# Cauchy Riemann $e^{z^2}$.

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.

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

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.