Final Exam - MATH 630: Solutions

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1 Final Exam - MATH 630: Solutions Problem. Find all x R satisfying e xeix e ix. Solution. Comparing the moduli of both parts, we obtain e x cos x, and therefore, x cos x 0, which is possible only if x 0 or x π +πk, k Z. Both parts of the equation equal at x 0, and equal i at x π + πk when k is even. On the other hand, the left-hand side is i, and the right-hand side is i at x π + πk when k is odd. Finally, the solutions are x 0 and x π + πn, n Z. Problem. Sketch each of the following sets and determine whether the given set is a domain: (a) Re + Im > 0; (b) < i 3; (c) 0: Arg 0; (d). Solution. (a) A domain; (b) Not a domain not open; (c) A domain; (d) Not a domain not connected. Problem 3. Find all the points x+iy, if any, where the function f() u(x, y) + iv(x, y) is differentiable. Find f () at these points. (a) f() x iy ; (b) f() x i y ; (c) Is any of these functions analytic? If so, find the points where f is analytic. Solution. (a) f() u(x, y) + iv(x, y), where u(x, y) x and v(x, y) y. Then u x x, u y 0, v x 0, v y y. The Cauchy Riemann equations u x v y, u y v x in this case are x y, 0 0. They are satisfied on the line y x. Therefore, f is differentiable at any point on this line. Its derivative at such a point is f () u x + iv x x + i 0 x.

2 (b)f() u(x, y) + iv(x, y), where u(x, y) and v(x, y). x y Then u x, u x y 0, v x 0, v y. The Cauchy Riemann y equations u x v y, u y v x in this case are, 0 0. They x y have no solutions, thus f is nowhere differentiable. (c) In part (a), there are no points of analyticity of f, because the points where f is differentiable do not have a neighborhood where f is differentiable. In part (b), there are no points of analyticity of f, because f is nowhere differentiable. Problem 4. Find the Maclaurin expansions of the given functions and the radii of convergence of the resulting series: (a) ( ) ; (b) ( ) (4 + ) ; (c) cos. Solution. (a) Since ( ) (( ) ), and ( ) n, with the disk of convergence <, by the theorem on differentiation of power series, ( ) n n n (n + ) n in the same disk, and with the same radius of convergence, R. (b) ( )(4 + ) 5( ) + 5(4 + ) 5( ) + 0( + (/) ) n + ( ) n n n 0 ( ) 5 + ( )n n, 0 4 n with the radius of convergence R (the smallest of the radii of convergence for the two partial fractions, or the distance from 0 to the closest singular point).

3 (c) cos + cos + cos + ( ) n n n (n)! ( ) n n n+ +, (n)! n where we used the Maclaurin expansion of cos, with the appropriate subsequent substitution. Since the function is entire, R. Problem 5. Find the radius of the convergence disk for the following series: (a) ) n n! n ; n ( n e (b) ( + cos(nπ)) n n. Solution. (a) R lim a n a n+ ( n lim e ) ( n e ) n+ n! n + (n + )! e lim ( n n + ) n. 3 (b) By the Cauchy Hadamard formula, R m lim sup am lim sup m lim k n a n (k) k. Problem 6. Does a function f() analytic in some neighborhood of 0 and satisfying ( ) f ( )n n n for all natural numbers n starting with some n 0, exist? Justify your answer.

4 4 Solution. Suppose such a function f and a neighborhood Γ exist. Let S {0,,,...,,...}. Since 4 k 0 lim k k and f( ) for all k k k S k Γ, by the uniqueness theorem f() for all Γ. Similarly, let S {0,,,...,,...}. Since 3 k+ 0 lim k k + and f( k+ ) k+ for all k k+ S Γ, by the uniqueness theorem f() for all Γ. Since only possible if 0, we get a contradiction. Therefore, such a function f cannot exist. Solution. Suppose such a function f exists. Then it must be continuous at 0 and We then have ( ) n f(0) lim f() lim f(/n) lim 0 n f (0) lim 0 f() f(0) lim f(/n) f(0) /n 0. lim ( ) n /n /n lim ( ) n which is impossible since the limit in the right-hand side does not exist. A contradiction. Problem 7. Find all isolated singular points of the given functions, determine their type, and find the residues at those points: (a) ( 3 ) ; (b) ( + )e ; (c) (e ). Solution. (a) The isolated singular points of ( 3 ) ( )(+) are 0,, and, which are simple poles. We have Res 0 ( )( + ) ( )( + ), 0 Res ( )( + ) Res ( )( + ) ( + ), Res ( )( + ) ( ). (b) ( + )e is an entire function, i.e., it has no singular points.

5 (c) (e ) has poles at eros of the denominator e, which occur at k kπi, k Z. These eros are simple because (e ) e is never ero. We have Res kπi e d d (e ). kπi ekπi Problem 8. Evaluate the integrals: (a) d, where C is the positively oriented circle C e { C: 4}; (b) 0. (x +) Solution. (a) The singular points of the integrand (poles) are the eros of the denominator e, i.e., (kπi) /, k Z. However, inside C there are only eros 0 and (kπi) /, k ±, ± (otherwise kπi > 4 6). We have Therefore, e Res 0 n n n! n n n! ( + + ). e Res 0 ( + + ). For all other 8 singular points k (we have four pairs of square roots), which are simple poles, we have C Res k e k k e k. Therefore, by the residue theorem, ( d e πi + 8 ) 0πi. 5 (b) The function has poles of order at i and i, ( +) however only i lies in the upper half plane. If C R is the upper semicircle of radius R > centered at 0 with the counterclockwise orientation, then by the Cauchy residue theorem R R (x + ) + C R d ( + ) πires i ( + ).

6 6 We have C R d ( + ) πr (R ) 0 as R. Taking into account that the function ( +) is even, we obtain 0 (x + ) lim R R (x + ) P.V. (x + ) R (x + ) πires i ( + ) πi d [ ] πi d ( + i) i ( (i) 3 ) π 4. Problem 9. Find the general form of a linear fractional transformation that maps 0 to w, and the first quadrant of the -plane onto the upper half of the unit disk in the w-plane. Solution. Let w a+b, with ad bc 0, be the linear fractional c+d transformation that is sought. Since w(0), we obtain that b d. Moreover, ad bc 0 implies that d b 0. Since 0 is at the intersection of positive real and imaginary semiaxes which are boundary lines for the st quadrant, w is at the intersection of the line segment [, ] and the unit upper semicircle which are boundary lines for the upper half of the unit disk, and since linear fractional transformations preserve the boundary orientation, we conclude that the image of the positive imaginary semiaxis is [, ] and the image of the positive real semiaxis is the unit upper semicircle. Then w( ), which implies that a c. Moreover, ad bc 0 implies that a c 0. So, our transformation has the form w a+b, or a+b w +B +B where B b a 0 Im w(iy) i. Next, for every y R we have ( iy + B iy + B iy + B ) iy + B (B + B)y iy + B, which is possible only if B + B 0, i.e., B it with t R. Thus, our transformation has the form w + it + it, t R. Moreover, for every y > 0 we must have w(iy) y + t y + t <.

7 7 Since for y > t we have y + t y + t y + t y t <, this implies that t < 0. On the other hand, if t < 0, then y + t < y t for every y > 0, so that w(iy) <. Therefore, our transformation must have the form w + it + it, t < 0. On the other hand, it is easy to verify that for every t < 0, the image of the positive real semiaxis under such a transformation is the unit upper semicircle. Indeed, for every x R we have and if x > 0 then Im w(x) i w(x) x + it x + it, ( x + it x + it x it ) xt x it x + t > 0. Problem 0. Let a be an isolated singular point of a function f(). Prove that if Re f() > 0 in some neighborhood of a then a is a removable singular point. Solution. First of all, by Picard s theorem, a cannot be an essential singular point of f because f does not take values with the negative real part in some neighborhood of a. Suppose a is a pole. Then f() g() in some neighborhood of a where g is analytic, g() 0, ( a) m Re f() > 0, and m N. Moreover, for any ɛ > 0 this neighborhood can be chosen such that arg g(a) ɛ < arg g() < arg g(a) + ɛ with some choice of a branch of arg. Let a+re iθ be in the chosen neighborhood of a. Then arg f() arg g() mθ can be made arbitrary, which is impossible. To make this argument more precise, let ɛ < π/4. Choose θ arg g(a) π m. Then arg f() arg g() arg g(a) + π ( 3π 4, 5π 4 which contradicts to the condition Re f() > 0. This contradiction implies that a is not a pole, and therefore, a is a removable singular point of f. ),