Answer to Question #234108 in Complex Analysis for smi

Question #234108

Show that f(z)=z2(ez-1) is differentiable at zo=0. Check whether f is analytic at O.


1
Expert's answer
2021-09-07T18:44:13-0400

The derivative of the function f(z)f(z) is: f(z)=2z(ez1)+z2ez=ez(z2+2z)2zf'(z)=2z(e^z-1)+z^2e^z=e^z(z^2+2z)-2z.

f(0)=0f(0)=0. A necessary conditions for the function to be analytic are: f(z,zˉ)zˉ=0\frac{\partial f(z,\bar{z}) }{\partial{\bar{z}}}=0. The latter holds. Rewrite the function f(z)f(z) as: f(z)=(x+iy)2(ex+iy1)=(x2y2+2xiy)(exeiy1)=(x2y2)exeiy(x2y2)+2xiyexeiy2xyi=(x2y2)ex(cos(y)+isin(y))+2xyiex(cos(y)+isin(y))2xiyf(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)f(z)=u(x,y)+iv(x,y) it is obvious that partial derivatives ux\frac{\partial u}{\partial x}, uy\frac{\partial u}{\partial y}, vx\frac{\partial v}{\partial x}, vy\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.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment