복소해석학 솔루션

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Exercises 2.1 (I suggest you solve #3, #4, #5, #6, #7, #8, #9, #12)
(In the sequel, we denote by A0 the set of all limit points of A.)
1. Prove that a neighborhood of a point on the real line (an open interval) is an open set in
R.
Solution. Let (x0 ¡ ²; x0 + ²) be any neighborhood of a point x0 2 R. For any y 2
(x0 ¡ ²; x0 + ²), let ± = ² ¡ jy ¡ x0j. Then ± > 0 and (y ¡ ±; y + ±) ½ (x0 ¡ ²; x0 + ²).
(* If z 2 (y ¡ ±; y + ±) then jz ¡ x0j · jz ¡ yj + jy ¡ x0j < ± + jy ¡ x0j = ².) ¤
2. Show that a set A of complex numbers is bounded if and only if, given z0 2 C, there exists
a real number M such that z 2 N(z0;M) for every z 2 A. Can M be chosen in dependent
of z0?
Solution. (i) ()) If A is bounded, there exists a number R > 0 such that A ½ N(0;R).
Then it is easy to check that A ½ N(z0;R + jz0j) (* jz ¡ z0j · jzj + jz0j < R + jz0j).
(() If A ½ N(z0;M) then A ½ N(0;M + jz0j). Therefore A is bounded.
(ii) M depends on jz0j. ¤
3. Show that a set of complex numbers is bounded if and only if both the sets of its real and
imaginary parts are bounded.
Solution. For any point z = x + iy 2 A, there holds
jxj; jyj · jzj · jxj + jyj: (1)
Let A1 = fx j z = x + iy 2 Ag and A2 = fy j x + iy 2 Ag.
()) If A is bounded, A ½ N(0;R) for some constant R > 0. (1) implies that jxj · R and
jyj · R for every x 2 A1 and y 2 A2. Namely, A1; A2 ½ N(0;R).

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