Solution to Homeowrk                         Sec 2.2, page 63 
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#9.  Show lim as z -> -i of 1/z is i.
     Pf.  

     |1/z - i| = |1/z + 1/i|, since 1/i = -i 
     
	       = |z+i| / |z|, after getting a common denominator.

     Now, 
     
     |z - (-i)| < delta  =>  1 - |z| < delta, since by the triangle  
     
     inequality |A| - |B| \leq |A - B| \leq |A| + |B|, where \leq

     means "less than or equal to."
     
     => |z| > 1-delta, after solving the last inequality for |z|

     => 1 / |z| < 1 / (1-delta), by taking reciprocals.

     Thus,

     |1/z - i| < delta/(1-delta).  Therefore, if we choose 

     delta = epsilon/(1+epsilon) we are done.
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#12. This problem was done in class. Refer to your lecture notes for
     details.
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