notes_wk2 - COMPLEX ANALYSIS III MELISSA TACY We would now like to build up some classes of holomorphic functions(after all a class of functions with no

notes_wk2 - COMPLEX ANALYSIS III MELISSA TACY We would now...

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COMPLEX ANALYSIS III MELISSA TACY We would now like to build up some classes of holomorphic functions (after all a class of functions with no or very few members is unlikely to be that interesting or useful). We already know that f ( z ) = z m is holomorphic for any integer m and that linear combinations of holomorphic functions are also holomorphic. Therefore any polynomial P ( z ) = N X n =0 c n z n must also be a holomorphic function. It then makes sense to take that one step further and consider infinite sums. From single variable real calculus remember that we may sometimes write a function f ( x ) as f ( x ) = X n =0 c n x n such a representation is called a power series. Reminder 0.1. For any power series f ( x ) = X n =0 c n x n there exists an R 0 called the radius of convergence. For | x | < R the series is absolutely convergent and for | x | > R the series diverges. Inside the radius of convergence, | x | < R , we may integrate and differentiate the series term by term and preserve the same radius of convergence We often refer to the region | z | < R as the disc of convergence . One of the major tools we use to determine the radius of convergence for a real power series is the ratio test Reminder 0.2. (Ratio Test) A series n a n converges absolutely if lim n →∞ | a n +1 | | a n | = c < 1 and diverges if lim n →∞ | a n +1 | | a n | = c > 1 . We can construct complex power series simply by replacing x with z . That is if f ( x ) = X n =0 c n x n is a power series valid for | x | < R we can define f ( z ) by f ( z ) = X n =0 c n z n for | z | < R . We will show that we automatically have that f ( z ) is a holomorphic function. One very important example is the complex exponential (1) e z = X n =0 z n n ! . Now lim n →∞ | z | n +1 ( n + 1)! · n ! | z | n = lim n →∞ | z | n + 1 = 0 1
2 MELISSA TACY so the power series (1) is absolutely convergent for all z C . This tells us that the function f ( z ) = e z is holomorphic on the whole complex plane, we call functions with this property entire . We can bring across some more entire functions from the real variable calculus for example sin( z ) = X n =0 ( - 1) n z 2 n +1 (2 n + 1)! cos( z ) = X n =0 ( - 1) n z 2 n (2 n )! and recall that we can link the trig and exponential functions by Euler’s formula cos( z ) = e iz + e - iz 2 sin( z ) = e iz - e - iz 2 i . Of course not all power series have infinite radius of convergence. For example from the geometric summation formula we have that 1 1 - x = X n =0 x n the power series is only convergent for | x | < 1 so when we move across to the complex setting we have 1 1 - z = X n =0 z n | z | < 1 . Now the function f ( z ) = 1 1 - z is in fact defined everywhere on the complex plane except at z = 1, from the power series representation we can conclude that it is holomorphic for | z | < 1 and, when we check with the Cauchy-Riemann equations (exercise), we find that f ( z ) is also holomorphic in the region C \ { 1 } .

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