 Solution

ID: past-exam-of-the-algebra-analysis-and-differential-equations-1-course-of-the-university-of-oxford/2013/analysis/4/b/solution

At any

z \neq = 0

, take sufficiently small

h

z + h \neq = 0

\frac{( z + h ) ^{- 1} - z ^{- 1}}{h} = - \frac{1}{z ( z + h )} ⟶ - \frac{1}{z ^{2}} .

Hence the complex derivative exists at every point of the punctured plane, and

f^{'} (z) = - z^{- 2}, f is holomorphic on C ∖ {0} .

The excluded point is essential to the domain statement: the reciprocal has a simple pole there.

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