Show that f(z)=z2(ez-1) is differentiable at zo=0. Check whether f is analytic at O.
The derivative of the function "f(z)" is: "f'(z)=2z(e^z-1)+z^2e^z=e^z(z^2+2z)-2z".
"f(0)=0". A necessary conditions for the function to be analytic are: "\\frac{\\partial f(z,\\bar{z}) }{\\partial{\\bar{z}}}=0". The latter holds. Rewrite the function "f(z)" as: "f(z)=(x+iy)^2(e^{x+iy}-1)=(x^2-y^2+2xiy)(e^xe^{iy}-1)=(x^2-y^2)e^xe^{iy}-(x^2-y^2)+2xiye^xe^{iy}-2xyi=(x^2-y^2)e^x(cos(y)+i\\,sin(y))+2xyie^x(cos(y)+i\\,sin(y))-2xiy"
In case we split real and imaginary part as: "f(z)=u(x,y)+iv(x,y)" it is obvious that partial derivatives "\\frac{\\partial u}{\\partial x}", "\\frac{\\partial u}{\\partial y}", "\\frac{\\partial v}{\\partial x}", "\\frac{\\partial v}{\\partial y}" are continuous. The latter are sufficient conditions for the function to be analytic. Thus, the function is analytic. In particular, at point 0.
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