Second Order Filter - Power Dissipation

When the angular frequency is infinite, the equivalent circuit is just the input voltage dropped over one of the resistors. This is the maximum power condition.

Sweep over the angular frequency and store the value of the angular frequency which results in a total power dissipation equal to half of the maximum possible value. The answer turns out to be $550.25 \, \text{rad/s}$ .

$ZC = - \frac{j}{\omega C} \\ Z_1 = R + Z_C \\ Z_2 = \frac{Z_1 \, Z_C}{Z_1 + Z_C} \\ V_1 = V_{in} \frac{Z_2}{R + Z_2} \\ V_{out} = V_1 \frac{R}{R + Z_C} \\ V_{R1} = V_{in} - V_1 \\ P = \frac{|V_{R1}|^2}{R} + \frac{|V_{out}|^2}{R}$

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import math

Vin = complex(100.0,0.0)

R = 1.0 * (10.0**3.0)
C = 1.0 * (10.0**-6.0)

dw = 0.01

w = dw

Pmax = (abs(Vin)**2.0)/R

P_minres = 9999999.0

while w <= 10.0**4.0:

    ZC = complex(0.0,-1.0/(w*C))

    Z1 = R + ZC
    Z2 = Z1*ZC/(Z1+ZC)

    V1 = Vin*Z2/(R+Z2)

    Vout = V1*R/(R+ZC)

    VR1 = Vin - V1

    P = (abs(VR1)**2.0)/R + (abs(Vout)**2.0)/R

    P_res = math.fabs(P - Pmax/2.0)

    if P_res < P_minres:
        P_minres = P_res
        w_store = w  

    w = w + dw

print ""
print ""
print P_minres
print w_store

Neglecting terms that act as merely scaling factors for the expression, the variation of active power with angular frequency is:

Judging by the monstrosity of the expressions obtained, this approach is a rather tedious one. Feedback would be appreciated