Mixed proofs
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Exercise:
Take f:mathbbCrightarrow mathbbC holomorphic and injective with f. abcliste abc Let r . Show that there exists a epsilon such that |fz| geq epsilon for all zin mathbbC with |z|geq r. abc Show that the isolated singularity in z of the function g:mathbbCbackslash rightarrow mathbbC gz fleftfraczright is a pole. abc Show that fz is a polynom and that fzcz for a cin mathbbCbackslash . abcliste
Solution:
abcliste abc Let r . Since f is holomorphic the open mapping theorem implies that fB_r is open. f implies that there exists an epsilon such that B_epsilon subset fB_r. Since f is injective for all z such that |z|geq r it holds that |fz|geq epsilon. abc Suppose that z is an essential singularity of g. Casaroti-Weierstrass implies that there exists a sequence of distinct pos z_krightarrow such that ffracz_krightarrow . We take k_ such that for all kgeq k_ it holds that |z_k| R r. Then |fz_k|geq epsilon which gives us a contradiction. If z were a removable singularity then we would have lim_zrightarrow fleftfraczright c This implies that for all |z| big enough we would have fz bounded. Liouville implies that f would be constant. So z is a pole. abc There exists a k positive eger such that gzfleftfraczrightz^k is holomorphic. Thus g is bounded around zero and we can conclude that |fz| leq c|z|^k. Cauchy's inequalities imply that f is a polynomial. Since it is injective and f we must have fzcz for cin mathbbCbackslash . abcliste
Take f:mathbbCrightarrow mathbbC holomorphic and injective with f. abcliste abc Let r . Show that there exists a epsilon such that |fz| geq epsilon for all zin mathbbC with |z|geq r. abc Show that the isolated singularity in z of the function g:mathbbCbackslash rightarrow mathbbC gz fleftfraczright is a pole. abc Show that fz is a polynom and that fzcz for a cin mathbbCbackslash . abcliste
Solution:
abcliste abc Let r . Since f is holomorphic the open mapping theorem implies that fB_r is open. f implies that there exists an epsilon such that B_epsilon subset fB_r. Since f is injective for all z such that |z|geq r it holds that |fz|geq epsilon. abc Suppose that z is an essential singularity of g. Casaroti-Weierstrass implies that there exists a sequence of distinct pos z_krightarrow such that ffracz_krightarrow . We take k_ such that for all kgeq k_ it holds that |z_k| R r. Then |fz_k|geq epsilon which gives us a contradiction. If z were a removable singularity then we would have lim_zrightarrow fleftfraczright c This implies that for all |z| big enough we would have fz bounded. Liouville implies that f would be constant. So z is a pole. abc There exists a k positive eger such that gzfleftfraczrightz^k is holomorphic. Thus g is bounded around zero and we can conclude that |fz| leq c|z|^k. Cauchy's inequalities imply that f is a polynomial. Since it is injective and f we must have fzcz for cin mathbbCbackslash . abcliste