Complex Analysis, Problem session week 10Preben Alsholm, July 30, 2008with(plots):If you have the DMat package (available from the homepage) then you can do the following with appropriate modifications.libname:="J:/DMat/libDMat",libname;with(DMat);Exercise ASum((z-1)^j,j=0..infinity): %=value(%);Convergent in the open disk with center 1 and radius 1.Sum((z-1)^(-j),j=0..infinity): %=value(%);Convergent outside the above mentioned circle.Sum((z-1)^(-j),j=1..infinity): %=value(%);Convergence as above.Sum(exp(j*z),j=0..infinity): %=value(%);Convergent in the open left half plane.Sum(((z-1)/(z+1))^j,j=0..infinity): %=value(%);Convergent in the open right half plane.Exercise BSum(1/j!/z^j,j=1..infinity): %=value(%);Absolutely convergent for all LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYmLUkjbWlHRiQ2JVEiekYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy1JI21vR0YkNi1RKyZOb3RFcXVhbDtGJy9GM1Enbm9ybWFsRicvJSZmZW5jZUdRJmZhbHNlRicvJSpzZXBhcmF0b3JHRj0vJSlzdHJldGNoeUdGPS8lKnN5bW1ldHJpY0dGPS8lKGxhcmdlb3BHRj0vJS5tb3ZhYmxlbGltaXRzR0Y9LyUnYWNjZW50R0Y9LyUnbHNwYWNlR1EsMC4yNzc3Nzc4ZW1GJy8lJ3JzcGFjZUdGTC1JI21uR0YkNiRRIjBGJ0Y5Rjk=. This is a Laurent series, not a power series.Sum(j/(j+1)*(2*z)^j,j=1..infinity): %=value(%);Absolutely convergent for abs(z) < 1/2. Radius of convergence is 1/2.Sum(j!*(z/j)^j,j=1..infinity): %=value(%);Ratio test:subs(j=j+1,j!*(z/j)^j)/(j!*(z/j)^j);simplify(%);simplify(subs(z=R,%)) assuming j::posint;limit(%,j=infinity);Absolutely convergent for abs(z) < e. (Radius of convergence is e).Exercise Cf:=z->1/(3-z);1/3*Sum((z/3)^j,j=0..infinity);Radius of convergence 3.Check:value(%);The coefficients area[j]=(1/3)^(j+1);1/2*Sum(((z-1)/2)^j,j=0..infinity);Radius of convergence is 2.Check:value(%);Coefficients:a[j]=(1/2)^(j+1);Exercise Dg:=z->z/(1+exp(z));h:=z->sin(z)/(1+z^2);Sum(a(n)*z^n,n=0..infinity)*convert(1+exp(z),Sum)=z;subs(infinity=3,%);value(%);collect(%,z);remove(s->evalb(degree(s,z)>3),lhs(%))=z;solve(identity(%,z),[a(0),a(1),a(2),a(3)]);Check:taylor(g(z),z=0,4);Sum(a(n)*z^n,n=0..infinity)=convert(sin(z),Sum)*Sum((-z^2)^n,n=0..infinity);subs(infinity=10,%);value(%);map(collect,%,z):lhs(%)=remove(s->evalb(degree(s,z)>10),rhs(%));solve(identity(%,z),[seq(a(k),k=0...10)]);res:=a(2*k+1)=(-1)^k*Sum(1/(2*j+1)!,j=0..k);seq(value(res),k=0..4);Check:taylor(h(z),z=0,11);Exercise ESum((3/(z-I))^j,j=0..infinity): %=value(%);The domain of convergence is the exterior to the circle with center i and radius 3.Convergence is uniform on the exterior of any circle with center i and radius greater than 3.By uniform convergence we can interchange summation and integration.All terms but one disappears since they all but one have integrands having antiderivatives.The surviving term corresponds to j = 1, so the result is2*Pi*I*3;