Are The Following Analytic/Differentiable Functions?

Question

1.$f(z)=x+y^3i$

2.$f(z)=z+Re(z)$

3.$f(z)=x^3+y^5$

4.$f(z)=(z-1)(Re(z))^2$

1. we get

$u_x=1$ $v_y=3y^2$

$u_y=0$ $-v_x=0$

So we have $3y^2=1\iff y=\pm \frac{1}{\sqrt{3}}$

So the function is differentiable at $(x,\frac{1}{\sqrt{3}})$ and $(x,-\frac{1}{\sqrt{3}})$? but not analytic as a line is not an open set?

2. if it is analytic so $Re(z)=f(z)-z$ is analytic, contradiction

3.we get

$u_x=3x^2$ $v_y=5y^4$

$u_y=0$ $-v_x=0$

So $3x^2=5y^4\iff x=y=0$

So the function is differentiable at $(0,0)$ but not analytic as a point is not an open set

4. How to attack this?

Answer 1

None of your functions is analytic because, for each of them, there is no point of $w\in\mathbb C$ such that the function is differentiable in some disk centered at $w$. You can check this easily using the Cauchy-Riemann equations.