Holomorphic, entire functions and the Cauchy-Riemann equations

Question

1. Where is $f\colon\mathbb C\to\mathbb C,~z\mapsto |z|^2$ $\mathbb C$-differentiable? Is there a restriction of $f$ that is holomorphic?
2. Is there an entire function $f$ with $\operatorname{Re}(f(x+\mathrm iy))=2xy$? If so, write it in the form of $f(z)$.
3. For which $\alpha\in\mathbb R$ does an entire function $f$ exist with $\operatorname{Re}(f(x+\mathrm iy))=x^2+\alpha y^2$? If there is such a function write it in the form of $f(z)$.

1. We can write $f(x+\mathrm i y)=u(x,y)+\mathrm iv(x,y)$ with $u(x,y)=x^2+y^2$ and $v(x,y)=0$. The Cauchy-Riemann equations yield $$\begin{align*}\frac{\partial u}{\partial x} = 2x &= 0 = \frac{\partial v}{\partial y}\\\frac{\partial u}{\partial y} = 2y &= 0 =-\frac{\partial v}{\partial x}.\end{align*}$$ This is only true for $x=y=0$ thus only for $z=0$. Hence I assume that only the restriction to $\{0\}$ would be holomorphic. Alternatively we know that $\overline z$ is not $\mathbb C$-differentiable thus $f(z)=z\overline z$ isn't either for all $z\neq 0$.
2. If $f$ is entire and thus holomorphic the Cauchy-Riemann equations hold hence we are looking for a function $$f(x+\mathrm i y) = 2xy + \mathrm i v(x,y)$$ which suffices $$\frac{\partial u}{\partial x}=2y,\qquad \frac{\partial u}{\partial y}=2x.$$ Applying the C-R equations yields $$\frac{\partial v}{\partial x}=-2x,\qquad \frac{\partial v}{\partial y}=2y.$$ Therefore we get $$v(x,y)=\int\frac{\partial v}{\partial x}\mathrm dx+g(y) = -x^2+c_1+g(y)$$ for a function $g(y)$ that has to be found and an arbitrary additive constant $c_1$. Since $$\frac{\partial(-x^2+c_1+g(y))}{\partial y}=g'(y)=2y=\frac{\partial v}{\partial y}$$ we know that $$g(y)=\int g'(y)\mathrm dy = y^2+c_2$$ for another arbitrary constant $c_2$ hence $v(x,y)=-x^2+y^2+(c_1+c_2)$ and finally $$f(x+\mathrm iy)=2xy +\mathrm i(-x^2+y^2+(c_1+c_2)).$$
3. Similiarly to 2. we have $$f(x+\mathrm iy)=(x^2+\alpha y^2)+\mathrm iv(x,y)$$ which suffices $$\begin{align*}\frac{\partial u}{\partial x} = 2x &= 2x = \frac{\partial v}{\partial y}\\\frac{\partial u}{\partial y} = 2\alpha y &= 2\alpha y =-\frac{\partial v}{\partial x}.\end{align*}$$ Once again we obtain $$v(x,y)=\int\frac{\partial v}{\partial x}\mathrm dx + g(y) = -2\alpha yx+c_1+ g(y)$$ where partial differentiation of the expression yields $$\frac{\partial (-2\alpha yx+c_1+ g(y))}{\partial y} = -2\alpha x+g'(y)=2x=\frac{\partial v}{\partial y}\implies g'(y)=2x(1+\alpha).$$ Therefore we obtain $$g(y)=2x(1+\alpha)\int \mathrm dy = 2xy(1+\alpha)+c_2\implies v(x,y)=2xy+(c_1+c_2).$$ A partial derivation for $x$ yields $\alpha =-1$ and finally $$f(x+\mathrm iy)=(x^2-y^2)+\mathrm i(2xy+(c_1+c_2))$$ as the only function to suffice the given properties.

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I am looking for feedback on my current solutions - are they even correct? And what can be optimized?

EDIT: Apparently I should rewrite my solutions for 2. and 3. as polynomials $f(z)$ in $z$ where i have to find appropriate coefficients. How may I do so?