Solution
$\operatorname{Re}\left(\mathrm{a} \mathrm{z}^{2}+\mathrm{bz}\right)=\mathrm{a} \Rightarrow \mathrm{az}^{2}+\mathrm{bz}+\mathrm{a} \overline{\mathrm{z}}^{2}+\mathrm{b} \overline{\mathrm{z}}=2 \mathrm{a} \Rightarrow \mathrm{a}\left(\mathrm{z}^{2}+\overline{\mathrm{z}}^{2}\right)+\mathrm{b}(\mathrm{z}+\overline{\mathrm{z}})=2 \mathrm{a}$
\[
\operatorname{Re}\left(\mathrm{b} \mathrm{z}^{2}+\mathrm{az}\right)=\mathrm{b} \Rightarrow \mathrm{~b} \mathrm{z}^{2}+\mathrm{az}+\mathrm{b} \overline{\mathrm{z}}^{2}+\mathrm{a} \overline{\mathrm{z}}=2 \mathrm{~b} \Rightarrow \mathrm{~b}\left(\mathrm{z}^{2}+\overline{\mathrm{z}}^{2}\right)+\mathrm{a}(\mathrm{z}+\overline{\mathrm{z}})=2 \mathrm{~b}
\]
\[
\begin{aligned}
& (1) \times b-(2) \times(a) \Rightarrow\left(b^{2}-a^{2}\right)(z+\bar{z})=0 \Rightarrow(z+\bar{z})=0\left(a^{2} \neq b^{2}\right) \\
& (1) \times a-(2) \times(b) \Rightarrow\left(a^{2}-b^{2}\right)(z+\bar{z})=2\left(a^{2}-b^{2}\right)\left(a^{2} \neq b^{2}\right) \\
& z^{2}+\bar{z}^{2}=2 \Rightarrow(z+\bar{z})^{2}-2 z \bar{z}=2 \Rightarrow z \bar{z}=-1 \Rightarrow 1+1^{2}=-1 \Rightarrow \text { No solution But when } a=-b, \operatorname{Re}\left(a z^{2}-a z\right)=a \\
& \Rightarrow \operatorname{Re}\left(a\left(x^{2}-y^{2}+i 2 x y\right)-a(x+i y)\right)=a \Rightarrow a\left(x^{2}-y^{2}\right)-a x=a \Rightarrow x^{2}-y^{2}-x=1 \Rightarrow x^{2}-x-1=y^{2}
\end{aligned}
\]
For any real values of y there two values of x, hence infinite complex numbers are possible.
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