What are the roots of the equation $z^{40} - z^{20} - a(a+1)$?

Question

I found this question in a book. The answer given is that the roots are $(a+1)^{1/20}\exp\left({\frac{i2k_1\pi}{20}}\right)$ and $(a+1)^{1/20}\exp\left({\frac{i(2k_1+1)\pi}{20}}\right)$. How do I begin this problem? I only know the basics of how to express a complex number in terms of Euler's number. Would someone please give me a simple explanation?

Answer 1

Hint:

Since $$z^{40}-z^{20}-a(a+1)=\left(z^{20}+a\right)\left[z^{20}-(a+1)\right]$$ you need to solve $$z^{20}+a=0\qquad\text{and}\qquad z^{20}-(a+1)=0$$ i.e. finding the $20$th's roots of $-a$ and $a+1$.

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If also it is given that $a$ is a real and positive number we have $$-a=a(\cos \pi+i\sin\pi)\qquad\text{and}\qquad a+1=(a+1)(\cos 0+i\sin 0)$$

Answer 2

HinT: Substituting $$z^{20}=t$$ then we get $$t^2-t-a(a+1)=0$$ to solve.