Complex Analysis, Problem session week 11Preben 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):
<Text-field style="Heading 1" layout="Heading 1">Exercise A</Text-field>f:=z->1/(3-z);asympt(f(z),z);-1/z*Sum((3/z)^j,j=0..infinity);combine(%);simplify(%);subs(u=z-1,asympt(f(u+1),u));-1/(z-1)*Sum((2/(z-1))^j,j=0..infinity);combine(%);simplify(%,power);The answer to the 3rd question is simple: There is one term, viz. -1/(z-3).
<Text-field style="Heading 1" layout="Heading 1">Exercise B</Text-field>f:=z->z^2*sin(1/z^2);convert(f(z),Sum);combine(%);The integral along the circle abs(z) = 3 (counterclockwise) is 0, since the coefficient of z^(-1) is 0.g:=z->exp(z)/(z+1)^2;g(zeta-1);combine(exp(1)*expand(%));convert(%,Sum);combine(exp(-1)*%);subs(zeta=z+1,%);The integral along the circle abs(z) = 3 (counterclockwise) is (1/e)*2*Pi*I, since the coefficient of (z+1)^(-1) is 1/e.
<Text-field style="Heading 1" layout="Heading 1">Exercise C</Text-field>f:=z->(I*z+1)/(z-2);limit((z-2)*f(z),z=2);2 er en simpel pol.normal(f(1/w));limit(%,w=0);f is analytic at infinity.f:=z->z/(z^2+1)^3;limit((z-I)^3*f(z),z=I);limit((z+I)^3*f(z),z=-I);f has poles of order 3 at I and -I.f:=z->(cos(z)-1)/z^2;limit(f(z),z=0);f has a removable singularity at 0.f:=z->sin(1/z);f(1/p/Pi);f(1/(p*Pi+Pi/2));f has an essential singularity at 0.sin(z) therefore has an essential singularity at infinity.f:=z->(exp(z)-1)/z^2;limit(z*f(z),z=0);f has a simple pole at 0.f:=z->1/sin(1/z);f has singularities at 1/(p*Pi). Thus 0 is not an isolated singularity.
<Text-field style="Heading 1" layout="Heading 1">Exercise D</Text-field>f:=cot:_EnvAllSolutions:=true:solve(f(z)=0,z);singular(f(z),z);limit((z-p*Pi)*f(z),z=p*Pi) assuming p::integer;The singularities are simple poles.series(f(z),z=0);limit(diff(z*f(z),z),z=0);limit(diff(z*f(z),z,z),z=0)/2;We compute a few of the coefficients in the Laurent series expansion valid in the annulus with radii Pi and 2*Pi.There must be infinitely many nonvanishing terms corresponding to positive and also to negative exponents ( since convergence breaks down at z = Pi and z = 2*Pi).The coefficients are computed using the formula (1) on p. 269. These integrals can be computed by using the residue theorem, but of course also numerically.p:=t->3*Pi/2*exp(I*t);
A:=j->1/(2*Pi*I)*Int(f(p(t))/p(t)^(j+1)*diff(p(t),t),t=0..2*Pi);value(A(-1));By the residue theorem the above is the sum of the 3 residuesmap2(residue,f(z),[z=0,z=Pi,z=-Pi]);value(A(0));seq(add(residue(f(z)/z^(j+1),z=z0),z0=[0,-Pi,Pi]),j=-5..5);evalf(%);evalf(seq(A(j),j=-5..5));simplify(fnormal([%]));We compute a few of the coefficients in the Laurent series expansion valid in the annulus with radii 2*Pi and 3*Pi.There must be infinitely many nonvanishing terms corresponding to positive and also to negative exponents ( since convergence breaks down at z = 2*Pi and z = 3*Pi.p:=t->5*Pi/2*exp(I*t);
A:=j->1/(2*Pi*I)*Int(f(p(t))/p(t)^(j+1)*diff(p(t),t),t=0..2*Pi);value(A(-1));By the residue theorem the above is the sum of the 5 residuesmap2(residue,f(z),[z=0,z=Pi,z=-Pi,z=2*Pi,z=-2*Pi]);By the periodicity of f clearly these residues are all the same (as is also seen above).seq(add(residue(f(z)/z^(j+1),z=z0),z0=[0,-Pi,Pi,-2*Pi,2*Pi]),j=-5..5);evalf(%);evalf(seq(A(j),j=-5..5));simplify(fnormal([%]));
<Text-field style="Heading 1" layout="Heading 1">Exercise E</Text-field>f:=z->(4*z^2-Pi^2)/cos(z);f has removable singularities at Pi/2 and -Pi/2. The next singularities are at 3*Pi/2 and -3*Pi/2. Thus the radius of convergence is 3*Pi/2.The distance from -1-I to 0 is LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYmLUkjbWlHRiQ2JVElc3FydEYnLyUnaXRhbGljR1EmZmFsc2VGJy8lLG1hdGh2YXJpYW50R1Enbm9ybWFsRictSShtZmVuY2VkR0YkNiQtRiM2JC1JI21uR0YkNiRRIjJGJ0YyRjJGMi1GLDYjUSFGJ0Yy which is therefore the radius of convergence asked for in question 2.
<Text-field style="Heading 1" layout="Heading 1">Exercise F</Text-field>No computation.