polynomials question (1 Viewer)

[ninetypercent]

EDIT: dw about this question. scroll down to Q2. need help with that

Thanks

 

[ninetypercent]

2. consider the polynomial P(z) = z^5 + az^3 + bz - c, where a, b and c are positive real numbers.
(i) show that P(z) has exactly one real root
(ii) hence show that P(z) has four complex roots with at least two with negative real parts

 

[shaon0]

2. consider the polynomial P(z) = z^5 + az^3 + bz - c, where a, b and c are positive real numbers.
(i) show that P(z) has exactly one real root
(ii) hence show that P(z) has four complex roots with at least two with negative real parts

i) P'(z)=5z^4+3az^2+b
=5[z^4+3az^2/5+b/5]
=5[(z^2+3a/10)^2+b/5-9a^2/100]
>=5[9a^2/100+b/5-9a^2/100]
=5(b/5)=b
>0

ie. P'(z)>0 Hence, P(z) is increasing function to infinity since its a poly.
Now, P(0)=-c <0. Thus, P(z) has only one real root as it can only cross x-axis once.

ii)If there is only one real root there must be 4 other complex roots. The complex roots are in the form z=x+iy,A+iB. But according to complex conjugate root theorem. conj(z)=x-iy and A-iB are roots of P(z). Now, conj(z)=x+i(-y) and a+i(-B) which means Im(z)<0 assuming y,B E R+.

I don't think the above suffices but that's all i can type up with the time i have. Any questions are welcome but may not have time to answer them.

 

[Rezen]

for ii) the question refers to real parts, not imaginary parts.

continueing from shaon0, using the sum of roots we get

x+iy+x-iy+A+iB+A-iB+R=0, where R is the real root.
2x+2A=-R
from i) we know p(0)=-c<0, therefore, R>0 and -R<0
It follows then that either x, A or both are <0.

 

[dasicmankev]

@ shaon0, why does

5[(z^2+3a/10)^2+b/5-9a^2/100] >= 5[9a^2/100+b/5-9a^2/100] ?

You're suggesting that (z^2 + 3a/10)^2 >= (3a/10)^2 but that is not necessarily true as z^2 may well be negative (since a complex number squared may be smaller than zero).
​

 

[shaon0]

@ shaon0, why does

5[(z^2+3a/10)^2+b/5-9a^2/100] >= 5[9a^2/100+b/5-9a^2/100] ?

You're suggesting that (z^2 + 3a/10)^2 >= (3a/10)^2 but that is not necessarily true as z^2 may well be negative (since a complex number squared may be smaller than zero).
​

lol, I'm verifying for only a real root. So say z is a real root the inequality holds.

 

[shaon0]

for ii) the question refers to real parts, not imaginary parts.

continueing from shaon0, using the sum of roots we get

x+iy+x-iy+A+iB+A-iB+R=0, where R is the real root.
2x+2A=-R
from i) we know p(0)=-c<0, therefore, R>0 and -R<0
It follows then that either x, A or both are <0.

Yeah, sorry about that. Didn't have much time to think about it but your methods what i would have used.

 

[Trebla]

@ shaon0, why does

5[(z^2+3a/10)^2+b/5-9a^2/100] >= 5[9a^2/100+b/5-9a^2/100] ?

You're suggesting that (z^2 + 3a/10)^2 >= (3a/10)^2 but that is not necessarily true as z^2 may well be negative (since a complex number squared may be smaller than zero).
​

What shaon0 did looks fine. The polynomial is monotonically increasing therefore it must have only one real root.
When you sketch the polynomial P(z) against z, you always take z as a real variable, hence its first derivative must be in terms of a real variable.
However, the SPECIFIC value of z that is non-real occurs for example, when it is a root, i.e. it satisfies P(z) = 0. Although it suggests z could be non-real this has NOTHING to do with the graph of P(z) because the graph is in terms of real variables only. We never see the complex value of z that gives P(z) = 0 on a graph because it is simply not a real number. Hence when we use a graphical argument we tend to ignore the possibility of z being non-real because it is irrelevant.

 

[dasicmankev]

My mistake, thanks for clarifying Trebla. Didn't occur to me for a sec that we consider z to be real when graphing.