# Proving that a function can't be real part of an analytic function

Given the function \begin{align*}u:\mathbb R^2&\to\mathbb R\\(x,y)&\mapsto e^x\sinh(y),\end{align*} I want to find out whether or not $u$ can be the real part of an analytic function \begin{align*}f:\mathbb R^2&\to\mathbb C\\(x,y)&\mapsto u(x,y)+\mathrm iv(x,y).\end{align*}

I know a theorem stating it is possible iff $u$ is harmonic, which is not the case here. However, I want to find a way without using this theorem and am getting a little stuck:

Using the Cauchy-Riemann equations, we find that if there is indeed such a function $v$, then $\partial_y v=e^x\sinh(y), \partial_x v=-e^x\cosh(y)$. Integrating yields $$v(x,y)=e^x\cosh(y)+C_1(x)=-e^x\cosh(y)+C_2(y).$$ How can we prove that this is not possible, i. e. there is no function $v$ such that $v(x,y)-e^x\cosh(y)$ only depends on $x$ and $v(x,y)+e^x\cosh(y)$ only depends on $y$?

• The systematic way of trying to find $v$ is not to integrate both equations, but rather to integrate one equation, differentiate the result, and compare with the second equation. See this answer, or this, for example. – Hans Lundmark Oct 9 '17 at 16:51

Let $V(x, y)$ be the vector field on $\Bbb R^2$ given by

$\vec V(x, y) = \begin{pmatrix} -e^x\cosh y \\ e^x \sinh y \end{pmatrix}; \tag 1$

note that the components of $\vec V(x, y) = (V_x(x, y), V_y(x, y))$ are the two functions

$V_x(x, y) = -e^x\cosh y \tag 2$

and

$V_y(x, y) = e^x \sinh y \tag 3$

which would be the $x$- and $y$-components of $\nabla v(x, y)$, the gradient of $v(x, y)$, provided such a function $v(x, y)$ were to exist. So we seek to know if $\vec V(x, y)$ is the gradient of some function. This may be quite easily resolved in the present case by calulating $\nabla \times \vec V(x, y)$, since we know that a vector field is a gradient if and only if its curl vanishes. We have

$\nabla \times \vec V(x, y) = \det \left (\begin{bmatrix} \mathbf i & \mathbf j & \mathbf k \\ \partial_x & \partial_y & \partial_z \\ V_x(x, y) & V_y(x, y) & 0 \end{bmatrix} \right )$ $= (\partial_y(0) - \partial_z V_y) \mathbf i + (\partial_z V_x - \partial_x(0))\mathbf j + (\partial_x V_y - \partial_y V_x)\mathbf k = (\partial_x V_y - \partial_y V_x)\mathbf k ; \tag 4$

now,

$\partial_x V_y(x, y) - \partial_y V_x(x, y) = e^x \sinh y - (-e^x\sinh y) = 2e^x \sinh y \ne 0; \tag 5$

we conclude $\vec V(x, y)$ is not a gradient and hence no function $v(x, y)$ such that $\vec V(x, y) = \nabla v(x, y)$ exists. Thus $u(x, y)$ has no harmonic conjugate, and thus cannot be the real part of an analyic function.

If one wants to follow through with the approach suggested by our OP Mophotla, note that

$e^x\cosh(y)+C_1(x)=-e^x\cosh(y)+C_2(y) \tag 6$

yields

$C_2(y) - C_1(x) = 2e^x \cosh y; \tag 7$

taking $x = 0$ in (7),

$C_2(y) = C_1(0) + 2\cosh y, \tag 8$

whereas with $y= 0$ we find

$C_1(x) = -2e^x + C_2(0); \tag 9$

thus,

$2\cosh y + 2e^x + (C_1(0) - C_2(0)) = 2e^x \cosh y; \tag{10}$

since $e^x\cosh y \ne 0$ for any $x, y$, we may write

$2e^{-x} + \dfrac{2}{\cosh y} + \dfrac{C_1(0) - C_2(0)}{e^x \cosh y} = 2; \tag{11}$

now letting $x, y \to \infty$ together, we see that $e^x, \cosh y, e^x \cosh y \to \infty$, and (11) yields the contradiction

$0 = 2; \tag{12}$

thus functions such as $C_1(x)$, $C_2(y)$ cannot exist, and this finishes off the proof, according to our OP Mophotla's suggested method, that $u(x, y)$ is not the real part of an analytic function.