How do you write a polynomial function in standard form with real coefficients whose zeros include #1#, #9i#, and #-9 i#?
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The simplest such polynomial is:
#f(x) = x^3-x^2+81x-81#
Given zeros #1#, #9i#, #-9i#.
Any polynomial in #x# with these zeros will have linear factors including #(x-1)#, #(x-9i)# and #(x+9i)#.
So let:
#f(x) = (x-1)(x-9i)(x+9i)#
#color(white)(f(x)) = (x-1)(x^2-(9i)^2)#
#color(white)(f(x)) = (x-1)(x^2+81)#
#color(white)(f(x)) = x^3-x^2+81x-81#
So this #f(x)# is a suitable example and any polynomial in #x# with these zeros will be a multiple (scalar or polynomial) of this #f(x)#.
#color(white)()#
Footnote
If only the zeros #1# and #9i# had been requested, we would still require #-9i# to be a zero too in order to get real coefficients.
One way to think of this is to recognise that #i# and #-i# are essentially indistinguishable from the perspective of the real numbers. So if #a+ib# is a zero of a polynomial with real coefficients then so is #a-ib#.
