By Alain Guichardet

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**Example text**

1. If f and g are diﬀerentiable at a point z0 , then (i). (f ± g) (z0 ) = f (z0 ) ± g (z0 ), (ii). (cf ) (z0 ) = cf (z0 ) (c is a constant), (iii). (f g) (z0 ) = f (z0 )g (z0 ) + f (z0 )g(z0 ), f g(z0 )f (z0 ) − f (z0 )g (z0 ) (z0 ) = if g(z0 ) = 0, and g (g(z0 ))2 (v). (f ◦ g) (z0 ) = f (g(z0 ))g (z0 ), provided f is diﬀerentiable at g(z0 ). (iv). 2. If f is diﬀerentiable at a point z0, then f is continuous at z0 . A function f of a complex variable is said to be analytic (or holomorphic, or regular) in an open set S if it has a derivative at every point of S.

If f is continuous at z0 , then given > 0 there exists a δ > 0 such that |z1 − z0 | < δ/2 ⇒ |f (z1 ) − f (z0 )| < /2 and |z2 − z0 | < δ/2 ⇒ |f (z2 ) − f (z0 )| < /2. But then |z1 − z2 | ≤ |z1 − z0 | + |z0 − z2 | < δ ⇒ |f (z1 ) − f (z2 )| ≤ |f (z1 ) − f (z0 )| + |f (z2 ) − f (z0 )| < . For the converse, we assume that 0 < |z−z0 | < δ, 0 < |z −z0 | < δ; otherwise, we can take z = z0 and then there is nothing to prove. Let zn → z0 , zn = z0 , and > 0. There is a δ > 0 such that 0 < |z − z0 | < δ, |z − z0 | < δ implies |f (z) − f (z )| < , and there is an N such that n ≥ N implies 0 < |zn − z0 | < δ.

2 − 3i)(−2 + i), (c). (1 − i)(2 − i)(3 − i), 4 + 3i 1+i i 1 + 2i 2 − i (d). , (e). + , (f). + , 3 − 4i i 1−i 3 − 4i 5i √ −10 (g). (1 + 3 i) , (h). (−1 + i)7 , (i). (1 − i)4 . 2. Describe the following loci or regions: (a). |z − z0 | = |z − z 0 |, where Im z0 = 0, (b). |z − z0 | = |z + z 0 |, where Re z0 = 0, (c). |z − z0 | = |z − z1 |, where z0 = z1 , (d). |z − 1| = 1, (e). |z − 2| = 2|z − 2i|, z − z0 (f). = c, where z0 = z1 and c = 1, z − z1 (g). 0 < Im z < 2π, Re z (h). > 1, Im z < 3, |z − 1| (i).