For u we have that b u a on D Assume that u z c a for seme z 2 D Then by the

# For u we have that b u a on d assume that u z c a for

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For " u we have that " b ! " u ! " a on @D . Assume that " u ( z 0 ) = c 0 > " a for seme z 0 2 D . Then by the maximum principle " u / c 0 on D , but this contradicts that " u is continuously extended to @D with " u ! " a on @D . So " u ! " a on D , equivalently a ! u ! b on D . This show that a ! u ! b on D . 29 III.5.2 Fix n 4 1 , r > 0 and W = Ve i' . What is the maximum modulus of z n + W over the disk fj z j ! r g ? Where does z n + W attain its maximum modulus over the disk? Solution (A. Kumjian) We claim that the maximum modulus of z n + W over the disk f z 2 C : j z j ! r g is r n + V and this is attained at X = re i8 where / = (2 k, + ' ) =n for k = 0 ; 1 ; : : : ; n " 1 . By the maximum modulus principle the maximum modulus occurs on the boundary f z 2 C : j z j = r g . For z 2 C with j z j = r we have by the triangle inequality j z n + W j ! j z n j + j W j = r n + V = 5 5 ( r n + V ) e i' 5 5 = j X n + W j : so the claim is proven. 30 III.5.3 Use the maximum principle to prove that the fundamental theorem of algebra, that any polynomial p ( z ) of degree n 4 1 has a zero, by applying the maximum principle to 1 =p ( z ) on a disk of large radius. Solution It is su¢cient to show that any p ( z ) has one root, for by division we can then write p ( z ) = ( z " z 0 ) g ( z ) , with g ( z ) of lower degree. Note that if p ( z ) = a n z n + a n ! 1 z n ! 1 + # # # + a 0 ; then as j z j ! 1 , j p ( z ) j ! 1 . This follows as p ( z ) = z n 5 5 5 a n + a n ! 1 z + # # # + a 0 z n 5 5 5 : Assume p ( z ) is non-zero everywhere. Then 1 p ( z ) is bounded when j z j 4 R . Also, p ( z ) 6 = 0 , so 1 p ( z ) is bounded for j z j ! R by continuity. Thus, 1 p ( z ) is a bounded entire function, which must be constant. Thus, p ( z ) is constant, a contradiction which implies p ( z ) must have a zero. Solution (K. Seip) If p ( z ) has no zeros, then 1 =p ( z ) is an entire function. Also, if we denote by m ( R ) the maximum of 1 =p ( z ) on the circle j z j = R , then m ( R ) ! 0 when R ! 1 , unless p ( z ) is a constant. By the maximum principle j 1 =p ( z ) j ! m ( R ) when j z j ! R , which means 1 =p ( z ) = 0 . This is impossible, and so 1 =p ( z ) is not an entire function. 31 III.5.4 Let f ( z ) be an analytic function on a domain D that has no zeros on D . (a) Show that if j f ( z ) j attains its minimum on D , then f ( z ) is constant. (b) Show that if D is bounded, and if f ( z ) extends con- tinuously to the boundary @D of D , then f ( z ) attains its minimum on @D . Solution (a) If j f ( z ) j attains its minimum on D , then j 1 =f ( z ) j attains its maximum on D , which can only happens if f ( z ) is a constant, by the maximum principle. (b) Since D [ @D is compact and f ( z ) is continuous it follows that j f ( z ) j attains both maximum and minimum value on D [ @D . Assume that j f ( z ) j does not attain its minimum on @D . Then j f ( z ) j attains its minimum on D and it follows by (a) that f is constant so j f ( z ) j attains its minimum on @D . 32 III.5.5 Let f ( z ) be a bounded analytic function on the right half-plane. Suppose that f ( z ) extends continuously to the imaginary axis and satisÖes j f ( iy ) j ! M for all points iy on the imaginary axis. Show that j f ( z ) j ! M for all z in the right half-plane.  #### You've reached the end of your free preview.

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• Fall '07
• Lim
• Complex number, Logarithm, -2, Complex Plane, Branch point, Z Öxed
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