Welcome to our community

Be a part of something great, join today!

Complex stuff questions (5)

Markov

Member
Feb 1, 2012
149

1) Let $a>0.$ Prove that $p(z)=z^3+3a^2z-1$ has a simple zero on $(0,1)$ and it doesn't have more real roots.

2) Consider a meromorphic function $f:\mathbb C\to\mathbb C$ so that $|f(z)|\xrightarrow[|z|\to\infty]{}\infty.$ Prove that $f$ is a rational function.

Attempts:

1) I'd need to use Rouché's Theorem here, so I think I need to split it in two cases, when $|z|=0$ and $|z|=1.$ For first case let $|p(z)-1|=|z^3+3a^2z|\le|z^3|+3a^2|z|=0\le1\le|p(z)|+1,$ since $p(z)$ and $1$ don't have zeroes for $|z|=0,$ then by Rouché's Theorem $p(z)$ has no zeros for $|z|>0.$ For $|z|=1$ let $|p(z)-3a^2z|=|z^3-1|\le 1<3a^2=3a^2|z|,$ here's my problem, I think it should be $a>1$ so $|p(z)-3a^2z|\le 3a^2|z|\le |p(z)|+3a^2|z|,$ since $p(z)$ and $3a^2z$ don't have zeroes for $|z|=1$ by Rouché's Theorem $p(z)$ has a single zero on $|z|<1.$ In conclusion $p(z)$ has only a root for $0<|z|<1$ and the zero is simple because $p'(z)=3z^2+3a^2\ne0.$

2) I think I can write $g(z)=p(z)f(z)$ where $p(z)$ is a monic polynomial whose roots are the poles of $f,$ then $g(z)$ is a entire function and has convergent Taylor series, so $g(z)=\displaystyle\sum_{j=0}^\infty\frac{g^{(j)}(0)}{j!}z^j,$ so the only thing I need to prove is that $g$ is a polynomial, but I don't see how to use the info. of the limit.
 
Last edited:

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,707
1) Let $a>0.$ Prove that $p(z)=z^3+3a^2z-1$ has a simple zero on $(0,1)$ and it doesn't have more real roots.
You don't need anything as advanced as Rouché for this.

1. $p(0) = -1 <0$, $p(1) = 3a^2>0$: Intermediate value theorem (there is a root in $(0,1)$).

2. $p'(z) = 3z^2+3a^2>0$ for all real $z$: Rolle's theorem (there cannot be more than one real zero, or a repeated zero).
 

Markov

Member
Feb 1, 2012
149
Oh, do those work for complex anyway? I didn't know it. So it was quite simple, thanks.

Could you help me with second problem please?
 

Markov

Member
Feb 1, 2012
149
Can anybody check my work for 2)? Am I on the right track?
 

AlexYoucis

New member
Jan 26, 2012
19
Can anybody check my work for 2)? Am I on the right track?
I mean, are you aware of the statement that an entire function with a pole at infinity is a polynomial?
 

Markov

Member
Feb 1, 2012
149
Yes, I know that fact, you mean the solution is short?
 

AlexYoucis

New member
Jan 26, 2012
19
Yes, I know that fact, you mean the solution is short?
You tell me, is it obvious that $fp$ has a pole at infinity?
 

Markov

Member
Feb 1, 2012
149
$f_p$ ?
 
Last edited by a moderator:

AlexYoucis

New member
Jan 26, 2012
19

Markov

Member
Feb 1, 2012
149
$z=\infty$ is a pole of $f,$ which means that $z=0$ is a pole of $g(z)=f\left(\dfrac1z\right).$ Since the poles are isolated, exists $R>0$ so that $|z|<R,$ then $g(z)$ has no poles (except at $z=0).$ Making $z\mapsto\dfrac1z$ we have $|z|>\dfrac1R,$ then $f$ has no poles. On the other hand on $|z|\le\dfrac1R$ we have infinite poles since is compact, and if they were infinite, then we'll have an accumulation point, contradicting the fact that they are isolated. This shows that $f$ has finite poles, so we can write $g(z)=p(z)f(z),$ so I need to prove that $g$ is a polynomial.

If you see my work on post #1, then I have
$g(z)=\displaystyle\sum_{j=0}^\infty\frac{g^{(j)}(0)}{j!}z^j,$ so if I can prove that $g$ is a polynomial I'm done. I was trying to use Cauchy's integral formula but I don't have a bound for $f,$ how would you prove that $g$ is a polynomial?
 

CaptainBlack

Well-known member
Jan 26, 2012
890
You don't need anything as advanced as Rouché for this.

1. $p(0) = -1 <0$, $p(1) = 3a^2>0$: Intermediate value theorem (there is a root in $(0,1)$).

2. $p'(z) = 3z^2+3a^2>0$ for all real $z$: Rolle's theorem (there cannot be more than one real zero, or a repeated zero).
It happens that Descartes rule of signs shows that there is exactly one positive real root (so the one in (0,1) is the only positive root) and no negative real roots.

The positive root could in principle have multiplicity 1 or 3, but as there is no \(z^2\) term the cubic cannot be a cube of a linear term, so the root in (0,1) is of multiplicity 1.

CB
 
Last edited: