Question:
In the neighborhood of z = 1, the function f(z) has a power series expansion of the form f(z) = 1+(1-z)+(1-z)2 + .... then f(z) is
In the neighborhood of z = 1, the function f(z) has a power series expansion of the form f(z) = 1+(1-z)+(1-z)2 + .... then f(z) is
Updated On: Mar 12, 2025
- \(\frac{1}{z}\)
- \(\frac{-1}{z-2}\)
- \(\frac{1}{2z-1}\)
- \(\frac{z-1}{z+1}\)
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The Correct Option is A
Solution and Explanation
The correct answer is(A): \(\frac{1}{z}\)
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