I have to solve an equation but I don't know if it's analytic or not, but I suppose it is.

It would be easy if it was $e^z$ without the exponent because I would only need to transform it to $e^x(\cos y+i\sin y)$ but this one has an exponent of $2$.

If I solve the polynomial it would be $e^{(x^2-y^2+2ixy)}$ and if I use the function exponential of complex number like the previous example, it would be $e^{(x^2-y^2)}(\cos 2xy+i\sin 2xy)$ and when I derive it, they won't be the same ($U_x$ is not $V_y$, $U_y$ is not $-V_x$).

As I say, I don't know if it's analytic or not.

Thanks for the answer.

  • $\begingroup$ Welcome to MSE. Please use MathJax to typeset math on this site. $\endgroup$
    – saulspatz
    Oct 6, 2019 at 16:02
  • 1
    $\begingroup$ Are you really sure that $u_x\neq v_y$ and $u_y\neq -v_x$? $\endgroup$ Oct 6, 2019 at 16:02
  • 1
    $\begingroup$ It's certainly analytic, since both $z^2$ and $e^z$ are entire functions. $\endgroup$
    – saulspatz
    Oct 6, 2019 at 16:04
  • $\begingroup$ I mean we have $$\exp(z^2)=1+\sum_{n=1}^\infty \frac{z^{2n}}{n!}$$ $\endgroup$ Oct 6, 2019 at 16:05

1 Answer 1


Upon careful scrutiny it is revealed that Cauchy-Riemann does in fact hold for the function $e^{z^2}$; with

$z = x + iy, \tag 0$

we have

$z^2 = x^2 - y^2 + 2ixy, \tag{0.1}$

as is in fact both easily calculated and well-known. Then

$e^{z^2} = e^{x^2 - y^2 + 2ixy} = e^{x^2 - y^2}(\cos 2xy + i \sin 2xy), \tag 1$

and thus

$U = e^{x^2 - y^2}\cos 2xy, \tag 2$

$V = e^{x^2 - y^2}\sin 2xy; \tag 3$

we differentiate:

$U_x = 2xe^{x^2 - y^2}\cos 2xy - 2ye^{x^2 - y^2}\sin 2xy, \tag 4$

$U_y = -2ye^{x^2 - y^2}\cos 2xy - 2x e^{x^2 - y^2}\sin 2xy, \tag 5$

$V_x = 2x e^{x^2 - y^2}\sin 2xy + 2y e^{x^2 - y^2}\cos 2xy; \tag 6$

$V_y = -2y e^{x^2 - y^2}\sin 2xy + 2x e^{x^2 - y^2}\cos 2xy. \tag 7$

Inspection of (4)-(7) shows that

$U_x = V_y, \; U_y = -V_x, \tag 8$

as we would expect since $e^{z^2}$ is the composition of the two holomorphic functions $z^2$ and $\exp(\cdot)$, hence itself holomorphic.


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