2
$\begingroup$

Consider the three functions $\displaystyle e^{\frac{r}{\ln r}}$, $\displaystyle e^r$, and $\displaystyle e^{r\ln r}$, where $r = |z|$. Note that these are not constant functions.

Can someone explain to me why all these three functions are not entire, especially the second one?

$\endgroup$

closed as unclear what you're asking by Travis Willse, egreg, Semiclassical, user98602, Jonas Meyer Jul 17 '15 at 21:00

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ These functions are not differentiable in $z$. $\endgroup$ – Alex S Jun 26 '15 at 17:36
  • $\begingroup$ Are these supposed to be constant functions? $\endgroup$ – Travis Willse Jun 26 '15 at 17:38
  • $\begingroup$ If they're function of $r$, they're obviously not entire, since you define them only for $r>0$. $\endgroup$ – egreg Jun 26 '15 at 17:44
  • $\begingroup$ @Travis: They are not constant functions. $\endgroup$ – cantor_paradise Jun 26 '15 at 18:19
2
$\begingroup$

If the second function were entire, then it would satisfy the Cauchy Riemann equations for all $z$. Let $z=x+iy$ so that $|z| = \sqrt{x^2 + y^2}$. Now $e^r = e^{\sqrt{x^2+y^2}}$. Now check the Cauchy Riemann equations. Note that in this case $f(z) = e^{\sqrt{x^2+y^2}} + i \cdot 0$.

A similar approach will work for the other problems as well.

$\endgroup$
  • 1
    $\begingroup$ For convenience, let $f(z)=u(x,y)+iv(x,y)$, where $u(x,y)=e^{\sqrt{x^2+y^2}}$ and $v(x,y)=0$. If $f(z)$ were entire, then we would have that $f^{\prime}(z)$ exists at an arbitrary point $z_0=x_0+iy_0$. At such a point, the Cauchy-Riemann equations would also have to hold, i.e. $u_x=v_y$ and $u_y=-v_x$. For the function $e^r$, it is clear that $\frac{x}{|z|}e^{|z|}=u_x\neq v_y=0$ and $\frac{y}{|z|}e^{|z|}=u_y\neq -v_x=0$. Hence, the function $e^{|z|}$ is not entire. $\endgroup$ – cantor_paradise Jun 26 '15 at 18:41
  • $\begingroup$ Yep that's the idea. @cantor_paradise $\endgroup$ – Joel Jun 26 '15 at 19:04
1
$\begingroup$

For any entire function $f,$ $f$ is either constant or $f(\mathbb C)$ is open in $\mathbb C.$ Your functions are not constant, and in all cases $f(\mathbb C)\subset \mathbb R,$ hence $f(\mathbb C)$ cannot be open, so none of them can be entire.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.