Find the number of squareroots of z7 4z3 2 0 in the regio

Find the number of squareroots of z^7 - 4z^3 - 2 = 0 in the region 1

Solution

Consider the equation z⁷ - 4z³ - 2 = 0

First, let us start by seeing how many roots of the polynomial lie in |z| < 2.

Let f(z) = z⁷ - 4z³ - 2 and g(z) = z⁷

On the boundary of the domain | z | = 2 , we have | f(z) - g(z) | = | - 4z³ - 2 | = | - ( 4z³ + 2 ) | = | 4z³ + 2 |

So that | f(z) - g(z) | 4 |z|³ + 2 = 4(2)³ + 2 = 34 and | g(z) | = | z |⁷ = (2)⁷ = 128

Hence on the boundary |z| = 2 we have , | f(z) - g(z) | | g(z) |

Since the domain is bounded with a smooth boundary and f(z) and g(z) are both analytic on D U D

Rouche\'s Theorem gives us that f(z) has the same number of zeros as g(z) in |z| < 2.

Hence f(z) must have 7 zeros in |z| < 2 since g(z) has a zero of multiplicity seven at z = 0.

Now, let\'s consider how many roots of f(z) lie in |z| < 1.

Let us define h(z) = - 2 then on the boundary of unit circle |z| = 1

we have , | f(z) - h(z) | = | z⁷ - 4z³ | |z|⁷ - 4|z|³ = (1)⁷ - 4(1)³ = 1 - 4 = - 3

and on the boundary of unit circle |z| = 1 , we have |h(z)| = | - 2 | = 2 hence | f(z) - h(z) | < |h(z)| on |z| = 1

Hence, by Rouche\'s f(z) and h(z) must have the same number of roots in |z| < 1, namely zero.

Since h(z) = -2 has no zeros inside |z| < 1 and hence f(z) has ni zeros inside |z| < 1.And so, all the 7 roots of f(z) must lie in the annulus 1 < |z| < 2.