Let $\alpha,\beta $ be two complex numbers with $\beta \neq 0$ and $f(z)$ be a polynomial function on $\Bbb C$ such that $f(z)=\alpha $ whenever $z^5=\beta $.What can you say about degree of $f(z)$?

My effort: I am unable to find the number of roots of $z^5-\beta =0$ .As $\Bbb C$ is algebraically closed so any polynomial would split and hence $z^5-\beta=0$ has exactly $5$ roots counting multiplicity.

So $f-\alpha $ has degree atleast $5$ if $z^5-\beta =0$ has distinct $5$ roots

Is it correct?I am unable to provide a specific answer.

Please give some hint.


$z^5 − \beta=0$ has exactly 5 roots counting multiplicity

If it had roots with multiplicity $\gt 1$ then those would be roots of its derivative as well, but $(z^5 - \beta)' = 5 z^4$ which has $0$ as its only root, and $0$ is not a root of $z^5 - \beta$ since $\beta \ne 0$. Therefore $z^5 - \beta = 0$ must have $5$ distinct roots.

So $f−\alpha$ has degree at least 5

And, since $f(z) = z^5 - \beta + \alpha$ satisfies the requirements, degree $5$ is the actual lower bound.


There are exactly $n$ distinct complex roots to the equation $z^n-1=0$. They are given by $\cos \frac{2\pi k}n +i\sin \frac{2\pi k}n$, for $k=0,1,2,\ldots, n-1$.

Now you can use Lagrange interpolation formula to construct a polynomial that should take specific values at given points.


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