How to Solve a Polynomial Function with Complex Zeros?

[doktorwho]

Homework Statement

Given the polynomial function $x^4+x^3+2x^2+4=0$ solve it if you know that it has at least one complex zero whose real part equals the complex part.

Homework Equations

3. The Attempt at a Solution [/B]
My guess is that if this function has one complex zero it must have a conjugate complex zero as well so we need to find those and two more. I am not sure how to start except realising that $P(b-bi)=P(b+bi)=0$ but solving the equation for that seems a lot of work and I am sure there's an easir way. Should i use Viet's formulas or some other way to approach this?

 

[Mark44]

Homework Statement

Given the polynomial function $x^4+x^3+2x^2+4=0$ solve it if you know that it has at least one complex zero whose real part equals the complex part.

Homework Equations

3. The Attempt at a Solution [/B]
My guess is that if this function has one complex zero it must have a conjugate complex zero as well so we need to find those and two more. I am not sure how to start except realising that $P(b-bi)=P(b+bi)=0$ but solving the equation for that seems a lot of work and I am sure there's an easir way. Should i use Viet's formulas or some other way to approach this?

I haven't worked this problem, but here are my thoughts. If x = b + bi is a solution (implying that x = b - bi is also a solution), what will be the effect of the x³ term? Will it be possible for $x^4 + x^3 + 2x^2$ to be equal to the pure real number -4?

 

[epenguin]

I was going to say State what you understand by 'Real part equals the complex part' which, I don't know, sounded slightly loose language to me, but what I understood is what doktorwho has said. It is telling you something special about one quadratic factor of the polynomial.

And there is also something else a bit special about the coefficients, and hence the roots. So lay that out and you have a few things you can may be put together (I also haven't worked the problem).

 

[doktorwho]

I haven't worked this problem, but here are my thoughts. If x = b + bi is a solution (implying that x = b - bi is also a solution), what will be the effect of the x³ term? Will it be possible for $x^4 + x^3 + 2x^2$ to be equal to the pure real number -4?

I was going to say State what you understand by 'Real part equals the complex part' which, I don't know, sounded slightly loose language to me, but what I understood is what doktorwho has said. It is telling you something special about one quadratic factor of the polynomial.

And there is also something else a bit special about the coefficients, and hence the roots. So lay that out and you have a few things you can may be put together (I also haven't worked the problem).

It seems that the other two zeros are complex as well. So we have some $a+zi$ and $a-zi$ as zeros.
$P(x)=(a+zi)(a-zi)(b+bi)(b-bi)$
$P(x)=2b^2 (a^2 + z^2)$
Doesnt seem right..
I can also write it in this form but doesn't seem to give any ideas..
$(x^2 + x + 2) x^2 + 4 = 0$

 

[Mark44]

It seems that the other two zeros are complex as well. So we have some $a+zi$ and $a-zi$ as zeros.
$P(x)=(a+zi)(a-zi)(b+bi)(b-bi)$

You didn't follow up on the question I asked, which is, in essence, "Is it possible for b + bi to be a solution?"
In your work above, you are tacitly assuming that both b + bi and b - bi are solutions.

$P(x)=2b^2 (a^2 + z^2)$
Doesnt seem right..
I can also write it in this form but doesn't seem to give any ideas..
$(x^2 + x + 2) x^2 + 4 = 0$

Factoring part of one side of an equation is no help.

 

[doktorwho]

You didn't follow up on the question I asked, which is, in essence, "Is it possible for b + bi to be a solution?"
In your work above, you are tacitly assuming that both b + bi and b - bi are solutions.

Factoring part of one side of an equation is no help.

It can't be equal to -4. Maybe there's some other methid I am not noticing.. Well if one complex zero has the two parts equal shouldn't ($b+bi$) be the right way to state that?

 

[Ray Vickson]

Homework Statement

Given the polynomial function $x^4+x^3+2x^2+4=0$ solve it if you know that it has at least one complex zero whose real part equals the complex part.

Homework Equations

3. The Attempt at a Solution [/B]
My guess is that if this function has one complex zero it must have a conjugate complex zero as well so we need to find those and two more. I am not sure how to start except realising that $P(b-bi)=P(b+bi)=0$ but solving the equation for that seems a lot of work and I am sure there's an easir way. Should i use Viet's formulas or some other way to approach this?

I think the problem is wrong: it is impossible to have a root of the form $x = a + i a$ for real $a$. It is true that one of the roots has a real part that is near the imaginary part, but they are not equal exactly.

You should be able to show this explicitly: expand out $P(a + ia)$ to find its real and imaginary parts. If you equate both of those parts to zero you will see that there is no solution for $a$.

 

[haruspex]

Homework Statement

Given the polynomial function $x^4+x^3+2x^2+4=0$ solve it if you know that it has at least one complex zero whose real part equals the complex part.

Are you sure it isn't $x^4+x^3+2x^2+2x+4=0$?

 

[Ray Vickson]

Are you sure it isn't $x^4+x^3+2x^2+2x+4=0$?

That does work.

 

[ehild]

Wolframalpha says that there are no roots with equal real and imaginary parts.

 

[Mark44]

Given the polynomial function $x^4+x^3+2x^2+4=0$ solve it if you know that it has at least one complex zero whose real part equals the complex part.

Assuming that the equation above is the correct equation (and not as surmised in post #8), the part "if you know that it has at least one complex zero whose real part equals the complex part." is clearly false, so no work is required.

 

[doktorwho]

I just got a response saying that it was a typo, $2x$ term should be included and then $-1-i$ and $-1+i$ are solutions and the rest can be gotten by dividing the polynomial by $(x-(-1-i))*(x-(-1+1))$.