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Unformatted text preview: Complex Variables Notes for Math 703. Updated Fall 2011 Anton R. Schep CHAPTER 1 Holomorphic (or Analytic) Functions 1. Definitions and elementary properties In complex analysis we study functions f : S → C , where S ⊂ C . When referring to open sets in C and continuity of functions f we will always consider C (and its subsets) as a metric space with respect to the metric d ( z 1 ,z 2 ) =  z 1 z 2  , where  ·  denotes the complex modulus, i.e.,  z  = p x 2 + y 2 whenever z = x + iy with x,y ∈ R . An open ball with respect this metric will be also referred to as an open disc and denoted by B ( a,B ( a,r ) = { z ∈ C :  z a  < r } , where a is the center and r > 0 is the radius of the open ball. The closed disc with center a and radius r is denoted by B ( a,r ), so B ( a,r ) = { z ∈ C :  z a  ≤ r } . Recall that G ⊂ C is called open if for all a ∈ G there exists r > 0 such that B ( a,r ) ⊂ G . If z = x + iy , then the conjugate z of z is defined by z = x iy . Now z z =  z  2 , so that 1 z = z  z  2 for z 6 = 0. Elementary properties of complex numbers are given by: (1) The real part Re z of z satisfies Re z = 1 2 ( z + z ), while the imaginary part Im z of z is given by Im z = 1 2 i ( z z ). (2) For all z 1 ,z 2 ∈ C we have z 1 + z 2 = z 1 + z 2 and z 1 z 2 = z 1 z 2 . (3) For all z 1 ,z 2 ∈ C we have  z 1 z 2  =  z 1  z 2  . 2. Elementary transcendental functions Recall also that if z = x + iy 6 = 0, then, using polar coordinates, we can write z = r cos θ + ir sin θ . In this case we write arg z = { θ + 2 kπ : k ∈ Z } . By Arg z we will denote the principal value of the argument of z 6 = 0, i.e. θ = Arg z ∈ arg z if π < θ ≤ π . Note that if z 1 =  z 1  (cos θ 1 + i sin θ 1 ) and z 2 =  z 2  (cos θ 2 + i sin θ 2 ), then we have z 1 z 2 =  z 1  z 2  (cos θ 1 cos θ 2 sin θ 1 sin θ 2 + i (sin θ 1 cos θ 2 + cos θ 1 sin θ 2 )) =  z 1 z 2  (cos( θ 1 + θ 2 ) + i (sin( θ 1 + θ 2 )). Hence we have arg ( z 1 z 2 ) = arg z 1 + arg z 2 . Define now e z = e x (cos y + i sin y ). Then  e z  = e x and arg e z = y + 2 kπ . In particular e 2 πi = 1 and the function e z is 2 πiperiodic, i.e., e z +2 πi = e z e 2 πi = e z for all z ∈ C . We want now to define log w such that w = e z where z = log w , but we can not define it as just the inverse of e z as e z is not onetoone. Consider therefore the equation w = e z for a given w . We must assume that w 6 = 0 as e z 6 = 0 (and thus log 0 is not defined). Then  w  =  e z  = e x and y = Arg w +2 kπ ( k ∈ Z ). Hence { log  w  + i (Arg w + 2 kπ ) : k ∈ Z } is the set of all solutions z of w = e z . We write log w for any w in the set { log  w  + i (Arg w + 2 kπ ) : k ∈ Z } ....
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 Fall '11
 Schep
 Topology, Derivative, Sets, Continuous function, Open set, g.

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