f(z) = e^x+iy = e^x.e^iy

= e^x(cos y + i sin y)     (∵ e^iy =  cos y + i sin y)

= e^x cos y + i e^x sin y

By comparing with f(z) = u(x, y) + i v(x, y), we get

u(x, y) = e^x cos y and

v(x, y) = e^x sin y

\(\frac{\partial u}{\partial x}\) = e^x cos y which is continuous on TR²

\(\frac{\partial u}{\partial y}\) = - e^x sin y which is continuous on TR²

\(\frac{\partial v}{\partial x}\) = e^x sin y which is continuous on TR²

\(\frac{\partial v}{\partial y}\) = e^x cos y which is continuous on TR²

By observation, we obtain

\(\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}\) and \(\frac{\partial u}{\partial y}= -\frac{\partial v}{\partial x}\)\(\)

Hence, function f satisfies the cauchy-riemann equations.

Since, \(\frac{\partial u}{\partial x}\),  \(\frac{\partial u}{\partial y}\), \(\frac{\partial v}{\partial x}\) and \(\frac{\partial v}{\partial y}\) are continuous on TR² and satisfy the cauchy-riemann equations.

\(\therefore\) f(z) = e^x+iy is an analytical function.