Find the laurent series for (1-exp2z)/z
f(z)=∑n=0∞cn(z−z0)nf(z)=\displaystyle{\sum_{n=0}^{\infin}}c_n(z-z_0)^nf(z)=n=0∑∞cn(z−z0)n
cn=f(n)(z0)n!c_n=\frac{f^{(n)}(z_0)}{n!}cn=n!f(n)(z0)
z0=0z_0=0z0=0
1−e2z=−2z+2z2−4z3/3+2z4/3−4z5/15+...1-e^{2z}=-2z+2z^2-4z^3/3+2z^4/3-4z^5/15+...1−e2z=−2z+2z2−4z3/3+2z4/3−4z5/15+...
1−e2zz=1z(1−e2z)=−2+2z−4z2/3+2z3/3−4z4/15+...\frac{1-e^{2z}}{z}=\frac{1}{z}(1-e^{2z})=-2+2z-4z^2/3+2z^3/3-4z^4/15+...z1−e2z=z1(1−e2z)=−2+2z−4z2/3+2z3/3−4z4/15+...
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