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The complex roots of a complex polynomial $P_n(z)=z^n+a_{n-1}z^{n-1}+\cdots+a_1z+a_0$ are $\alpha_i$, $i=1,2,...,n$. Calculate the product $(\alpha_1+1)(\alpha_2+1)\cdots(\alpha_n+1)$

By the fundamental theorem of algebra, polynomial of degree $n$ has $n$ roots. Could someone show how to find this product?

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  • $\begingroup$ Can you write $P_n(z)$ as a product of linear factors? $\endgroup$ – Daniel Fischer Aug 27 '15 at 12:13
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Observe that $$ P_n(z)=(z-\alpha_1)(z-\alpha_2)...(z-\alpha_n)$$ giving $$(\alpha_1+1)(\alpha_2+1)...(\alpha_n+1)=(-1)^nP_n(-1) $$ that is $$(\alpha_1+1)(\alpha_2+1)...(\alpha_n+1)=(-1)^n\left((-1)^n+a_{n-1}(-1)^{n-1}+...-a_1+a_0\right). $$ Hoping this helps.

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Since $P_n$ is a monic polynomial, $$ P_n(z) = \prod_{i=1}^{n}(z-\alpha_i)\tag{1} $$ hence by evaluating at $z=-1$, $$ (-1)^n P_n(-1) = \prod_{i=1}^{n}(1+\alpha_i).\tag{2}$$

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