# 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?

Mar 13, 2017

$f \left(z\right) = \left(2 x + \frac{x}{{x}^{2} + {y}^{2}}\right) + 1 i$

#### Explanation:

By the very definition of $| z |$ we have $0 \le | z | < \infty$, so the only restriction is to ensure that $| z | \ne 0$

Suppose that we define:

$f \left(z\right) = U \left(x , y\right) + V \left(x , y\right) i$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = \left(2 x + \frac{x}{{x}^{2} + {y}^{2}}\right) + V \left(x , y\right) i$

Where $V \left(x , y\right)$ is an arbitrary real function

Then we require:

$| {U}^{2} \left(x , y\right) + {V}^{2} \left(x , y\right) | > 0$
$\therefore | \left(2 x + \frac{x}{{x}^{2} + {y}^{2}}\right) + V \left(x , y\right) i | > 0$
$\therefore \sqrt{{\left(2 x + \frac{x}{{x}^{2} + {y}^{2}}\right)}^{2} + {V}^{2}} > 0$
$\therefore {\left(2 x + \frac{x}{{x}^{2} + {y}^{2}}\right)}^{2} + {V}^{2} > 0$

Clearly:

$0 \le {\left(2 x + \frac{x}{{x}^{2} + {y}^{2}}\right)}^{2} < \infty$

So we can choose any function $V \left(x , y\right) > 0$

Thus one such function is:

$f \left(z\right) = \left(2 x + \frac{x}{{x}^{2} + {y}^{2}}\right) + 1 i$

Mar 13, 2017

See below.

#### Explanation:

If $f \left(z\right) = u \left(x , y\right) + i v \left(x , y\right)$ is analytic then

$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$
$\frac{\partial v}{\partial x} = - \frac{\partial u}{\partial y}$

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 $u \left(x , y\right) = 2 x + \frac{x}{{x}^{2} + {y}^{2}}$ and

$\frac{\partial u}{\partial x} = 2 - \frac{{x}^{2} - {y}^{2}}{{x}^{2} + {y}^{2}} ^ 2 = \frac{\partial v}{\partial y}$
$\frac{\partial u}{\partial y} = - \frac{2 x y}{{x}^{2} + {y}^{2}} ^ 2 = - \frac{\partial v}{\partial x}$

but

${v}_{1} \left(x , y\right) = {\int}_{0}^{x} \frac{2 x y}{{x}^{2} + {y}^{2}} ^ 2 \mathrm{dx} + \xi \left(y\right) = \frac{y}{{x}^{2} + {y}^{2}} + \xi \left(y\right)$

and also

${v}_{2} \left(x , y\right) = {\int}_{0}^{y} \left(2 - \frac{{x}^{2} - {y}^{2}}{{x}^{2} + {y}^{2}} ^ 2\right) \mathrm{dy} + \zeta \left(x\right) = 2 y - \frac{y}{{x}^{2} + {y}^{2}} + \zeta \left(x\right)$

but ${v}_{1} \left(x , y\right) - {v}_{2} \left(x , y\right) = 2 y - \xi \left(y\right) + \eta \left(x\right) = 0$

then $\eta \left(x\right) = 0$ and $\xi \left(y\right) = 2 y$

so finally

$v \left(x\right) = 2 y + \frac{y}{{x}^{2} + {y}^{2}}$

Concluding,

$f \left(z\right) = 2 x + \frac{x}{{x}^{2} + {y}^{2}} + i \left(2 y + \frac{y}{{x}^{2} + {y}^{2}}\right)$