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16 Theorems on Limits 49 Let us now start with the assumption that limit (1) holds. With that assumption, we know that for each positive number ε, there is a positive number δ such that |(u + iv) − (u0 + iv0 )| < ε (5) whenever 0 < |(x + iy) − (x0 + iy0 )| < δ. (6) But |u − u0 | ≤ |(u − u0 ) + i(v − v0 )| = |(u + iv) − (u0 + iv0 )|, |v − v0 | ≤ |(u − u0 ) + i(v − v0 )| = |(u + iv) − (u0 + iv0 )|, and |(x + iy) − (x0 + iy0 )| = |(x − x0 ) + i(y − y0 )| = (x − x0 )2 + (y − y0 )2 . Hence it follows from inequalities (5) and (6) that |u − u0 | < ε and |v − v0 | < ε whenever (x − x0 )2 + (y − y0 )2 < δ.

Evidently, a point z0 is not an accumulation point of a set S whenever there exists some deleted neighborhood of z0 that does not contain at least one point of S. Note that the origin is the only accumulation point of the set zn = i/n (n = 1, 2, . ). 10/29/07 3:32pm 32 Brown-chap01-v3 sec. 11 Exercises 10/29/07 33 EXERCISES 1. Sketch the following sets and determine which are domains: (a) |z − 2 + i| ≤ 1; (b) |2z + 3| > 4; (c) Im z > 1; (d) Im z = 1; (e) 0 ≤ arg z ≤ π/4 (z = 0); (f) |z − 4| ≥ |z|.

2 branch. Suppose that (x, y) is on the branch lying in the first quadrant. Then, since y = c2 /(2x), the first of equations (1) reveals that the branch’s image has parametric representation c2 (0 < x < ∞). u = x 2 − 22 , v = c2 4x Observe that lim u = −∞ and x→0 x>0 lim u = ∞. x→∞ Since u depends continuously on x, then, it is clear that as (x, y) travels down the entire upper branch of hyperbola (3), its image moves to the right along the entire horizontal line v = c2 . Inasmuch as the image of the lower branch has parametric representation u= and since c22 − y2, 4y 2 v = c2 lim u = −∞ and y→−∞ (−∞ < y < 0) lim u = ∞, y→0 y<0 it follows that the image of a point moving upward along the entire lower branch also travels to the right along the entire line v = c2 (see Fig.