Power Series and Laurent Series Questions and Answers

Answer: C

The coefficient of \(\frac{1}{z-1}\) is 0.

Put \(w = z - 1\), so \(z = w + 1\) and the region is \(|w| \gt 1\). There \(|1/w| \lt 1\), so expand in powers of \(1/w\):

\[ \begin{aligned} f &= \frac{1}{w(w+1)} = \frac{1}{w^{2}}\cdot\frac{1}{1 + 1/w} \\ &= \frac{1}{w^{2}}\left(1 - \frac{1}{w} + \frac{1}{w^{2}} - \cdots\right) \\ &= \frac{1}{w^{2}} - \frac{1}{w^{3}} + \frac{1}{w^{4}} - \cdots \end{aligned} \]

Every term has power \(-2\) or lower, so there is no \(1/w\) term. The value 1 is the residue at \(z = 1\), the coefficient in the other expansion, valid for \(0 \lt |z-1| \lt 1\), where \(\frac{1}{w(1+w)} = \frac{1}{w}(1 - w + \cdots)\); that is the wrong region here, and \(-1\) and \(-2\) arise from neither expansion.

Answer: A

P is TRUE but Q is FALSE.

Use Rouche's theorem with the dominant term \(g(z) = -25z^{3}\). On \(|z| = 1\), \(|g| = 25\) and

\[ \begin{aligned} |f - g| &\le 1 + \frac{1}{5!} + \frac{1}{7!} + \frac{1}{9!} + \frac{1}{11!} \\ &\lt 2 \end{aligned} \]

so f has the same number of zeros as \(z^{3}\) in the unit disc: three. P is true.

On \(|z| = \tfrac{1}{2}\), \(|g| = 25/8 = 3.125\) while the remaining terms add up to less than 0.51, so f also has three zeros in \(|z| \lt \tfrac{1}{2}\). Hence all three zeros lie inside the smaller disc and none in the annulus \(\tfrac{1}{2} \lt |z| \lt 1\): Q is false. (Indeed \(f(z) \approx z(1 - 25z^{2})\) has zeros near 0 and \(\pm\tfrac{1}{5}\).)

Answer: A

f is analytic on \(\{z \in \mathbb{C} : |\text{Im}\, z| \lt \ln 5\}\).

For \(z = x + iy\), \(|\cos(nz)|\) grows like \(\frac{1}{2} e^{n|y|}\). The terms therefore behave like

\[ 5^{-n} e^{n|y|} = \left( \frac{e^{|y|}}{5} \right)^{n} \]

a geometric series that converges when \(e^{|y|} \lt 5\), that is \(|y| \lt \ln 5\), and diverges when \(|y| \gt \ln 5\). On any strip \(|\text{Im}\, z| \le r \lt \ln 5\) the convergence is uniform (Weierstrass M-test), and a uniform limit of analytic functions is analytic.

Outside the strip the series diverges, and conditions on \(\text{Re}\, z\) are irrelevant because \(\cos(nz)\) is bounded in the real direction but not in the imaginary one.

Answer: C

The Taylor series is \(z - z^{3} + z^{5} - \ldots\).

For \(|z| \lt 1\), expand \(1/(1 + z^{2})\) as a geometric series and multiply by \(z\):

\[ \begin{aligned} \frac{z}{1 + z^{2}} &= z(1 - z^{2} + z^{4} - \ldots) \\ &= z - z^{3} + z^{5} - \ldots \end{aligned} \]

The radius of convergence is 1, the distance from 0 to the poles at \(\pm i\).

\(z - z^{3}/3! + z^{5}/5! - \ldots\) is \(\sin z\); \(z - z^{3}/3 + z^{5}/5 - \ldots\) is \(\tan^{-1} z\), which is the integral of \(1/(1 + z^{2})\), not \(z/(1 + z^{2})\); and \(z + z^{3} + z^{5} + \ldots\) is \(z/(1 - z^{2})\).

Answer: D

The radius of convergence is 2.

With \(a_{n} = \dfrac{1}{(n^{2} + 1)2^{n}}\), the root test gives

\[ |a_{n}|^{1/n} = \frac{1}{2(n^{2} + 1)^{1/n}} \to \frac{1}{2} \]

since \((n^{2} + 1)^{1/n} \to 1\). So \(R = 1/(1/2) = 2\). The polynomial factor \(n^{2} + 1\) helps convergence on the boundary circle \(|z| = 2\), where the series converges absolutely, but it does not change the radius.

1/2 is the reciprocal mistake (taking the limit itself as the radius), 1 ignores the \(2^{n}\), and \(\infty\) would need the coefficients to shrink faster than any geometric sequence.

Answer: D

The radius of convergence is \( \frac{1}{4} \).

By the ratio test, with \( a_{n} = \frac{(2n)!}{(n!)^{2}} \):

\[ \begin{aligned} \frac{a_{n+1}}{a_{n}} &= \frac{(2n+2)(2n+1)}{(n+1)^{2}} \\ &\to 4 \text{ as } n \to \infty \end{aligned} \]

The series converges when \( 4|z - 3i| \lt 1 \), that is within a distance \( R = 1/4 \) of the centre \( 3i \). The centre does not affect the radius.

Infinity would need the ratio to tend to 0, as for \( \sum z^{n}/n! \). \( 1/2 \) would be the radius if the series were in powers of \( (z - 3i)^{2} \), and 2 is that value inverted; neither fits a series in \( (z - 3i)^{n} \).

Answer: A

The expansion is \(1 - (z - 1) + (z - 1)^{2} - (z - 1)^{3} + \cdots\).

Write \(z = 1 + (z - 1)\) and use the geometric series, valid when \(|z - 1| \lt 1\):

\[ \begin{aligned} \frac{1}{z} &= \frac{1}{1 + (z - 1)} \\ &= \sum_{n \ge 0} (-1)^{n}(z - 1)^{n} \end{aligned} \]

The series with all plus signs would be \(1/(2 - z)\), and the two options in powers of \(z\) are expansions about 0 of \(1/(1 + z)\) and \(1/(1 - z)\), not of \(1/z\) about 1.

Answer: C

The zero set S is closed.

S is the inverse image of the closed set {0} under the continuous function \(f\), so it is closed. More is true: because \(f\) is not identically zero, its zeros are isolated, so S is a discrete closed set with no limit point in C.

The other descriptions fail. S cannot contain an open set, since then \(f\) would vanish identically. It cannot be infinite and bounded, since a bounded infinite set has a limit point, which would force \(f \equiv 0\). And a discrete subset of C is countable. S may even be empty, as for \(e^{z}\).

Answer: D

The radius of convergence is \(\frac{1}{2}\).

The series is geometric in \(-4z^{2}\):

\[ \begin{aligned} &1 - 4z^{2} + 16z^{4} - \cdots \\ &= \sum_{n=0}^{\infty} (-4z^{2})^{n} = \frac{1}{1 + 4z^{2}} \end{aligned} \]

It converges exactly when \(|4z^{2}| \lt 1\), that is, \(|z| \lt 1/2\). Equivalently, the function \(1/(1 + 4z^{2})\) has poles at \(z = \pm i/2\), and the radius of convergence about 0 is the distance to the nearest singularity, 1/2.

The value 2 is the reciprocal error (treating 4 as the radius squared rather than its inverse), 1 ignores the coefficients altogether, and \(\infty\) would require an entire function.

Answer: B

The radius of convergence is \(e\).

With \(a_{n} = n!/n^{n}\), the ratio test gives

\[ \begin{aligned} \frac{a_{n}}{a_{n+1}} &= \frac{n!}{n^{n}} \cdot \frac{(n+1)^{n+1}}{(n+1)!} \\ &= \frac{(n+1)^{n}}{n^{n}} = \left(1 + \frac{1}{n}\right)^{n} \to e \end{aligned} \]

so \(R = \lim |a_{n}/a_{n+1}| = e\). Stirling's formula \(n! \approx \sqrt{2\pi n}\,(n/e)^{n}\) confirms it: \(|a_{n}|^{1/n} \to 1/e\).

The option \(1/e\) is the limit of \(|a_{n}|^{1/n}\), not its reciprocal. The radius is not 1, because the factorial and the power do not cancel, and not \(\infty\), because \(n!/n^{n}\) decays only like \(e^{-n}\) (times \(\sqrt{2\pi n}\)), not faster than every geometric rate.