Question

d o ille Tunctions on to the effect that every bounded harmonic function on the whole complex plane has to be a constant function. (11) (Fundamental Theorem of Algebra) Let P be a polynomial given by de the unit circle P(x) = do +212 + a2z2 + ... + anz", ZEC, where do, 21, 22, ...,an, are complex constants and n is a positive in- teger. Prove that P has a zero in C in the sense that there exists a complex number zo such that P(20) = 0. orphic function co Sunded entire furti Zositive munter

Answer 1

Now, Let P(2) = a + a2 + Qz 2 ² + ... + an-12h tan.ah where Zec and as, a, a, ..., an are complex Constants. and a 70. and n is a positiv Suppose that P(2) does not have a zero in c. Then P(2) is never zero in c, which means that I is analytic everywhere. ie is an function. (2) entire Now, we see that a 0 ***** Ples zu Panterat 0.91 P() This implies that I l< for sufficiently large 2, say 1217R. sine every continuous function on a closed and bounded Set is bounded, it follows that in the remaining 2 lie in 1215R, I is bounded Hene P(2) is bounded for all 2EC. Then by Lioville's theorem, 1 is constant and hence P(2) is constant which is a contradiction as nz 1. [= P(2) is a polynomial] Thus the polynomial P (2) has at least one 2000. Say 20..i e P (20) =0. Thus, the polynomial P (2) has at least one 2000 P(2) inc.