2.15

2.15 Prove that if \(f(z)\) is given by a polynomial in \(z\), then \(f(z)\) is entire.
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Let \(z_0 \in \mathbb{C}\) and \(f(z)=a_0+a_1z+a_2z^2+...+a_nz^n\) be an arbitrary polynomial in \(z\). Let \(f_i(z)\) be a function of \(z\) given by \(f_i(z)=a_iz^i\) for some \(i \in \lbrace0,1,2,...,n\rbrace\). Then, \(f(z)=f_0(z)+f_1(z)+f_2(z)+...+f_n(z).\) Then by Proposition 2.15 on page 30, \(f'(z)=(f_0(z)+f_1(z)+...+f_n(z))'=f'_0(z)+f'_1(z)+f'_2(z)+...+f'_n(z)\). Again by Proposition 2.15, for every \(f_i(z)\), we have \(f'_i(z)=(a_iz^i)'=a_i(z^i)'=ia_iz^{i-1}\). Thus, \(f'(z)=a_1+2a_2z+3a_3z^2+...+na_nz^{n-1}\). Thus \(f'(z_0)=(a_iz_0^i)'=a_1+2a_2z_0+3a_3z_0^2+...+na_nz_o^{n-1}\). Because the derivative exists for \(z_0 \in \mathbb{C}, f(z)\) is entire.