My question is about the function $f(z)=e^{-z^2}$. Is it everywhere continous? Holomorphic?




Holomorphicity is a much stronger condition than continuity. If you can show that a function is holomorphic then it is continuous, and then some!

Let's examine the Cauchy-Riemann equation. If $f(z) = e^{-z^2}$ then consider $f(x,y) = e^{-(x+iy)^2}.$ Notice that $(x+iy)^2 = (x^2-y^2) + 2ixy$ and so we have

$$e^{-(x+iy)^2} = e^{(y^2-x^2)-2ixy} = e^{y^2-x^2} \cdot e^{-2ixy} = e^{y^2-x^2}(\cos(2xy)-i\sin(2xy)) \, . $$

Thus $U(x,y) = e^{y^2-x^2}\cos(2xy)$ while $V(x,y)=-e^{y^2-x^2}\sin(2xy).$ We need $U_x = V_y$ and $U_y = -V_x.$ Direct computation shows that:

$$U_x = -2e^{y^2-x^2}(x\cos(2xy)+y\sin(2xy)) \, , $$ $$U_y = 2e^{y^2-x^2}(y\cos(2xy)-x\sin(2xy)) \, , $$ $$V_x = -2e^{y^2-x^2}(y\cos(2xy)-x\sin(2xy)) \, , $$ $$V_y = -2e^{y^2-x^2}(x\cos(2xy)+y\sin(2xy)) \, . $$

Thus, the C-R equations hold for all $x$ and $y$ and so $f(z)$ is holomorphic on all of $\mathbb{C}$, i.e. $f$ is an "entire function". A more enlightening explanation is that $f$ is holomorphic if and only if $df/d\overline{z} = 0$. Using the product, chain and quotient rules you can easily show that the sum, product, quotient and composite of holomorphic functions are again holomorphic.


Compositions of continuous functions are continuous, and compositions of holomorphic functions are holomorphic (and since holomorphic functions are continuous, you get continuity here for free). $f$ is just $z\mapsto z^2 \mapsto -(z^2) \mapsto e^{(-(z^2))}$, so it is holomorphic (and thus continuous) on the entire complex plane.

Alternate argument: A holomorphic function is one with a power-series representation. You can plug $-z^2$ into the exponential power series and show that it converges everywhere.

This function is not holomorphic, or even continuous on the Riemann sphere, since it takes every complex number infinitely often near $\infty$. In fact, a function which is holomorphic on the entire Riemann sphere is necessarily constant.


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