin\theta/(1 – 2r\cos\theta + r^{2}),\end{aligned} provided $$r < 1$$. If we change $$\theta$$ into $$\theta + \pi$$, 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 $$\phi(n) = r^{n} \cos n\theta$$ converges to $$0$$ when $$r < 1$$ and to $$1$$ when $$r = 1$$ and $$\theta$$ is a multiple of $$2\pi$$. Prove further that if $$r = 1$$ and $$\theta$$ is not a multiple of $$2\pi$$, then $$\phi(n)$$ oscillates finitely; if $$r > 1$$ and $$\theta$$ is a multiple of $$2\pi$$, then $$\phi(n) \to +\infty$$; and if $$r > 1$$ and $$\theta$$ is not a multiple of $$2\pi$$, then $$\phi(n)$$ oscillates infinitely.

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

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

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

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