Region of roots

is that from Mathematica, or any reasoning?

sir are there 4 roots ?
since the sum of roots is 0
is one root is z then -z is another root
_ _
also if z is one root then z is also a root and - z
so four roots are there

Divide the given eqn on both sides by z^2 to have

z^5+4+\frac{11}{z^2}=0

Taking modulo on both sides we have \left|z^5+4 \right|=\frac{11}{|z^2|}\le |z|^5+4 from triangle's inequality, from which we get the obvious reasoning that |z|>1.

Similarly we have |z|^2=\left|\frac{11}{z^5+4} \right|\le \frac{11}{|z|^5+4}

From this we have that |z|<2....thus any z satisfying the given eqn must have its modulo between 1 and 2.

@theprophet
now that you mentioned it, this is what Mathematica gives

Not much useful.

As far as I am concerned, I used the Rouche's theorem (which should be familiar if any one does a course on complex analysis) which gives easily the required number of zeroes in annulus 1<|z| <2.

Rouche's theorem doesnt (directly) give the number of zeroes in an annulus.

You will first need soumik's result that we need |z|>1.

Soumik, your second inequality is incorrect. Would you like to try some more?

More or less its direct by Rouche's.
For |z|=2,
|4z²+11| ≤ 4|z|² + 11 =27 < 128 =|z⁷|
So in the disc |z|<2, the polynomial z⁷ + 4z² +11 has the same number of roots as z⁷, namely, 7 (counting multiplicities).

On the other hand, on the unit circle |z|=1, the least value of |4z²+11| is 7 which is greater than 1=|z⁷|. As such on |z|=1,
|z⁷|<|4z²+11|

Hence, in the disc |z|<1, the polynomial z⁷ + 4z² +11 has the same number of roots as 4z²+11, namely, 0.

And, obviously there are no roots with |z|=1.

Hence, all the roots of z⁷ + 4z² +11 lie in the annulus 1 < |z| < 2.

I know - I messed up the triangle's inequality....[4]

Dividing by z³ we have

z^4+\frac{4}{z}+\frac{11}{z^3}=0

Taking modulo on both sides we have |z|^4=\left|\frac{4}{z}+\frac{11}{z^3} \right|\le \frac{4}{|z|}+\frac{11}{|z|^3}

That gives |z|<2....

I think this is free from errors....

That's a 7th degree polynomial equation, so there are seven roots (counting multiplicities).

@soumik. Nice work, that was the approach I was looking for