If all 3 roots of ##az^3+bz^2+cz+d=0## have negative real part

[cr7einstein]

Homework Statement

Hi all!

The problem is - 'find the condition that all roots of $$f(z)=az^3+bz^2+cz+d=0$$ have negative real part, where $$z$$ is a complex number'.

The answer - $$a,b,d$$ have the same sign.

Homework Equations

The Attempt at a Solution

Honestly, I have no clue about how to proceed. Here is what I tried- $$ f'(z)=3az^2+2bz+c$$, which at extrema gives the roots as $$z=\frac{-2b+/-\sqrt{(4b^2-12ac)}}{6a}$$. If the real part is negative, then $$\frac{-2b}{6a}<0$$, which implies that $$a,b$$ have the same sign. I am not sure if what I have done is right, and have no idea about proving the rest of it. Please help.
Thanks in advance!

 

[HallsofIvy]

I don't see that "extrema" have anything to do with this. Are we to assume that the coefficients are real? You don't say that but I suspect it is assumed. In that case, roots come in complex complementary pairs. We can write them as u+ iv, u- iv, and w with u and w negative. So the equation must be of the form a(z- u- iv)(z- u+ iv)(z- w)= a((z-u)^2+ v^2)(z- w)= a(z^2- 2uz+ u^2+ v^2)(z- w)= az^3- (aw+ 2au)z^2+ (au^2+ av^2+ 2auw)z- (au^2w+ av^2w)= az^3+ bz^2+ cz+ d.

 

[cr7einstein]

Ok, I think I have it.

Since $$z$$ is a complex number, $$f(z)$$ must have a complex root. As they occur in conjugate pairs, 2 out of 3 roots of f(z) must be imaginary. The remaining one shall be purely real (and obviously rational). Now, all three have negative real parts. So, let's name the roots-$$\alpha=-x+iy, \beta=-x-iy, \gamma=-k$$, where $$x,y,k>0$$. Now,since $$\alpha+\beta+\gamma=-b/a$$; we get (after substituting and a trivial simplification),

$$-(2x+k)=-b/a$$. As both x and k are positive, we get $$b/a>0$$ i.e. $$b$$ and $$a$$ are of same sign.(Which can also be arrived at with differentiation, as in my question)

Now, $$\alpha\beta\gamma=-d/a$$, which after substitution and simplification gives-
$$(-k)(x^2+y^2)=-d/a$$.

As $$x^2+y^2$$ is always positive, multiplying with negative number $$(-k)$$ gives the LHS as negative. Thus, $$-d/a<0$$, implying $$d/a>0$$, and hence, $$a,d$$ have same sign.

Combining above two results, $$a,b,d$$ have the same sign.

 

[Samy_A]

Ok, I think I have it.

Since $$z$$ is a complex number, $$f(z)$$ must have a complex root. As they occur in conjugate pairs, 2 out of 3 roots of f(z) must be imaginary. The remaining one shall be purely real (and obviously rational).

It's not relevant to your solution, but why is it obvious that the real root is rational?

 

[Mark44]

Minor complaint about your layout: you are overusing LaTeX here, which is why you have so many lines with just a single character on them. For many of the things you wrote, inline LaTeX is a big improvement over standalone LaTeX. For inline, use ## at the beginning and end, instead of $$.
Making this change, your second sentence below becomes:

Since $z$ is a complex number, $f(z)$ must have a complex root.

Isn't that an improvement?

Ok, I think I have it.

Since $$z$$ is a complex number, $$f(z)$$ must have a complex root. As they occur in conjugate pairs, 2 out of 3 roots of f(z) must be imaginary. The remaining one shall be purely real (and obviously rational). Now, all three have negative real parts. So, let's name the roots-$$\alpha=-x+iy, \beta=-x-iy, \gamma=-k$$, where $$x,y,k>0$$. Now,since $$\alpha+\beta+\gamma=-b/a$$; we get (after substituting and a trivial simplification),

$$-(2x+k)=-b/a$$. As both x and k are positive, we get $$b/a>0$$ i.e. $$b$$ and $$a$$ are of same sign.(Which can also be arrived at with differentiation, as in my question)

Now, $$\alpha\beta\gamma=-d/a$$, which after substitution and simplification gives-
$$(-k)(x^2+y^2)=-d/a$$.

As $$x^2+y^2$$ is always positive, multiplying with negative number $$(-k)$$ gives the LHS as negative. Thus, $$-d/a<0$$, implying $$d/a>0$$, and hence, $$a,d$$ have same sign.

Combining above two results, $$a,b,d$$ have the same sign.

 

[Ray Vickson]

It's not relevant to your solution, but why is it obvious that the real root is rational?

It might not be, if $a$, $b$, $c$ and/or $d$ are irrational.

 

[Samy_A]

It might not be, if $a$, $b$, $c$ and/or $d$ are irrational.

Thought so, thanks.