Complex Analysis Problem session week 3Preben Alsholm, June 26, 2008with(plots):If you have the DMat package (available from the homepage) then you can do the following with appropriate modifications.libname:="L:/DMat/libDMat",libname;with(DMat):Exercise Af:=z->6*z^3+I*z+10;diff(f(z),z);f:=z->(z^2-3*I)^2;diff(f(z),z);f:=z->(6*z^3+I*z+10)/(z^2-3*I)^2;singular(f(z),z);diff(f(z),z);normal(%);f:=z->z^(-6);singular(f(z),z);diff(f(z),z);Exercise B1f:=z->cos(Re(z))*cosh(Im(z))-I*sin(Re(z))*sinh(Im(z));u:=evalc(Re(f(x+I*y))); v:=evalc(Im(f(x+I*y)));Diff(u,x): %=value(%);Diff(v,y): %=value(%);Diff(u,y): %=value(%);-Diff(v,x): %=value(%);The function is analytic everywhere.Derivative:Diff(u,x)+I*Diff(v,x): %=value(%);convert(f(z),exp);combine(expand(%));subs(Re(z)=z-I*Im(z),%);Thus f(z) is justsimplify(%);Check:expand(cos(x+I*y));and its derivativeexpand(-sin(x+I*y));2f:=z->exp(conjugate(z));u:=evalc(Re(f(x+I*y))); v:=evalc(Im(f(x+I*y)));Diff(u,x): %=value(%);Diff(v,y): %=value(%);Diff(u,y): %=value(%);-Diff(v,x): %=value(%);The function is differentiable nowhere, since cos(y) and sin(y) cannot both vanish. 3f:=z->Re(z)^3+I*Im(z)^3;u:=evalc(Re(f(x+I*y))); v:=evalc(Im(f(x+I*y)));Diff(u,x): %=value(%);Diff(v,y): %=value(%);Diff(u,y): %=value(%);-Diff(v,x): %=value(%);The function is differentiable on the lines y = x and y = -x and nowhere else. It is not analytic anywhere.The derivative on the line y = x and y = -x isDiff(u,x)+I*Diff(v,x): %=value(%);Exercise CNo computation.Exercise Df:=z->(a*z+b)/(c*z+d);Limit(f(z),z=infinity): %=value(%);Cheating some:Limit(f(z),z=-d/c): %=value(%) assuming a*d-b*c>0;This problem is done much better by hand. Exercise ER:=z->(z+I)/(z-I)^2;P:=z->z^3-3*z-4;f:=unapply(R(z)+P(z),z);subs(w=z-I,expand(f(w+I)));mtaylor(P(z),z=I);s:=0:
d:=2:
limit(1/s!*diff((z-I)^d*R(z),[z$s]),z=I);s:=1:
d:=2:
limit(1/s!*diff((z-I)^d*R(z),[z$s]),z=I);convert(R(z),parfrac);series(f(z),z=I);Although the result contains no O-term, it does have type series:whattype(%);It can be converted (the term polynom is not mathematically correct):convert(%%,polynom);Using the formula on page 106 on all of f gives:for s from 0 to 5 do A[s]:=limit(1/s!*diff((z-I)^d*f(z),[z$s]),z=I) end do;add(A[s]*(z-I)^(s-2),s=0..5);Exercise Fq:=z^4-4*I*z^2-3;qf:=factor(q);map2(map,abs,qf);subs(abs(z)=2,qf);The inequality in the other direction is not obtained elegantly in Maple, but can be done this way:subs({z=2,I=1},qf);p:=z^3-1:map(abs,p);subs(abs(z)=2,%);subs({z=2,I=1},p);f:=unapply(p/q,z);The estimates given in the problem description are not bad, but they could be improved:complexplot([f(2*exp(I*t)),3*exp(I*t),1/5*exp(I*t)],t=0..2*Pi);plot([abs(f(2*exp(I*t))),3,1/5],t=0..2*Pi,0..3);numapprox:-infnorm(abs(f(2*exp(I*t)))-3,t=0..2*Pi,'tt');tt;abs(f(2*exp(I*tt)));plot([abs(f(2*exp(I*t))),1/5],t=tt-.1..tt+.1);animate(complexplot,[f(2*exp(I*t)),t=0..T],T=0..2*Pi);