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Susanne217
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Homework Statement
Let [tex]p(z) = \sum_{j=0}^{n} a_{n-j}z^j[/tex] be a polynomial of at least degree 1 thus
[tex]n \geq 1[/tex].
Show that if [tex]z\neq 0[/tex] then 1/z is a root of the polynomial p.
Homework Equations
Fundamental theorem of Algebra
The Attempt at a Solution
If a expand the polynomial above then
[tex]p(z) = \sum_{j=0}^{n} a_{n-j}z^j = a_n + a_{n-1}z + a_{n-2}z^2 + \cdots + a_0 z^n[/tex]
to test if 1/z is a root of p then this must be equal too
[tex]p(1/z) = \sum_{j=0}^{n} a_{n-j}z^j = a_n + a_{n-1}(1/z) + a_{n-2}(1/z)^2 + \cdots + a_0 (1/z)^n = 0[/tex]
Then as I see it its my task to show that the new sum
[tex]\sum_{n=0}^{\infty} a_n (\frac{1}{z})^n[/tex] converges to zero. Then by the ratio-test
[tex]\frac{a_{n+1}(\frac{1}{z})^{n+1}}{a_n (\frac{1}{z})^{n}} \rightarrow 0[/tex] as [tex]z \rightarrow \infty[/tex] therefore 1/z is a root of p.
Am I on the right track here?
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