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 \(-\frac{1}{2}\).


$\leftarrow$85-86. Limits of complex functions and series of complex terms Main Page MISCELLANEOUS EXAMPLES ON CHAPTER IV $\rightarrow$
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