In my textbook I found a text where it says that $e^z$ (z is a complex number) is analytic everywhere.
But $e^x=e^{\mathrm{Re}\,z}$ is not.
How can I prove that about $e^x$ and what is the difference?
In my textbook I found a text where it says that $e^z$ (z is a complex number) is analytic everywhere.
But $e^x=e^{\mathrm{Re}\,z}$ is not.
How can I prove that about $e^x$ and what is the difference?
Hint: Use the Cauchy-Riemann Equations to check the differentiability
Let $f(z)=e^x=u(x,y)+iv(x,y)$
Check the condition $u_x=v_y,u_y=-v_x$
There is a lot in the implicit notation here let us be explicit:
The map $$\exp : \begin{cases}\mathbb{C} \to \mathbb{C} \\ z \mapsto e^z \end{cases} $$ is analytic; for example as it is given by the powerseries $\sum_{n=0}^{\infty} \frac{z^{n}}{n!}$
The map $$\exp \circ \Re : \begin{cases}\mathbb{C} \to \mathbb{C} \\ z \mapsto e^{\Re z} \end{cases} $$ is not analytic, for example as the Cauchy-Riemann equations do not hold, (they do not even hold at a single point). It is of course a power series in $\Re z$, yet not in $z$ which is what is relevant.
Yet the map $$\exp : \begin{cases}\mathbb{R} \to \mathbb{C} \\ x \mapsto e^x \end{cases} $$ is of course still (real) analytic. It is the restriction of the first to the reals, that happens to be equal to the second there.