Complex Analysis, Problem session week 6Preben Alsholm, July 24, 2008with(plots):If you have the DMat package (available from the homepage) then you can do the following with appropriate modifications.libname:="F:/DMat/libDMat",libname;with(DMat):Describe(Conformal3D);Exercise Af:=z->(I*z+1)/(z+I);Here we define 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:f(-I):=infinity;Here we define LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYmLUkjbWlHRiQ2I1EhRictRiM2JkYrLUYjNiYtRiw2JVEiZkYnLyUnaXRhbGljR1EldHJ1ZUYnLyUsbWF0aHZhcmlhbnRHUSdpdGFsaWNGJy1JI21vR0YkNi1RMCZBcHBseUZ1bmN0aW9uO0YnL0Y6USdub3JtYWxGJy8lJmZlbmNlR1EmZmFsc2VGJy8lKnNlcGFyYXRvckdGRC8lKXN0cmV0Y2h5R0ZELyUqc3ltbWV0cmljR0ZELyUobGFyZ2VvcEdGRC8lLm1vdmFibGVsaW1pdHNHRkQvJSdhY2NlbnRHRkQvJSdsc3BhY2VHUSYwLjBlbUYnLyUncnNwYWNlR0ZTLUkobWZlbmNlZEdGJDYkLUYjNiQtRiw2JVEoJiM4NzM0O0YnRjZGOUZARkBGQEYrRkBGK0ZA:f(infinity):=I;L:=0,1,-1,I,-I,infinity;Lf:=op(map(f,[L]));L[1..4];LLf:=[L[1..4],Lf[1..4]];Instead of the labeling A = f(a), B = f(b), ... we use colors:C:=[red,green,blue,cyan,red,green,blue,cyan]:
SYM:=[solidcircle$4,soliddiamond$4]:for k to 8 do p[k]:=complexplot([LLf[k]],style=point,symbol=SYM[k],symbolsize=30,color=C[k]) end do: p1:=display(seq(p[k],k=1..4)):p2:=display(seq(p[k],k=5..8)):display(Array([p1,p2]));The 3 consequtive points on the unit circle map intof(1),f(I),f(-1);which lie on the real line. Thus C1 maps onto the extended real line. The interior of the unit circle maps onto the lower half-plane.conformal(z,z=0..0.8+I*2*Pi,coords=polar,numxy=[150,150],scaling=constrained):
conformal(f(z),z=0..0.8+I*2*Pi,coords=polar,numxy=[150,150],scaling=constrained):
display(p1,%%):
display(p2,%%):
display(Array([%%,%]));animate(conformal,[z,z=0..a+I*2*Pi,coords=polar,numxy=[150,150],scaling=constrained],a=0..0.9,background=p1):
animate(conformal,[f(z),z=0..a+I*2*Pi,coords=polar,numxy=[150,150],view=[-5..5,-9..0],scaling=constrained],a=0..0.9,background=p2):
display(Array([%%,%]));Conformal3D(z,z=0..1+I*2*Pi,coords=polar,transparency=0.6,axes=normal):
Conformal3D(f(z),z=0..1+I*2*Pi,coords=polar,transparency=0.6,axes=normal):
display(Array([%%,%]));Sincef(-1),f(0),f(1);the extended real line maps onto the unit circle. The upper half-plane maps onto the interior of the disk
.conformal(z,z=-2..2+4*I,scaling=constrained):
conformal(f(z),z=-2..2+4*I,scaling=constrained):
display(Array([%%,%]));animate(conformal,[z,z=-a..a+2*a*I,scaling=constrained],a=0..3):
animate(conformal,[f(z),z=-a..a+2*a*I,scaling=constrained,numxy=[50,50]],a=0..3):
display(Array([%%,%]));Sincef(0),f(I),f(infinity);f maps the extended imaginary axis onto itself and the right half-plane onto itself. conformal(z,z=-2*I..2-1.1*I):
conformal(f(z),z=-2*I..2-1.1*I,numxy=[150,50]):
display(Array([%%,%]));conformal(z,z=-0.9*I..2+4*I):
conformal(f(z),z=-0.9*I..2+4*I,numxy=[150,50]):
display(Array([%%,%]));conformal(z,z=-2*I..2+4*I):
conformal(f(z),z=-2*I..2+4*I,view=[0..10,-5..5],numxy=[50,200]):
display(Array([%%,%]));Conformal3D(z,z=-2*I..2+4*I,numxy=[50,50],transparency=.6,axes=normal):
Conformal3D(f(z),z=-2*I..2+4*I,numxy=[50,50],transparency=.6,axes=normal):
display(Array([%%,%]));Exercise Bcomplexplot([1+exp(I*t),2+2*exp(I*t)],t=0..2*Pi,thickness=3);p1:=%:f:=z->(4-z)/z;f(0):=infinity;Thus both circles map onto lines.f(1-I),f(2);The smaller circle maps onto the extended line Re(w) = 1, and its interior is mapped onto the half plane to the right, the exterior therefore to the left half. f(2-2*I),f(4);The larger circle maps onto the extended imaginary axis, and its interior is mapped onto the half plane to the right. k:='k':complexplot([seq(r+r*exp(I*t),r=[(1+k/10)$k=1..9])],t=0..2*Pi,color=blue):
p2:=display(p1,%,scaling=constrained):
complexplot([f(1+exp(I*t)),f(2+2*exp(I*t))],t=0..2*Pi,view=[0..1,-5..5],thickness=3):
complexplot([seq(f(r+r*exp(I*t)),r=[(1+k/10)$k=1..9])],t=0..2*Pi,view=[0..1,-5..5],color=blue):
p3:=display(%%,%,scaling=constrained):
display(Array([p2,p3]));animate(complexplot,[r+r*exp(I*t),t=0..2*Pi,color=blue,scaling=constrained],r=1..2,trace=25,background=p1):
complexplot([f(1+exp(I*t)),f(2+2*exp(I*t))],t=0..2*Pi,view=[0..1,-5..5],thickness=3):
animate(complexplot,[f(r+r*exp(I*t)),t=0..2*Pi,view=[0..1,-5..5],color=blue],r=1..2,background=%,trace=25):
display(Array([%%%,%]));C:='C': k:='k':psi:=z->A*z+B;eq1:=psi(0)=0;eq2:=psi(1)=k;solve({eq1,eq2},{A,B});subs(%,psi(f(x+I*y)));evalc(Re(%));phi:=unapply(%,x,y);phi(1+cos(t),sin(t));simplify(%);simplify(phi(2+2*cos(t),2*sin(t)));Using the knowledge (the maximum principle) that the extrema are attained at the boundary makes the graph easy to draw using the view option:plot3d(subs(k=1,phi(x,y)),x=0..4,y=-2..2,view=0..1,axes=boxed);contourplot3d(subs(k=1,phi(x,y)),x=0..4,y=-2..2,contours=[seq(k/10,k=0..10)],view=0..1,axes=boxed,filledregions=true);Exercise Cf:=z->(I*z+1)/(z+I);D(f)(I);f(I);The result now follows from Theorem 4, The Riemann mapping theorem.normal(diff(f(z),z));Exercise Dphi:='phi':f:=z->(I*z+1)/(z+I);F:=z->f( (exp(-I*phi)*z)^(Pi/theta));F(z);F(r*exp(I*phi));simplify(%);F(r*exp(I*(phi+theta)));simplify(%);simplify(%) assuming theta>0,theta<Pi,r>0;simplify(%,symbolic);Exercise Ef:=z->exp(I*theta)*(z-alpha)/(conjugate(alpha)*z-1);f(I/2)=0;solve(%,alpha);f((1+I)/2)=0;alpha1:=I/2:
f1:=z->(z-alpha1)/(conjugate(alpha1)*z-1);alpha2:=(1+I)/2:
f2:=z->(z-alpha2)/(conjugate(alpha2)*z-1);f2inv:=unapply(solve(f2(z)=w,z),w);h:=unapply(simplify(f2inv(f1(z))),z);conformal(z,z=0..1+2*Pi*I,coords=polar,numxy=[5,50],scaling=constrained):
conformal(h(z),z=0..1+2*Pi*I,coords=polar,numxy=[50,150],scaling=constrained):
display(Array([%%,%]));