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Write the following functions $f(z)$ in the forms $f(z)=u(x, y)+i v(x, y)$ under Cartesian coordinates with $u(x, y)=\operatorname{Re}(f(z))$ and $v(x, y)=\operatorname{Im}(f(z))$ :
(a) $f(z)=z^{3}+z+1$
(b) $f(z)=z^{3}-z$;
(c) $f(z)=\frac{1}{i-z}$;
(d) $f(z)=\overline{\exp \left(z^{2}\right)}$
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Solution. (a)
\begin{aligned} f(z) &=(x+i y)^{3}+(x+i y)+1 \\ &=(x+i y)\left(x^{2}-y^{2}+2 i x y\right)+x+i y+1 \\ &=x^{3}-x y^{2}+2 i x^{2} y+i x^{2} y-i y^{3}-2 x y^{2}+x+i y+1 \\ &=x^{3}-3 x y^{2}+x+1+i\left(3 x^{2} y-y^{3}+y\right) \end{aligned}
(b)
\begin{aligned} f(z) &=z^{3}-z=(x+y i)^{3}-(x+y i) \\ &=\left(x^{3}+3 x^{2} y i-3 x y^{2}-y^{3} i\right)-(x+y i) \\ &=\left(x^{3}-3 x y^{2}-x\right)+i\left(3 x^{2} y-y^{3}-y\right) \end{aligned}

(c)
\begin{aligned} f(z) &=\frac{1}{i-z}=\frac{1}{-x+(1-y) i} \\ &=\frac{-x-(1-y) i}{x^{2}+(1-y)^{2}} \\ &=-\frac{x}{x^{2}+(1-y)^{2}}-i \frac{1-y}{x^{2}+(1-y)^{2}} \end{aligned}

(d)
$f(z)=\overline{\exp \left(z^{2}\right)}=\overline{\exp \left((x+y i)^{2}\right)}$
$=\overline{\exp \left(\left(x^{2}-y^{2}\right)+2 x y i\right)}$
$=\overline{e^{x^{2}-y^{2}}(\cos (2 x y)+i \sin (2 x y))}$
$=e^{x^{2}-y^{2}} \cos (2 x y)-i e^{x^{2}-y^{2}} \sin (2 x y)$
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