Yes, there is a connection. Cauchy's integral formula assumes that the function in question is analytic. A function is analytic if and only if it satisfies the Cauchy-Riemann equations:
If f(z)=u(x,y)+iv(x,y), then
\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}
\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}
Since the function is analytic, then u and y have continuous partial derivatives of all orders, so we may differentiate the above expressions to obtain:
\frac{\partial^2 u}{\partial x^2}=\frac{\partial^2 v}{\partial x \partial y}
\frac{\partial^2 u}{\partial y^2}=-\frac{\partial^2 v}{\partial y \partial x}
Since these derivatives are continuos, then:
\frac{\partial^2 v}{\partial y \partial x}=\frac{\partial^2 v}{\partial x \partial y}
Therefore:
\frac{\partial^2 u}{\partial x^2}=-\frac{\partial^2 u}{\partial y^2}
\rightarrow \frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=0
Which is Laplace's equation. It can be proven similarly that the imaginary part of f also satisfies Laplace's equation.
