How to Factorize a Complex Polynomial with Roots of the Form z = ix?

[metgt4]

Homework Statement

The polynomial f(z) is defined by

f(z) = z⁵ - 6z⁴ + 15z³ - 34z² + 36z - 48

Show that the equation f(z) = 0 has roots of the form z = ix where x is real, and hence factorize f(z)

The Attempt at a Solution

So I know that you begin by factoring out (z-ix) from the function, but I'm not quite sure how to work that out. I can only figure out how to get the first and last terms in the first step:

f(z) = (z-ix)(z⁴ + ... - 48i/x)

How would you go about finding everything in between those two terms?

 

[sara_87]

The next term should be A*z^3, you have to find a.
Now, we want a value A such that: when we expand the brackets, we get -6 for the coefficient of z^4.
(z-ix)(z^4 + Az^3...)
when you expand to get the z^4 term, we have: -ix*z^4 + A*z^4=-6*z^4.
that means A=ix-6
agree?
now that you have A,
can you do this for the rest of the terms?
so what will you get?

 

[Mark44]

Evaluate f(ix) and then simplify all powers of i.
Rewrite f(ix) as a complex number: g(x) + h(x)*i.
Set f(ix) = 0. This implies that g(x) = 0 and h(x) = 0.
Factor g(x) and h(x). This gives you a number of values of x for which f(ix) = 0.

 

[metgt4]

Thanks to both of you! I evaluated f(ix) and that simplified things quite a bit!