Math 312 Fall 2013 Final Exam Solutions (2 + i)(i + 1) = (i 1)(i + 1) = 2i i2 + i. i 2 1
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1 . (a) We have 2 + i i Math 32 Fall 203 Final Exam Solutions (2 + i)(i + ) (i )(i + ) 2i i2 + i i 2 3i i.. (b) Note that + i 2e iπ/4 so that Arg( + i) π/4. This implies 2 log 2 + π 4 i.. (c) We have 2e iπ/3 2 cos(π/3) + i sin(π/3)] 3 + i (a) We find u x (x 0, y 0 ) 2(e 2y 0 +e ky 0 ) cos(2x 0 ) so that u xx (x 0, y 0 ) 4(e 2y 0 +e ky 0 ) sin(2x 0 ) and u y (x 0, y 0 ) (2e 2y 0 + ke ky 0 ) sin(2x 0 ) so that u yy (x 0, y 0 ) (4e 2y 0 + k 2 e ky 0 ) sin(2x 0 ) which implies u xx (x 0, y 0 ) + u yy (x 0, y 0 ) 0 if and only if k Thus, the required values of k are 2 and (b) If k { 2, 2} and f() u(x, y) + iv(x, y) is assumed to be analytic, then the auchy-riemann equations imply that v(x, y) satisfies v y (x 0, y 0 ) 2(e 2y 0 + e ky 0 ) cos(2x 0 ) and v x (x 0, y 0 ) (2e 2y 0 + ke ky 0 ) sin(2x 0 ). From the first equation, we obtain v(x, y) (e 2y + 2k ) eky cos(2x) + (x) and from the second equation, we obtain v(x, y) 2 (2e2y + ke ky ) cos(2x) + 2 (y) where is a function of x only and 2 is a function of y only. following. Hence, we obtain the If k 2, then v(x, y) (e 2y e 2y ) cos(2x), and if k 2, then v(x, y) 2e 2y cos(2x), 3. Observe that if x + iy, then f() f(0) 0 ( )2 2 ( x iy. We will now show x + iy f() f(0) that does not converge as 0 by considering two paths approaching 0. First 0 consider 0 along the real axis. Thus, ( ( x iy x iy ( x. 0, y0 x + iy x 0, y0 x + iy x 0 x Now consider 0 along the y x line. Since ( ( x iy x ix 0, xy x + iy x 0 x + ix x 0 we conclude that f() is not differentiable at 0. ( i + i ( i, + i (continued)
2 Remark. If we try to take 0 along the imaginary axis, we obtain 0, x0 ( x iy x + iy y 0, x0 ( x iy x + iy y 0 ( iy. iy Thus, this function has the property that the auchy-riemann equations ARE satisfied at 0, but the function is not differentiable at Observe that f() h 3 h 2 h () where h () +, h 2 () /, and h 3 () 2. If D { : < } and D h (D), then D { : < }. Let D 2 h 2 (D ). In order to determine D 2, suppose that D and w / u + iv. Hence, < /w < w < w (u +v 2 < u 2 +v 2 u > /2 and so D 2 { : Re() > /2}. Finally, let D 3 h 3 (D 2 ) f(d) so that f(d) { : Re() < 0}. 5. (a) Since e is entire, the auchy Integral Formula implies 5. (b) If we parametrie by (t) e it, 0 t 2π, then e 2π d e eit 2π ie it dt ie dt 2πei. 0 e it 0 5. (c) The Laurent series for f() e / valid for > 0 is e d 2πie0 2πi. e / j j! This implies that d e / f() d 2πi Res(f; 0) 2πi. 5. (d) Since f() e is entire, the auchy Integral Theorem implies 6. (a) Since f() sin( i) ( 2 + )( 2 9 sin( i) ( i)( + i)( 3 ( + 3, d 0. e we conclude that i is a removable singularity, 2 0 is a simple pole, 3 i is a simple pole, 4 3 is a pole of order two, and 5 3 is a pole of order two.
3 6. (b) Since only, 2, and 3 are inside, we conclude f() d 2πi Res(f; ) + Res(f; 2 ) + Res(f; 3 )]. Since i is a removable singularity, Res(f; ) 0. Moreover, sin( i) Res(f; 2 ) ( 2 + )( 2 9 sin( i) sin(i) and which implies that Res(f; 3 ) sin( i) ( i)( 2 9 i sin(2i) f() d 2πi sin(i) ] sin( 2i) ( i)( 2i)(i 2 9) sin(2i) Note that If >, then f() ( + 4 ) /4 + / 4 4 ( ) j 4j ( ) j 4 4j so that f() + 2 ( ) j 4 4j ( ) j 6 4j + ( ) j 5 4j ] ] If { } denotes the unit circle parametried by (θ) e iθ, 0 θ 2π, then 2π 0 + sin 2 θ dθ + ( / /(2i) d 4i 2 i Note that ( is the difference of perfect squares so that ( ( 2 2)( 2 + 2). ( d. We now write 2 2 ( )( 2 ) where + 2 and 2 2, as well as ( 3 )( 4 ) where and 4 2, and note that only
4 2 and 3 are inside. By the auchy Residue Theorem, ( 2 4 d 2 ( )( 2 )( 3 )( 4 ) d ] 2 2πi ( 2 )( 2 3 )( 2 4 ) + 3 ( 3 )( 3 2 )( 3 4 ) ] 2 2πi ( 2 2)(2 2 2)(2) ( 2)( )(2 2) 2πi 8 2 ] 8 2 πi 2 2 and so 2π 0 + sin 2 dθ 4i θ πi d 4i ( π. 9. The basic error with the reasoning in the problem has to do with the definition of square root of a complex variable. If x is a non-negative real number, then x /2 is defined to be the unique non-negative real number y such that y 2 x. In other words, we define x /2 x. If is any complex variable which is not purely real with non-negative real part, then /2 describes a set, namely the set of all complex variables w such that w 2. In fact, there are always two distinct such values. Thus, the error in the problem is that ( ) /2 is being used, on the one hand to represent one of its values, and on the other hand to represent its other value; that is, the problem incorrectly writes e iπ/2 ( ) /2 e iπ/2 and deduces the contradiction instead of writing ( ) /2 {e iπ/2, e iπ/2 }. 0. In order to prove that f() has an isolated singular point at 0, note that if 0, then f() e/ sin is the ratio of an analytic function to a non-ero analytic function and is 2 therefore analytic. Hence, f() is analytic everywhere except 0 implying that f() has an isolated singular point at 0. In order to classify the isolated singular point at 0, recall from class that a function f() has a pole of order 2 at 0 if and only if f() g() 2 for some analytic function g() satisfying g(0) 0. Since f() e/ sin 2 and since g() e / sin is not analytic at 0, we conclude immediately that f() does NOT have a pole at 0. This means that 0 is either a removable singlularity or an essential singularity. In order to determine which type it is, we must consider the Laurent series for
5 f(). If > 0, then sin 2 3! + 4 5! ( ) j 2j (2j + )! and so that e/ ! + 3 k! k+ f() k0 ] ( ) j 2j ]. (2j + )! k! k+ k0 Suppose now that the Laurent series for f() is given by f() n c n n. By definition, f() has a removable singularity at 0 if c n 0 for all n < 0. If c n 0 for only finitely many n < 0, then f() has a pole at 0, whereas if c n 0 for infinitely many n < 0, then f() has an essential singularity at 0. Thus, since we already know that f() does NOT have a pole at 0, if we can show c n 0 for at least one n < 0, then we can conclude that f() has an essential singularity at 0. We will show that c 0. Basically, one needs to multiply the two series together and keep track of which products give a contribution to the term. That is, Since c c 2 3! ( ) j (2j)!(2j + )! 2! ! 4! 5 ( ) j (2j)!(2j + )!. 0, we conclude that 0 is an essential singularity. Note that the expression for c is a special case of a Kelvin function, named after Lord Kelvin (of absolute ero temperature fame), and occurs in the study of cylindrical harmonics.
= 2πi Res. z=0 z (1 z) z 5. z=0. = 2πi 4 5z
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