Question

Let z be a complex number and f(z) = frac{e^{2z}}{(z-1)^2}. Then the Laurent series of f in 0 < |z - 1| < ? is e^2 [frac{2}{z-1} + frac{1}{(z-1)^2}] + sum_{n=0}^{?} frac{2^{n+2}e^2}{(n+2)!} (z-1)^n None of these the above e^{-2} [frac{2}{z-1} + frac{1}{(z-1)^2}] + sum_{n=0}^{?} frac{2^{n+2}e^{-2}}{(n+2)!} (z-1)^n e^{-2} [frac{2}{z-1} + frac{1}{(z-1)^2}] + sum_{n=0}^{?} frac{2^ne^{-2}}{(n+2)!} (z-1)^n

          Let z be a complex number and f(z) = frac{e^{2z}}{(z-1)^2}. Then the Laurent series of f in 0 < |z - 1| < ? is

e^2 [frac{2}{z-1} + frac{1}{(z-1)^2}] + sum_{n=0}^{?} frac{2^{n+2}e^2}{(n+2)!} (z-1)^n

None of these
the above
e^{-2} [frac{2}{z-1} + frac{1}{(z-1)^2}] + sum_{n=0}^{?} frac{2^{n+2}e^{-2}}{(n+2)!} (z-1)^n
e^{-2} [frac{2}{z-1} + frac{1}{(z-1)^2}] + sum_{n=0}^{?} frac{2^ne^{-2}}{(n+2)!} (z-1)^n

$Let z be a complex number and f(z) = frace^2z(z-1)^2. Then the Laurent series of f in 0 < |z - 1| < ? is e^2 [frac2z-1 + frac1(z-1)^2] + sumn=0^? frac2^n+2e^2(n+2)! (z-1)^n None of these the above e^-2 [frac2z-1 + frac1(z-1)^2] + sumn=0^? frac2^n+2e^-2(n+2)! (z-1)^n e^-2 [frac2z-1 + frac1(z-1)^2] + sumn=0^? frac2^ne^-2(n+2)! (z-1)^n$

Added by Erin M.

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