Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1996 based on Amplitude and Complex numbers.
Let P be the product of the roots of \(z^{6}+z^{4}+z^{2}+1=0\) that have a positive imaginary part and suppose that P=r(costheta+isintheta) where \(0 \lt r\) and \(0 \leq \theta \lt 360\) find \(\theta\)
Equations
Complex Numbers
Integers
Answer: is 276.
AIME, 1996, Question 11
Complex Numbers from A to Z by Titu Andreescue
here\(z^{6}+z^{4}+z^{2}+1\)=\(z^{6}-z+z^{4}+z^{2}+z+1\)=\(z(z^{5}-1)+\frac{(z^{5}-1)}{(z-1)}\)=\(\frac{(z^{5}-1)(z^{2}-z+1)}{(z-1)}\) then \(\frac{(z^{5}-1)(z^{2}-z+1)}{(z-1)}\)=0
gives \(z^{5}=1 for z\neq 1\) gives \(z=cis 72,144,216,288\) and \(z^{2}-z+1=0 for z \neq 1\) gives z=\(\frac{1+-(-3)^\frac{1}{2}}{2}\)=\(cis60,300\) where cis\(\theta\)=cos\(\theta\)+isin\(\theta\)
taking \(0 \lt theta \lt 180\) for positive imaginary roots gives cis72,60,144 and then P=cis(72+60+144)=cis276 that is theta=276.
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