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+Q1)
+
+$$f(z) = {4z-5\over z(z-1)}.$$
+
+To determine the residue of $|z|=2,$ we sum the residues of interior
+poles ($z=0$ and $z=1.$)
+
+Starting with $z=0,$
+$$f(z) = -{4z-5\over z}\cdot{1\over 1-z} = (5/z-4)(1+z+z^2+\ldots),$$
+giving a residue of $b_1 = 5.$
+
+Then, $z=1,$
+$$f(z) = {4(z-1)-1\over z-1}\cdot{1\over 1-(1-z)} = (4 -
+1/(z-1))(1-(z-1)+(z-1)^2+\ldots),$$
+giving a residue of $b_1 = -1.$
+
+The sum of these residues is $4,$ so the value of the given integral is
+$2\pi i\cdot4 = 8\pi i.$
+
+Q2)
+
+$$f(z) = {z^3 + 2z\over (z-i)^3}.$$
+
+The residue at $z=i$ is $g''(i)/2!$ where $g(z) = z^3 + 2z,$ giving
+$g''(i)/2! = (6i)/2! = 3i.$
+
+Q3)
+
+I'm going to sleep in, and I think I'll be able to actually relax on
+Wednesday.
+
+\bye