The point at K=4*I..1+5*I that appears a solution is at:

{K = (1/2)*sqrt(nu*Pi*(-9*Pi*nu+(8*I)*h^2))/nu};

The above is a solution to cosh(Q)=0. cosh(Q) appears in the denominator.

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A := (4*x + 1)/(diff(f(y), y)) = a;
b := (3*(4*x + 1))/((diff(f(y), y))*(3*x + 1));
frontend(algsubs, [A, b, diff(f(y), y)], [{Non}(function), {}]); 

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Do You mean to do something like this:
"numer(L = 2/3)";
f := e -> lcm((denom@lhs, denom@rhs)(e))*e;
f(L = 2/3);

 

collect(Your_stuff_goes_here, s, factor); #normal might do also

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You can plot this directly by applying the transform to the original inequalities:

  T[0] := [abs(z) < 3, 1 < abs(z - 1)];
  T[1] := z = x + I*y;
  T[2] := evalc(subs(T[1], T[0])) assuming real;
  T[3] := w = radnormal(subs(x1 = 9/2 - 1/2*sqrt(45), x2 = 9/2 + 1/2*sqrt(45), ((3 + sqrt(5))*(z - x1))/(2*(z - x2))));
  T[4] := z = solve(T[3], z);
  T[5] := radnormal(subs(T[4], T[0]));
  T[6] := w = I*v + u;
  T[7] := factor(expand(subs(T[6], T[5]))) assuming real;
  plots[inequal](T[2], x = -4 .. 4, y = -4 .. 4);
  plots[inequal](T[7], u = -3/2 .. 3/2, v = -3/2 .. 3/2);

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Look at:

Sol := [solve(x^3+(a-3)^3*x^2-x*a^2+a^3 = 0, x, real, parametric)];

Also consider this method:

Sol2 := [solve(x^3 + (a - 3)^3*x^2 - a^2*x + a^3 = 0, a, real, parametric)];
subs(Sol2[1][1], a);
plot(%, x = -4 .. 4);
#or
#Sol2 := [RealDomain:-solve(x^3 + (a - 3)^3*x^2 - a^2*x + a^3 = 0, {a})];
#subs(Sol2[1], a);
#plot(%[1], x = -4 .. 4);

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Notice your E procedure is missing x as input.

Simply use:
(f@@depth)(x);

You have to define the region of integration in a new way:

Int(Int(Int(1, y = 0 .. 1), tau = 0 .. z + q), z = 0 .. t); 
IntegrationTools[CollapseNested](%); 
subs([(tau = 0 .. z + q) = (z = piecewise(tau < q, 0, q <= tau, tau - q) .. t), (z = 0 .. t) = (tau = 0 .. q + t)], %); 
(expand(value(%%) = value(%)) assuming (0 < t, 0 < q)), (value(%%) = value(%));

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D[1, 2](f)(x, y) means diff(f(x,y), x, y). It means first derivative in first coordinate, first derivative in second coordinate. Your expression D[1, 2](1/x); implies x is a function of 2 coordinates.

These problems tend to be difficult for obvious reasons. See the following:

E0 := x^2 + y^2 - 10*x - 75 = 0;
E := (a*x + b*y + c)^2 + (d*x + e*y + f)^2 = r^2;
V := [[a, b, c], [d, e, f], r];

(lhs - rhs)(expand(E - E0));
C := {coeffs}(%, {x, y});
T := eliminate(C, {c, f, r});
S1 := simplify(T[1], T[2]);
S2 := [solve](T[2], {a, b, d, e});

#A solution:
applyop(evala, 2, subs(S1, S2[1], E));
subs(S1, S2[1], V);

#Another method:
map2(remove, evalb, evala({solve}(C, {a, b, c, d, e, f, r})));
select(proc(S) subs(S, r); is(0 < %) = true; end proc, %);
map(factor, map(allvalues, %));
subs(b=1, %);

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8*x*exp(-3*x^2) - 24*x^3*exp(-3*x^2);
factor(%);
solve(%, {x});

[sin(x), cos(x), tan(x), sec(x), csc(x), cot(x), sinh(x), cosh(x), tanh(x), sech(x), csch(x), coth(x)];
map(f -> f-convert(f, exp), %);
simplify(%);

Simplify does not reach across the exp & trig boundary. You are expected to introduce the relevant identities yourself.

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Here is a procedure tha draws a cylinder from a specified base, along a vector and of a given radius and number of sides:

C := proc(d::list, v::list, r, n::posint)
local x, y, z, f;
description `Draws a cylinder with base at d along vector v, radius r, having n sides. Attach additional options for the plottools[cylinder] command used.`;

use LinearAlgebra in applyop(Normalize, {2, 3}, GramSchmidt([Vector(v), op(NullSpace(v))]), 2); end;
f := unapply(convert(add((% *~ [z, x, y])) + Vector(d), list), [x, y, z]);
use plottools in transform(f)(cylinder([0, 0, 0], r, 1, strips = n, _rest)); end;
end proc;

plots[display](seq(C([0, 0, 1], [10, 3, 10], 1, 40, op(opts)), opts = 
[[color = COLOUR(RGB, 0.2, 0.4, 0.2), transparency = 0, style = contour, thickness = 0, linestyle = 1], 
[color = green, transparency = 0.8, style = patchnogrid]]), 
scaling = constrained, orientation = [160, 45, 30], contours = 20, labels = ['x, y, z'], lightmodel = light4);

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F:=f->taylor(f, x, 4);
F(ln(1+x));
F(ln(1+sin(x)));
F(%%-%);
 

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Like before:

A := cos(5*t)=a*cos(t)^5+b*cos(t)^3*sin(t)^2+c*cos(t)*sin(t)^4;
collect(combine((lhs-rhs)(A)), {sin, cos});
solve({coeffs}(%, indets(%, dependent(t))));

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