You do not have to convert these numbers into polar form, can immediately build these points:

z0 := sqrt(3)+I;
z1 := -1+I*sqrt(3);
z2 := -sqrt(3)-I;
z3 := 1-I*sqrt(3);

Points := seq(plots[pointplot]([Re(z||i), Im(z||i)], color = red, symbol = solidcircle, symbolsize = 15), i = 0 .. 3):
Names := seq(plots[textplot]([Re(z||i)+.2, Im(z||i), cat('z', i)], color = blue, font = [TIMES, ROMAN, 16]), i = 0 .. 3):

plots[display](Points, Names);

Of course, the same can be build by  plots[polarplot]  command:

restart; 

z0 := sqrt(3)+I: 
z1 := -1+I*sqrt(3): 
z2 := -sqrt(3)-I: 
z3 := 1-I*sqrt(3): 

Points := plots[polarplot]([seq([abs(z||i), argument(z||i)], i = 0 .. 3)], style = point, color = red, symbol = solidcircle, symbolsize = 15):

Names := seq(plots[textplot]([Re(z||i)+.2, Im(z||i), cat('z', i)], color = blue, font = [TIMES, BOLD, 16]), i = 0 .. 3):

plots[display](Points, Names);

Your body is limited to 5 surfaces:

1. y=x  and  y=-x  - two planes (green).

2. x^2+y^2+z^2 = 4  and  x^2+y^2+z^2 = 9  - two spherical surface (yellow).

3. z = sqrt(x^2+y^2)  -  conical surface (cyan).

All surfaces are defined by the parametric equations:

A:=plot3d([r*cos(t)/sqrt(2),r*cos(t)/sqrt(2),r*sin(t)], r=2..3, t=Pi/4..Pi/2, color=green):

B:=plot3d([-r*cos(t)/sqrt(2),r*cos(t)/sqrt(2),r*sin(t)], r=2..3, t=Pi/4..Pi/2, color=green):

C:=plot3d([r*cos(t),r*sin(t),r], r=2/sqrt(2)..3/sqrt(2), t=Pi/4..3*Pi/4, color=cyan):

E:=plot3d([2*cos(phi)*sin(theta),2*sin(phi)*sin(theta),2*cos(theta)], phi=Pi/4..3*Pi/4, theta=0..Pi/4, color=yellow):

F:=plot3d([3*cos(phi)*sin(theta),3*sin(phi)*sin(theta),3*cos(theta)], phi=Pi/4..3*Pi/4, theta=0..Pi/4, color=yellow):

plots[display](A,B,C,E, F, view=[-2.5..2.5,0..2.5,0..3.5],scaling=constrained, axes=normal, orientation=[25,75], labels=[x,y,z]);

You must use the geometric meaning of the definite integral. Your integral is precisely the area of a quarter of a circle of radius 4, because the plot of the function  f(x)=sqrt(16-x^2), x=-4..4, is the upper semicircle of the radius 4.

The name of a function or expression should be  name  type. But

type(omega^2, name); 

                     false

Instead of  omega^2  you can write  `omega^2`

type(`omega^2`, name);

                      true 

restart;

plots[implicitplot](x^3+y^3-3*x*y=0, x=-3..3, y=-3..3, thickness=2, gridrefine=4);

I believe this is the simple way:

[7,9,1]:

add(%[i]*10^(i-1), i=1..nops(L));

                       197

Should be  add  instead of  sum . Write as follows:

Omega := n-> R^n*P(n, z/R)*ln(r)+R^n*add((2*n-4*i+1)*P(n-2*i, z/R)/(i*(2*n-2*i+1)), i = 1 ..

(1/2)*(n-(n mod 2))) ;

restart;

ee := -ln(-a/(b-c)):

ff:=combine(ee) assuming a/(c-b)>0;

gg:=op(1, ff);

subs(gg=numer(gg)/denom(gg), ff);

In your expression you forgot to put the separators (colons or semicolons).

restart;

m := 2:

k := 1.0931:

a := k-m:

b := k+m-1:

z := k*m/10^(.1*10):

simplify((10^(.1*yo))^((b-a+2*p-1)*(1/2))*z^((b-a+2*p+1)*(1/2))*GAMMA((1/2)*(1-b+a-2*p))/(p!*GAMMA(p-a+1)*GAMMA(1+(1/2)*(1-b+a-2*p))));

[seq(limit(%, p = k), k = 0 .. 10)];

Example:

cat(op(0, A[1]), op(A[1]));

                  A1

The inequality  a<b<c  is equivalent to  a<b and b<c . The last entry may be used to manipulate in Maple with this inequality and for its solution.

Example:

x/3-1>1 and x/3-1<5;  #  It's equivalent to 1 < x/3-1 < 5

map(x->3*x, %);  #  Both inequalities are multiplied by 3

solve(%);

In older versions there are 2 ways. You can convert all constants to symbols 

`2`+`3`;

or use double quotes

``(2)+``(3) ;

In general case it is impossible. For a given matrix  C  there are infinitely many matrices A and B, such that C = AB . For example  C=I.C=(I/2).(2C)  and so on (here  I  is identity matrix). 

When all matrices have a size  2 times 2 and the matrix  C  is known, then to find the matrices A  and  B, such that C = AB, we must solve a nonlinear system of 4 equations with 8 unknowns. In general case, the set of solutions of the system is 4-dimensional manifold in the 8-dimensional space.

In one line:

map(t->t^2, [0,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1]);

Of course, your problem is easily solved in the usual brute force method:

N:=0:

for n from 1 to 2013 do

if (irem(n, 3)=0 and irem(n+1, 4)=0) or (irem(n, 4)=0 and irem(n+1, 3)=0) then N:=N+1 fi;

od:

N;

                                                                     336

More interesting to get the number of such pairs in any  RealRange(a, b) .

Here is the formula solving this problem:

restart;

N:=(a,b)->floor((b-4)/12)+floor((b-9)/12)-ceil((a-3)/12)-ceil((a-8)/12)+2;

Examples:

N(1,2014),  N(1, 1000000),  N(1000, 10000);

                              336, 166667, 1500