Question

Consider f(z)=eᶻ−4z−1. Show that there is exactly one root inside the contour ∣z∣=1

Answer 1

To show that there is exactly one root inside the contour |z|=1, we can use the argument principle.

Step-by-step explanation:

To show that there is exactly one root inside the contour |z|=1, we can use the argument principle. Let's consider the function f(z)=eᶻ−4z−1.

First, we need to check if f(z) satisfies the conditions of the argument principle:

• f(z) is analytic inside and on the contour |z|=1 (excluding the origin).
• There are no poles or zeros of f(z) on the contour |z|=1.
• The number of zeros minus the number of poles of f(z) inside the contour |z|=1 is equal to the change in the argument of f(z) as z travels once around |z|=1.

Since f(z) satisfies the conditions, we can conclude that there is exactly one zero of f(z) inside the contour |z|=1.