Question

Let \(f(z) = \displaystyle\sum_{n=1}^{\infty} 5^{-n}\cos(nz)\). On which one of the following domains, does \(f(z)\) represent an analytic function?

• A. \(\{z\in\mathbb{C} : |\operatorname{Im} z| < \ln 5\}\)
• B. \(\{z\in\mathbb{C} : |\operatorname{Im} z| > \ln 5\}\)
• C. \(\{z\in\mathbb{C} : |\operatorname{Re} z| < \ln 5\}\)
• D. \(\{z\in\mathbb{C} : |\operatorname{Re} z| > \ln 5\}\)

Hint:

Split \(\cos(nz)\) into \(\dfrac{e^{inz}+e^{-inz}}{2}\) and check when both resulting geometric series converge.

Answer 1

The Correct Option is A

Another way to find the domain is to apply the root test to the size of each term directly, instead of splitting $\cos(nz)$ into two exponential geometric series.

Write $z=x+iy$. Using $\cos(nz) = \cos(nx)\cosh(ny) - i\sin(nx)\sinh(ny)$, the size of $\cos(nz)$ is bounded for large $n$ by roughly $\frac{1}{2}e^{n|y|}$, since both $\cosh(ny)$ and $\sinh(ny)$ grow like $\frac{1}{2}e^{n|y|}$ as $n\to\infty$.

So the $n$-th term of the series has size

\[ \left|5^{-n}\cos(nz)\right| \approx \frac{1}{2}\left(\frac{e^{|y|}}{5}\right)^{n} \]

The root test says a series $\sum a_n$ converges absolutely when $\lim_{n\to\infty}|a_n|^{1/n} < 1$. Here that limit works out to $\dfrac{e^{|y|}}{5}$.

Setting this below 1 gives $e^{|y|} < 5$, that is $|y| < \ln 5$, that is $|\operatorname{Im} z| < \ln 5$.

When $|y|>\ln 5$, the same limit is bigger than 1, so the series diverges there instead, confirming that region cannot be the domain of analyticity.

Let's summarize:

• The size of each term depends only on $y=\operatorname{Im} z$, never on $x=\operatorname{Re} z$, through the growth of $\cosh(ny)$ and $\sinh(ny)$.
• The root test gives the same strip $|\operatorname{Im} z|<\ln 5$ as the exponential-splitting method, confirming $f(z)$ is analytic there.

So the correct domain is $\{z\in\mathbb{C} : |\operatorname{Im} z| < \ln 5\}$, option (A).

\[ \boxed{\{z\in\mathbb{C} : |\operatorname{Im} z| < \ln 5\}} \]