Complex Analysis exam December 2004Preben Alsholm December 2, 2008restart;computer:=ssystem("hostname"):
if computer=[0,"PC-PKA1"] then
libname:="C:/Documents and Settings/alsholm/Dokumenter/DMat/libDMat",libname
elif computer=[0,"pc-pka1"] then
libname:="J:/DMat/libDMat",libname
elif computer=[0,"alsholm-PC"] then
libname:="F:/DMat/libDMat",libname
else
print("Fremmed computer: Omdefin\303\251r selv libname.")
end if;with(DMat);with(plots):Problem 4 is omitted since we have covered this type of material recently.Problem 1ligning:=cos(z)=I*sqrt(3);_EnvAllSolutions:=true:solve(ligning ,z);res:=convert(%,ln);res2:=subs(_B1=b,_Z1=p,res);complexplot([seq(seq(res2,b=0..1),p=-2..2)],style=point,symbol=solidcircle,symbolsize=20);convert(ligning,exp);expand(%);subs(exp(I*z)=w,%);res:=solve(%,w);exp(I*(x+I*y))=PolarForm(res[1],useExp);applyop(expand,1,%);From this follows the one half of the results found above.The other half:exp(I*(x+I*y))=PolarForm(res[2],useExp);applyop(expand,1,%);Problem 2f:=z->1/z^2-1;g:=z->if not type(z,complex) then 'procname(z)' else Log(0,z) end if;f is analytic everywhere except at 0.g is analytic everywhere except at the nonnegative real axis.f(z)=w;isolate(%,z^2);solve(%,z);g(-2-2*I);g(f(z));g(f(2+I));The inverse image by f of the nonnegative real line is therefore [-1, 1], so that h = g o f is analytic outside that interval.conformal(f(z),z=-4-4*I..4+4*I,numxy=[50,50],view=[-3..1,-1..1]):
conformal(g(f(z)),z=-4-4*I..4+4*I,numxy=[50,50]):
display(Array([%%,%]));Problem 3f:=z->I/z;complexplot([I+exp(I*t),I/2+exp(I*t)/2],t=0..2*Pi,thickness=2,scaling=constrained):
FyldtPlot([2*sin(t),sin(t)],t=0..Pi,plotoptions=[coords=polar,color=gray]):
display(%%,%);Both circles are mappep onto lines since 0 is on both.f((1+I)/2),f(I);So the small circle C2 is mapped onto the line Re(w) = 1.f(1+I),f(2*I);So the big circle C1 is mapped onto the line Re(w) = 1/2.The interior of each circle is mapped onto the right hand side of the given line.conformal(I+z,z=0..1+2*Pi*I,coords=polar):
conformal(I/2+z,z=0..1/2+2*Pi*I,color=[blue,black],coords=polar):
p1:=display(%%,%,scaling=constrained):
conformal(f(I+z),z=0..1+2*Pi*I,coords=polar):
conformal(f(I/2+z),z=0..1/2+2*Pi*I,color=[blue,black],coords=polar):
p2:=display(%%,%,view=[0..2,-1..1],scaling=constrained):
display(Array([p1,p2]));psi:=(u,v)->A*u+B;psi(1/2,v)=1;psi(1,v)=0;solve({%%,%},{A,B});subs(%,psi(u,v));phi:=unapply(subs(u=evalc(Re(f(x+I*y))),%),x,y);plot3d(phi(x,y),x=-1..1,y=0..2,view=0..1,axes=boxed);Problem 5f:=z->(1/z-c/z^2)*exp(z);convert(f(z),Sum);expand(%);map(combine,%);applyop(SkiftVariabel,2,%,_k1=k+2);applyop(SkiftVariabel,1,%,_k1=k+1);-c*z^(-2)+subs((k=-2..infinity)=(k=-1..infinity),%);combine(%);map(factor,%);map(simplify,%);subs((k=-1..infinity)=(k=-2..infinity),select(has,%,Sum));res:=%:subs(z=1,op(1,res));a1:=eval(%,k=-1);2*Pi*I*a1;The integral is different from zero when c is not 1, thus the function f has no antiderivative in C\134{0}.If on the other hand c = 1 then the integral along all closed loops in C\134{0} is zro, thus f has an antiderivative there.A Laurent series for an antiderivative can be obtained by termwise integration:subs(c=1,res);applyop(int,1,%,z);