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Unformatted text preview: 148 0.5 0.25 0 1 -0.25 -0.5 -1 -1 0.5 0 -0.5 -0.5 0 0.5 1 -1 Figure 5.4: Graph of the function f1,1 (r, θ). Assume first that y (z0 ) > 0. Let f (z ) = sin(z − z0 ). Since f (z ) = −f (z ) and y (z ) satisfies (5.31), d [y (z )f (z ) − y (z )f (z )] = y (z )f (z ) − y (z )f (z ) dz = −y (z )f (z ) + e2z y (z )f (z ) = (e2z − 1)y (z )f (z ). Note that f (z ) > 0 for z between z0 and z0 + π . If also y (z ) > 0 for z between z0 and z0 + π , then y (z )f (z ) > 0 and z0 +π 0< z0 z +π (e2z − 1)y (z )f (z )dz = [y (z )f (z ) − y (z )f (z )]z0 0 = y (z0 + π )f (z0 + π ) − y (z0 )f (z0 ) = −y (z0 + π ) − y (z0 ) < 0, a contradiction. Hence our assumption that y (z ) be postive for z between z0 and z0 + π must be incorrect, y (z ) must change sign at some z between z0 and z0 + π , and hence y (z ) must be zero at some point in this interval. If y (z0 ) < 0, just apply the same argument to −y . In either case, we conclude that y (z ) must be zero at some point in any interval to the right...
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