How to rebuild the analytical function f(z) if given #U(x,y)=Re f(z)=2x+(x/(x^(2)+y^(2)))# #0<|z|<+oo#?

2 Answers
Mar 13, 2017

# f(z) = (2x+x/(x^2+y^2)) + 1i #

Explanation:

By the very definition of #|z|# we have #0 le |z| lt oo#, so the only restriction is to ensure that #|z| != 0#

Suppose that we define:

# f(z) = U(x,y) + V(x,y)i #
# \ \ \ \ \ \ \ = (2x+x/(x^2+y^2)) + V(x,y)i #

Where #V(x,y)# is an arbitrary real function

Then we require:

# |U^2(x,y)+V^2(x,y)| > 0 #
# :. |(2x+x/(x^2+y^2)) + V(x,y)i| > 0 #
# :. sqrt((2x+x/(x^2+y^2))^2 + V^2) > 0 #
# :. (2x+x/(x^2+y^2))^2 + V^2 > 0 #

Clearly:

# 0 le (2x+x/(x^2+y^2))^2 < oo #

So we can choose any function #V(x,y) > 0#

Thus one such function is:

# f(z) = (2x+x/(x^2+y^2)) + 1i #

Mar 13, 2017

See below.

Explanation:

If #f(z)=u(x,y)+iv(x,y)# is analytic then

#(partial u)/(partial x)=(partial v)/(partial y)#
#(partial v)/(partial x) = -(partial u)/(partial y)#

These equations occurred already in the 18th century in J.L. d'Alembert's and L. Euler's studies on functions of a complex variable.

Here #u(x,y)=2x+x/(x^(2)+y^(2))# and

#(partial u)/(partial x) = 2 - (x^2 - y^2)/(x^2 + y^2)^2 = (partial v)/(partial y)#
#(partial u)/(partial y) =-(2 x y)/(x^2 + y^2)^2=-(partial v)/(partial x) #

but

#v_1(x,y)=int_0^x(2 x y)/(x^2 + y^2)^2dx+xi(y) = y/(x^2 + y^2)+xi(y)#

and also

#v_2(x,y)=int_0^y (2 - (x^2 - y^2)/(x^2 + y^2)^2)dy+zeta(x)=2 y - y/(x^2 + y^2)+zeta(x)#

but #v_1(x,y)-v_2(x,y) = 2y-xi(y)+eta(x)=0#

then #eta(x)=0# and #xi(y)=2y#

so finally

#v(x)=2y+ y/(x^2 + y^2)#

Concluding,

#f(z)=2x+x/(x^(2)+y^(2))+i(2y+ y/(x^2 + y^2))#