could you please help me with number 5 Verify that the function f(a) is analytic

Question

could you please help me with number 5

Verify that the function f(a) is analytic on the anewalus 4 4al 5 and find the Laurent series expansion of f ror this antulus (centered at 1i). Shew the work (10 points) 2. Find principal part of the Laurent series of the function f(z) in o Irl oo. Use it to find the residue of the function at 0. Justiry your answers (10points) 3. Prove that ir f(z) has a pole of order at zo then f (c) has a pole of order n 1 at ze. Find the residue of the function f at Justify your answer. (10 points) 4. Evaluate the following integrals. Justify your calculations (10 points each) 8r 20 5. Use Rouché 's theorem to find the number of ronts of the polynomial as t za t 3 in the annul 1 Izl 2010 points)

Explanation / Answer

Let V be an open set and D be a closed disk contained in V . If f, g H(V ) satisfy |f(z) g(z)| < |f(z)| + |g(z)| on the boundary of D, then f and g have the same number of zeroes in the interior of D. The more classical Rouche’s theorem states the result for the stronger condition |f(z) g(z)| < |f(z)|

Let f(z) be the given polynomial. For |z| = 2, we have |f(z) z ^5 | = |z^3 + 3| |z^3| + 3 = 11 < 32 = |z^ 5 | . Therefore by Rouche’s theorem f(z) and z ^5 have the same number of zeros inside the disk |z| < 2. z^ 5 has five zeros (which are all 0, pun intended) in |z| < 2, so f has 5 zeros in |z| < 2. And for |z| = 1, |f(z) (z^3 + 3)| = |z ^5 | = 1 < 2 = ||z^3| | 3|| |z^3 (3)| = |z^3 + 3| . Therefore again by Rouche’s theorem f(z) and z^3 + 3 have the same number of zeros in |z| < 1. The only zero of z^3 + 3 is 3^(1/3) which is in the unit disk; thus f has a single zero in |z| < 1. Psychologically, the problem seems over here but there is a subtle point: We showed that f has four zeros in |z| < 2 and a single zero in |z| < 1. Hence, f has three zeros in 1 |z| < 2. To make the first inequality strict, we also need to show that f has no zeros on |z| = 1. Indeed, in this case |f(z)| = |z^3 + 3 (z ^5 )| ||z^3 + 3| | z ^5 || = ||z^3 + 3| 1| And just above we showed above that |z^3 + 3| > 2 on |z| = 1. Hence |f(z)| > 1 on |z| = 1. It might be worth noting that the functions we used to apply Rouche’s theorem to were all entire, i.e. in H(C), so we were able to use any disk we want.

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