# $\{z\in\mathbb{C} | \operatorname{Re}(z)=\operatorname{Im}(z)\}$ open set

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!!

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

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 $$0$$.
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.