Let f(z)=e1/z on C\0\ and R n=\x+iy: x <1/n, y <1/n\\0\. Bars denote closure in C. Which relation is stated correctly?

$\overline{f(R_n)\setminus f(R_{n+1})}=\overline{f(R_n\setminus R_{n+1})}$

$\overline{f(\bigcap_{n=1}^\infty R_n)}=\bigcap_{n=1}^\infty\overline{f(R_n)}$

GATE 2020 MA – Question 16

Complex Analysis · Power Series and Singularities · 1 mark · Multiple choice

Let $f(z)=e^{1/z}$ on $\mathbb C\setminus\{0\}$ and $R_n=\{x+iy:|x|<1/n,|y|<1/n\}\setminus\{0\}$. Bars denote closure in C. Which relation is stated correctly?

$\overline{f(R_{n+1})}\ne\overline{f(R_n)}$
$\overline{f(R_n)\setminus f(R_{n+1})}=\overline{f(R_n\setminus R_{n+1})}$
$\overline{f(\bigcap_{n=1}^\infty R_n)}=\bigcap_{n=1}^\infty\overline{f(R_n)}$
$\overline{f(R_n)}=\overline{f(R_{n+1})}$

Practise this question in The GATE Grind →

Show answer and explanation

Correct answer: (A) $\overline{f(R_{n+1})}\ne\overline{f(R_n)}$; (D) $\overline{f(R_n)}=\overline{f(R_{n+1})}$

Explanation

Every punctured neighbourhood of zero has image dense in C under $e^{1/z}$. Explicitly, for w ≠ 0 choose $z_k=1/(\log w+2\pi ik)$, tending to zero with f(zₖ) = w. Zero is also a limit along negative real z. Hence every displayed image closure is C, proving D and disproving A as printed. The appended official key accepts A OR D; both historical alternatives are preserved, while D is the mathematically valid printed relation. Since the nested domain intersection is empty, C is false.

GATE 2020 MA – Question 16

Explanation

More Power Series and Singularities questions