Answer to Question #252969 in Complex Analysis for Msa

Question #252969

Find the laurent series for (1-exp2z)/z


1
Expert's answer
2021-10-19T03:45:52-0400

f(z)=n=0cn(zz0)nf(z)=\displaystyle{\sum_{n=0}^{\infin}}c_n(z-z_0)^n


cn=f(n)(z0)n!c_n=\frac{f^{(n)}(z_0)}{n!}

z0=0z_0=0


1e2z=2z+2z24z3/3+2z4/34z5/15+...1-e^{2z}=-2z+2z^2-4z^3/3+2z^4/3-4z^5/15+...


1e2zz=1z(1e2z)=2+2z4z2/3+2z3/34z4/15+...\frac{1-e^{2z}}{z}=\frac{1}{z}(1-e^{2z})=-2+2z-4z^2/3+2z^3/3-4z^4/15+...


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