MA30056: Complex Analysis. Exercise Sheet 7: Applications and Sequences of Complex Functions

Transcription

1 MA30056: Complex Analysis Exercise Sheet 7: Applications and Sequences of Complex Functions Please hand solutions in at the lecture on Monday 6th March..) Prove Gauss Fundamental Theorem of Algebra. Hint: Suppose that a polynomial pz) has no zeroes and conclude that fz) pz) is bounded.) Solution: We proof this by contraposition, i.e., since we want to proof that a nonconstant polynomial has at least one zero, we actually show that a polynomial without a zero is constant. Suppose the polynomial pz) n k0 a kz k, a n 0, has no zeroes. Then fz) pz) defines an entire function. We wish to show that f is bounded: i) Since pz) a nz n + a n + + a 0 a n z a n as z, there is R > 0 so that z n z > R pz) a n z n. Therefore, we have R > 0 so that z > R fz) a n z n a n R n. ii) Since B R 0) is compact and f is continuous, there is M R so that z R fz) M. Thus f is bounded by max{ a n R n, M}). Hence, by Liouville s Theorem, f is constant and so is p..) Evaluate z e iαz z z 0 ) dz for α > 0 and i) z 0 <, ii) z 0 >. Solution: ii) If z 0 > then z eiαz z z 0 is holomorphic in the interior B ) 0) of the circle {z z }. Hence, by Cauchy s Theorem, e iαz dz 0. z z 0 ) z

2 i) If z 0 < then we apply Cauchy s formula for the first derivative to obtain e iαz πi d dz z z 0 )! dz eiαz i απ e iαz 0 απ e iαz 0. zz0 z 3.) Noting that for the multifunction log on the cut plane C \ R 0 we have log z log z + i Arg z + k πi and log z) log z + i Arg z + k + ) πi where k Z, find the mistake in the following argument: log z) ) log z ) log z) + log z) log z + log z log z) log z log z) log z. Solution: First, we note that log z log z + i Arg z + k πi for k Z is the sloppy way to actually say log z {log z + i Arg z + k πi k Z} so, instead of a single value as a function, a multifunction associates a whole set of values to a point). We look at a little example: let z e iπ/0. Then we have z e iπ/0 e iπ /0 e iπ 9/0, z e iπ/5 and z) e iπ /5 e iπ/5. So, we have the following logarithms note that all these numbers have modulus ): log z {iπ 0 } + πi k k Z, log z) + πi k k Z, and log z + πi k k Z log z). So, certainly the first line in the above argument holds. For the second line, we have to make sense of log z)+log z) and also log z+log z) in this multifunction case: The values of a sum are obtained by adding each value of one set to each value of the other set, i.e., log z) + log z) + πi k k Z + + πi l l Z { iπ } + πi k + l) k, l Z 0 ) { iπ 95 + πi m m Z } m m+ log z), {iπ 5 + πi m m Z } )

3 where in ) we make use of the fact that the sum of two integers is again an integer and that every integer can be written as sum of two integers, i.e., that Z + Z Z). A similar calculation shows that log z + log z log z, so everything is fine by going from the first to the second line. Now, look at the third line: log z) + πi k k Z { iπ 95 } + 4πi k k Z + πi n ) n Z, and comparing this set with the one in ) above, we see that only the values with odd m are in this set. Similarly, we have log z {iπ 0 } + πi k k Z + 4πi k k Z + πi ñ) ñ Z, and here only the values of ) with even m are present. So, the third line is false the sets on the left and right are not equal) and the mistake occurs in going from the second to the third line. This holds generally: The mistake in the above argument occurs in going from the second to the third line! The sum log z) + log z) cannot be replaced by log z), since the sum in question is obtained from the set of values log z) multifunction!) by adding each of these values to itself and to all the other values of the set log z), whereas the set log z) is obtained by simply doubling all the numbers log z), i.e., by adding each such number to itself only. Therefore log z) + log z) log z), and similarly, log z + log z log z. Lesson learnt: Be careful with multifunctions;-) Remark: The deal with multifunctions is that we look at all possible values that, e.g., an inverse function like the logarithm or some root might have) at once! The following applet by Terence Tao nicely visualises what that is about the principal branch is coloured red on the right watch out how the colour jumps if you go in circles around the origin) By the way, Tao has won the Field s medal in 006 the highest honor in mathematics), mainly for proving the existence of arbitrarily long arithmetic progressions of prime numbers. This fact is a converse to the Classical Dirichlet Theorem which claims that there are infinite numbers of primes in certain arithmetic progressions.

4 4.) Calculate all possible values of i i. What is the principal value of i i? Do the same for i 7/0. State without proof) under which conditions the power b a is single-valued, has finitely many values or has infinitely many values. Solution: i i e i log i e i log i +i argi)+πik) e i 0+i π +πi k) e π π k, where k Z note that these are infinitely many real values). The principal value is i i e i Logi) e π. i 7/0 e 7 log i/0 e 7 log i +i argi)+πik)/0 e 7 0+i π +πi k)/0 e 7iπ 0 e 7 πi k 0 { } where k Z. However, note that the set e 7 πi k 0 k Z is exactly the set of all 0th roots of unity and thus of cardinality 0), compare with Question on Exercise sheet. The principal value is e 7iπ i. Looking closer at these two examples and the examples given in the lecture and/or on Self-assessment sheet 7) suggest the following characterisation: For b 0, the power b a is single-valued iff a Z, has finitely many values iff a Q if a p/q is given in lowest terms, two such values are related via a factor of a qth root of unity), and has infinitely many values iff a C \ Q. For a proof of this claim, see [I. Stewart & D. Tall: Complex Analysis; Section 4.6]. 5.) Prove that the limit function of a locally uniformly convergent sequence of continuous functions is continuous i.e., Theorem IV..). Solution: Let f n : D C be continuous and suppose f n f locally uniformly on D. We want to show that f : D C is continuous. This is done by the usual M argument with a slight twist. Fix ε > 0. Fix z 0 D and ϱ > 0 so that f n f on B ϱ z 0 ) D by local uniform convergence). Then there is N N so that whenever n N. z B ϱ z 0 ) : f n z) fz) < ε 3 Since f N is continuous on B ϱ z 0 ) D there is δ > 0 so that for any z B ϱ z 0 ). Putting these together we obtain z z 0 < δ f N z) f N z 0 ) < ε 3 fz) fz 0 ) fz) f N z) + f N z) f N z 0 ) + f N z 0 ) fz 0 ) < 3 ε 3 ε whenever z z 0 < min{δ, ϱ}, showing that f is continuous at z 0 D.

5 Optional questions: 6.) Determine the inverse function arccos of the complex cosine cos z ei z + e i z ). Hence calculate all values arccos3/). Which one of these is the principle value? Solution: We want to a function fz) such that cos fz) z for all z C, i.e., z e i fz) + e i fz)). Multiplying the last equation by e i fz), we have to solve the equation ) e ifz) z e i fz) + 0 for fz). This yields e i fz) z + z ) note that the square root is a multifunction with in general) two values, otherwise you might prefer to write this as e i fz) z ± z ) ) and therefore fz) i log z + z ) ) where now also log is a multifunction). We check that fcos z) z cos fz) for all z C) wherefore this is indeed the complex arccos. We use Eq. ) to calculate arccos3/): 3 arccos3/) i log 3 ) + i log ) i 3 + log ± 5 ) ) ) ) i log + πik ) 3 ± i log ) 5 3 ± ) 5 + πk with k Z after the step ) the function log denotes the usual real logarithm, before that ) step the multi-valued complex logarithm). The principal value is i log i Remark: We continue our remark on multifunctions after Question 3 here. Instead of introducing multifunctions, one can define the complex logarithm and then also the roots, the above mentioned arccos, etc.) over an appropriate Riemann surface M; that is, instead of defining the logarithm log : C \ {0} C as )

6 multifunction, the logarithm is defined by log : M C over an appropriate Riemann surface M. For the complex logarithm, countably but infinitely many copies of C named C k with k Z in the upper part of the figure to the right) of the cut plane are pasted together at the respective cut to yield this Riemann surface M and allow the logarithm to vary continuously, see picture to the right. This figure is taken from [I. Stewart & D. Tall: Complex Analysis; Fig. 4.8]. 7.) Let g : D A be continuous and f n f locally uniformly on A. Prove that f n g f g is locally uniformly on D i.e., Lemma IV..3). Solution: Let z D and denote w gz); fix r > 0 so that f n f on B r w). Since g : D A is continuous Now take ε > 0. Since f n f locally uniformly on A ϱ > 0 w B ϱ z) : gw) B r w). N N n N ω B r w) : f n ω) fω) < ε. Therefore we have N N so that n N w B ϱ z) : f n gw)) fgw)) < ε, that is, we have ϱ > 0 so that f n g f g on B ϱ z). 8.) Show: f n z) converges pointwise but not uniformly on C \ N to the zero z n function. However, the sequence f n converges uniformly on B 0) to the zero function. Solution: The sequence of functions f n ) n converges pointwise to the zero function on C \ N since, for any ε > 0, we have z n < n z, if n > z, < ε, if n > ε + z.

7 It follows that lim n 0 if z C \ N. z n In other words, we have shown that it is possible to find a number N dependent on z) such that, for any ε > 0, we have /z n) < ε if n > N. However, for a given ε > 0, there is no N N such that z n ε if n > N for all z C \ N: For a given m N, the set C \ N contains points z for which z m < e.g., ε z m + ) and thus > ε. ε z m The sequence of functions f n ) n converges uniformly to the zero function on the open unit disk B 0), since, given any ε > 0, we have z n <, if n >, n < ε, if n > ε +. We can therefore choose N to be the smallest natural number greater than /ε)+. Notice that N depends on ε, but not on z.