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## Homework Statement

Given [tex]P(z)=4z^3-2z+1[/tex] where [tex]z=cost+isint[/tex], find the maximum and minimum modulus on the argand diagram for the graph as t moves from 0 to [itex]2\pi[/itex].

I want to check if my solution is valid, and if there is an easier approach to it because I do somewhat answer the question, but I skip all the nitty-gritty arithmetic at the end since it's too overwhelming for me to solve by hand.

## The Attempt at a Solution

Substituting z(t) into P(z) gives:

[tex]P(z)=4cos(3t)-2cost+1+i(4sin(3t)-1)[/tex]

[tex]P(z)=16cos^3t-14cost+1+isint.(10-16sin^2t)[/tex]

[tex]|z|=\left(\left(4cos(3t)-2cost+1\right)^2+\left(4sin(3t)-2sint\right)^2\right)^{\frac{1}{2}}[/tex] ---

**(1)**

[tex]\frac{d|z|}{dt}=\frac{1}{2}\left(\left(4cos(3t)-2cost+1\right)^2+\left(4sin(3t)-2sint\right)^2\right)^{\frac{-1}{2}}[/tex]

[tex]\left(2\left(4cos(3t)-2cost+1\right)\left(-12sin^3t+2sint\right)+2\left(4sin(3t)-2sint\right)\left(12cos(3t)-2cost\right)=0[/tex]We're only interested in the numerator to equal zero, therefore,

[tex]\left(4cos(3t)-2cost+1\right)\left(-12sin^3t+2sint\right)+\left(4sin(3t)-2sint\right)\left(12cos(3t)-2cost\right)=0[/tex]

Expanding and simplifying, which has nice cancelling and so makes this seemingly difficult task into something much easier.

Expanded: [tex]-48cos(3t)sin(3t)+8sintcos(3t)+24costsin(3t)-4costsint-12sin(3t)+2sint=0[/tex]

Simplified: [tex]=-12sin(3t)+16sin(2t)+2sint[/tex] ---

**(2)**

By De'Moivres Theorem: [tex]sin(3t)=-4sin^3t+3sint; sin(2t)=2sintcost[/tex]

From (2), converting into trig powers: [tex]=48sin^3t-36sint+32costsint+2sint[/tex]

[tex]2sint\left(24cos^2t-16cost-7\right)=0[/tex]

Hence, [tex]sint=0, t=2k\pi[/tex], [tex]k \in Z[/tex]

and, [tex]24cos^2t-16cost-7=0[/tex]

[tex]cost=\frac{4\pm \sqrt{58}}{12}[/tex]

Finding out the answer graphically (because it's easier on me),

[tex]t=cos^{-1}\left(\frac{4-\sqrt{58}}{12}\right)[/tex]

Hence, by graphically means again,

min |z| occurs when [tex]t=\pi(2k+1)[/tex]

max |z| occurs when [tex]t=(-1)^k\left(cos^{-1}\left(\frac{4-\sqrt{58}}{12}\right)-\frac{\pi}{2}\right)+\frac{\pi}{2}(2k+1)[/tex]

Now, substituting t into (1) to find min |z|, we get |z|=1

However, for max |z| the arithmetic is too tricky.

[tex]cost=\frac{4-\sqrt{58}}{12}[/tex]; [tex]sint=\frac{\sqrt{70+8\sqrt{58}}}{12}[/tex]

Therefore, [tex]|z|^2=\left(16\left(\frac{4-\sqrt{58}}{12}\right)^3-14\left(\frac{4-\sqrt{58}}{12}\right)+1\right)^2+\left(\frac{70+8\sqrt{58}}{144}\right)\left(10-\frac{70+8\sqrt{58}}{9}\right)^2[/tex]

I don't want to expand these trinomials, make an arithmetic error in the process early on, then finally after all the

sweat and tears find the answer is incorrect.

There must be an easier way!

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