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

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1. Dec 3, 2015

cr7einstein

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

3. 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.

2. Dec 3, 2015

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.

3. Dec 3, 2015

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, lets 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.

4. Dec 3, 2015

Samy_A

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

5. Dec 3, 2015

7. Dec 4, 2015

Samy_A

Thought so, thanks.