>>14242094Fix r.
Clearly picking the t that maximizes cos(4t) gives the max value for the given r.
cos(4t) is minimum for t = pi/4 and increases as t goes down to zero.
This gives either z in [0,1] or z in 1+i[0,1]
For t=0 obviously a bigger r gives a bigger value since all terms are positive.
So z is in 1 +i[0,1] or equivalently r = 1/cos(t)
Using this value for r and expanding cos(4t) = 8cos(t)^4 - 8cos(t)^2 + 1 gives
80 - 64/cos(t)^2 + 8/cos(t)^4 + 1/cos(t)^8
Since t is in [0,pi/4], 1/cos(t)^2 = q is in [1,2]
Now we just need to find the maximum of 80 - 64q + 8q^2 + q^4
Differentiating with respect to q gives 4q^3 + 16q - 64
This derivative is negative for q in[1,2) and equals 0 at 2.
This tells us q = 1 is the maximum. The maximum is 80-64+8+1=25
This corresponds to r=1, t=0
Which is the same as z=1
Which is the same as c=d=1, a=0, b=2
