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Is $f: \mathbb{C} \to \mathbb{C}, z \to\operatorname{Re}(z)^2+\operatorname{Im}(z)^3+i(\operatorname{Im}(z)^2-\operatorname{Re}(z)^3)$ holomorphic?

Cauchy-Riemann equations show: $$u_x=2x, u_y=3y^2, v_x=-3x^2, v_y=2y.$$ So $u_x=v_y, u_y=-v_x$ for $z\in U=\{z\in\mathbb{C} | \operatorname{Re}(z)=\operatorname{Im}(z)\}.$ Now I have shown that $f$ is complex differentiable, but to know if $f$ is holomorphic, I need to know if $U$ is an open set…

Thanks for any support!!

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  • $\begingroup$ Hint: $\Re(z)=\frac{z+\bar z}{2}$ and $\Im(z)=\frac{z-\bar z}{2i}$... $\endgroup$
    – Surb
    Apr 12 at 16:13
  • $\begingroup$ Can you describe the set geometrically? $\endgroup$
    – Umberto P.
    Apr 12 at 16:13
  • $\begingroup$ If you identify $\mathbb{C}=\mathbb{R}^2$ (sorry, I don´t know the isomorphic sign), then U is the identiy function $\endgroup$
    – sina1357
    Apr 12 at 16:15
  • $\begingroup$ How can I conclude that $f$ is not holomorphic by using Surb´s hint? $\endgroup$
    – sina1357
    Apr 12 at 16:19

2 Answers 2

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No, $U$ is not an open set. If $r>0$, then $r\in B_r(0)$, but $r\notin U$. So, $U$ contains no open disk centered at $0$.

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Geometrically, the set $$ U = \{ z = u + i v \in \mathcal{C} : \mbox{Re}(z) = \mbox{Im}(z) \} $$ is the straight line $v = u$ passing through the origin.

Obviously, $U$ is not an open set in the complex plane $\mathcal{C}$. Indeed, $U$ does not contain any open ball $B(\mathbf{0}, r)$ centered at the origin.

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