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1 United Arab Emirates University ollege of Sciences Department of Mathematical Sciences FINAL EXAM { SOLUTION omplex Analysis I MATH 5 SETION 0 RN 56 9:0 { 0:45 on Monday & Wednesday Date: Wednesday, January 6, 00 ID No: Solution Name: Solution Score: 70/70

2 omplex Analysis I FINAL EXAM { SOLUTION Fall, 009. (5 points) Express ( + i p ) 9 in the form of x + iy. Answer. The number + i p has the exponential form + i p j + i p je i e i ; ( + i p )9 9 e i 9 9 e i 9 ( ) 9 :. (5 points) Use the real and imaginary parts of f(z) e z to nd the real and imaginary parts of g(z) e f(z) exp (e z ). Answer. Euler's formula implies e z e x+iy e x e iy e x (cos y + i sin y) e x cos y + ie x sin y: That is, Re f(z) Re (e z ) e x cos y; Im f(z) Im (e z ) e x sin y: Since f(z) Re f(z) + i Im f(z), so we have Therefore, we get g(z) e f(z) e Re f(z)+i Im f(z) e Re f(z) i Im f(z) e e Re f(z) [cos (Im f(z)) + i sin (Im f(z))] e Re f(z) cos (Im f(z)) + ie Re f(z) sin (Im f(z)) e ex cos y cos (e x sin y) + ie ex cos y sin (e x sin y) : Re g(z) e ex cos y cos (e x sin y) ; Im g(z) e ex cos y sin (e x sin y) : Page of 7 Points Earned: 0/0

3 omplex Analysis I FINAL EXAM { SOLUTION Fall, 009. (Total 0 points) onsider f(z) z Log z, where Log z is the principal branch. ( + i)z (.) (5 points) Find all singular points. Answer. f(z) has the singular points at z ( + i)z z(z ( + i)) 0, i.e., z 0 or z ( + i) 0. We solve the latter one z + i. Using the exponential form, r e i z + i p e i( +n) 4 ; (n 0; ; ; : : :) r p ; 4 + n; r 4 ; 8 + n z 4 e i( 8 +n) ; c 0 4 e i 8 : i0:45509; c 4 e i 9 8 :09868 i0:45509: Thus, all singular points are z 0, z 4 e i 8 and z 4 e i 9 8. (.) (5 points) ompute f( + i). Answer. Since z ( + i)z z(z ( + i)), f( + i) Log( + i) ( + i)(( + i) ( + i)) Log( + i) ( + i) ( + i ) ln + i4 ln 4 i 8 : Log( + i) Page of 7 Points Earned: 0/0

4 omplex Analysis I FINAL EXAM { SOLUTION Fall, (5 points) Let S be the region [0; ] [0; ]. Find the image S 0 of S under the mapping w e zi and sketch the region S 0. Answer. The region S n z (x; y) j 0 x ; 0 y o is bounded by four lines x 0, x, y 0 and y. We look for the images of four points (0; 0), (0; ), (; 0) and (; ) under the map w e zi : w e zi e (x+iy)i e y+ix e y e ix. Point (0; 0): It is mapped to w e 0 e i0. Point (0; ): It is mapped to w e e i0 e. Point (; ): It is mapped to w e e i e cos 4 Point (; 0): It is mapped to w e 0 e i e i cos + ie sin + i sin e + i p e. + i p 5 The line segment from (0; 0) to (0; ) is mapped to w e y e i0 e y with 0 y, which is on the real axis. 6 The line segment from (0; ) to (; ) is mapped to w e e ix with 0 x, which is a part of the circle of radius e and centered at the origin. 7 The line segment from (; ) to (; 0) is mapped to w e y e i with 0 y, which is a ray of length and with angle. 8 The line segment from (; 0) to (0; 0) is mapped to w e 0 e ix e ix with 0 x, which is a part of the circle of radius and centered at the origin. S 0 w (u; v) j e p v u + v ; 0 tan u w e i j e ; 0. : y v i + i 4 0 e i S O 4 x 0 e +i O 0 e S 0 0 u 5. (5 points) Let a function f be analytic everywhere in a domain D. Prove that if f is real{valued for all z in D, then f must be constant throughout D. Proof. Since f(z) u(x; y) + iv(x; y) is real{valued, v(x; y) 0 for all z in D. Because f is analytic in D, it should satisfy the auchy{riemann equations in D, u x v y 0; u y v x 0; i:e:; u x 0 u y : It implies u(x; y) is constant and thus f is constant in D. Page of 7 Points Earned: 0/0

5 omplex Analysis I FINAL EXAM { SOLUTION Fall, (5 points) Prove that u(x; y) log(x + y ) is harmonic on f0g. Proof. Let f(x; y) x + y. Then u(x; y) log(x + y ) log(x + y ) log f(x; y): We need to deduce u xx + u yy 0. u x f x f ; u xx f xxf f x f xxf f x 4f f u y f y f ; u yy f yyf f y f yyf f y 4f f u xx + u yy f xxf + f yy f f x f y f (f xx + f yy) f f x f y f : Since f(x; y) x + y, so f x x and f y y and f xx f yy. Putting them into the last equation above, we get u xx + u yy (f xx + f yy ) f f f x y ( + ) f (x) (y) f f 4f 4 (x + y ) f 4f 4f f 0: 7. (5 points) Find the principal value of ( + i). Answer. We rewrite the number as ( + i). The principal value of ( + i) is obtained by P: V: ( + i) e Log(+i) e (ln j+ij+i Arg(+i)) e (ln +i 4 ) e 4 ln i 8 e 4 ln cos i sin cos i sin 0: i0:797: 8 8 Page 4 of 7 Points Earned: 0/0

6 omplex Analysis I FINAL EXAM { SOLUTION Fall, (5 points) Use the denition of the complex exponent, prove that (e z ) w e zw for complex numbers z and w. Proof. Let c e z e x+iy e x e iy. Then we have jcj je x e iy j e x ; and arg c arg e x e iy y + n: The left{hand side of the given equation becomes (e z ) w c w and by the denition of the complex exponent, (e z ) w c w e w log c e w(ln jcj+i arg c) e w(ln ex +i(y+n)) e w(x+i(y+n)) e wz e zw : ( i)( + e 9. ) (5 points) By solving the integral, show that e i Log z dz, where is the semicircle z e i, 0 and Log z is the principal branch. Proof. Putting z() e i with 0 into the integral, we have e i Log z dz A simple computation shows 0 e i Log ei ie i d i 0 e i+i Log ei d: Log e i ln je i j + i Arg e i ln + i i; i Log e i : So the integral above becomes e i Log z dz i e i+i Log ei d i 0 i 0 i i i i e i d 0 h e (i ) d i e (i )i i 0 h )0i i e (i ) e (i h e i e i h e i i( + e ) i( + i)( + e ) i i ( i)( + e ) : Page 5 of 7 Points Earned: 0/0

7 omplex Analysis I FINAL EXAM { SOLUTION Fall, (5 points) Let be the boundary of the triangle with vertices at the points 0, i and 4 oriented in the counterclockwise direction (see gure). Then show that (e z z) dz 60: Answer. We recall the formula that for a piecewise continuous f on, where M is the maximum modulus of f(z) for z on and L is the total length of. On the given, f(z) e z z has the maximum modulus jf(z)j je z zj je z j + jzj e x + jzj e : f(z) dz ML, Since the total length of is , thus by the formula, we have f(z) dz (e z z) dz 5 60: y i 4 O x. (5 points) Evaluate the contour integral! dz, where is the positively ori- z ented circle jzj. Answer. We observe z + f(z) z +! z + z z z + z (z ) + (z ) : The singularity of f lies at z which is outside the circle jzj. So f is analytic inside and on the simple closed contour jzj. The auchy{goursat Theorem implies f(z) dz z +! dz 0: z Page 6 of 7 Points Earned: 0/0

8 omplex Analysis I FINAL EXAM { SOLUTION Fall, 009. (5 points) Evaluate the contour integral jz ij. e z dz, where is the positively oriented circle (z i) Answer. Let f(z) e z. Then f is entire, i.e., analytic everywhere in the whole complex plane. Since z i is inside the simple closed contour jz ij, so the auchy Integral Formula (for Derivatives) implies e z We compute derivatives of f: (z i) dz f(z) i dz f () (i) if () (i): (z i)! f 0 (z) ze z zf; f 00 (z) (f + zf 0 ) f + z f + z f; f 00 (i) + i f(i) f(i) e i e : Therefore, we conclude e z (z i) dz if () (i) e i i e :. (5 points) Let f be an entire function such that jf(z)j for all z. Prove that f is constant on the whole complex plane. Proof. Since jfj for all z, so f 6 0 at any z. Let g, which is dened for all z. Then f we observe, for all z, jgj jfj ; because jfj. That is, g is bounded on the whole complex plane. Since f is entire, so g is also entire. By Liouville's Theorem, the bounded entire function g is constant on the whole complex plane. Since g f, thus f is also constant on the whole complex plane. Page 7 of 7 Points Earned: 0/0

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