Tricky complex / polynomial question (1 Viewer)

[kingkong123]

I got 6. There must be a root between 1 and 2 since the y values change sign, and because it is even there must be a root between -1 and -2 ; so thats 2/6 roots. Now i said that since it has a complex root (eg Z) and has real co-efficients, the complex conjugate root (Z'bar') must also be a root. Now since it is even, -Z and -Z'bar' must also be roots, hence it has 6 roots and the least degree is 6.

Im not sure about the last part. Could someone confirm my answer or explain how to get the actual answer?

Thanks

 

[deswa1]

I got 6. There must be a root between 1 and 2 since the y values change sign, and because it is even there must be a root between -1 and -2 ; so thats 2/6 roots. Now i said that since it has a complex root (eg Z) and has real co-efficients, the complex conjugate root (Z'bar') must also be a root. Now since it is even, -Z and -Z'bar' must also be roots, hence it has 6 roots and the least degree is 6.

Im not sure about the last part. Could someone confirm my answer or explain how to get the actual answer?

Thanks

I'm not so sure about this part. Its obvious that for an even function, all real roots have to exist in pairs (oppositely signed) but I don't know about complex parts. I just created a polynomial with roots 2i,-2i,3i,-3i which is equation : x^4+13x^2+36=0.

This is even and the roots do occur in complex conjugate pairs but note that the bit I bolded doesn't apply (i.e., if z=2i, z bar is a root but -z and -z bar aren't). This would suggest to me that the answer is 4 but hopefully someone else can confirm.

 

[funnytomato]

Consider the complex root z=a+bi

note:
z^2=(a+bi)^2=a^2+b^2+2abi
zbar^2=(a-bi)^2=a^2+b^2-2abi

So (a+bi)^2=(a-bi)^2 if and only if a=0 or b=0
i.e. z=a+bi is purely real or purely imaginary

when z is purely imaginary
that is, z=bi , then zbar=-bi
but then -z=-bi=zbar, -zbar=bi=z
so -z and -zbar are just zbar and z

therefore there must be roots:
z, zbar, alpha, -alpha
where alpha is real, z is complex

 

[Nooblet94]

Consider the complex root z=a+bi

note:
z^2=(a+bi)^2=a^2+b^2+2abi
zbar^2=(a-bi)^2=a^2+b^2-2abi

So (a+bi)^2=(a-bi)^2 if and only if a=0 or b=0
i.e. z=a+bi is purely real or purely imaginary

when z is purely imaginary
that is, z=bi , then zbar=-bi
but then -z=-bi=zbar, -zbar=bi=z
so -z and -zbar are just zbar and z

therefore there must be roots:
z, zbar, alpha, -alpha
where alpha is real, z is complex

But, as you said, that only applies to purely imaginary roots. The question simply says complex roots and if z=a+ib, zbar=a-ib, -z=-a-ib and -zbar=-a+ib. They're all different, so I'm leaning towards 6 roots, but I'm not 100% sure.

 

[funnytomato]

"P(x) is an even polynomial with real coefficients"

yeah, I wasn't really thinking clearly, OP's approach was right
(i.e. if z=rcis(theta) is a root, then -z=-rcis(theta) must also be a root since it's even)

we have:
one complex root z,
one real root a, 1< a <2

then zbar is a root as well because coeffs are real
and -z , -zbar, -a are 3 more zeros, since it's even

so we have 6 zeros, namely a , -a, z, -z, zbar, -zbar

But when z=bi , i.e. purely imaginary (hence complex)
z=-zbar=bi , -z=zbar=-bi
then we may have a polynomial that looks like:

P(x)=(x^2-a^2)(x^2+b^2)=x^4+(b^2-a^2)*x^2-a^2*b^2
where 1 < a < 2, b is real
And this polynomial is of degree 4