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\begin{equation*}
\begin{aligned}
(1) &\frac{2i}{1-i} \\ &= \frac{2i(1+i)}{(1-i)(1+i)} \\
&= \frac{-2+2i}{2}\\ &=-1+2i
\end{aligned}
\hspace{2cm}
\begin{aligned}
(2) &\frac{1-\sqrt{3}i} {1+ \sqrt{3}i} \\
&= \frac{(1-\sqrt{3}i)^2} {(1+ \sqrt{3}i)(1-\sqrt{3}i)}\\
&= \frac{1-2\sqrt{3}i+(\sqrt{3}i)^2}{1-(\sqrt{3}i)^2}\\
\end{aligned}
\quad \quad
\begin{aligned}
&=\frac{-2-2\sqrt{3}i}{1+3} \\ &= \frac{-2-2\sqrt{3}i}{4} \\
&= \frac{-1-\sqrt{3}i} {2}
\end{aligned}
\hspace{7cm}
\end{equation*}


Souce code
$\verb+\begin{equation*}+$
$\verb?\begin{aligned}\\?$
$\verb?(1) &\frac{2i}{1-i} \\ &= \frac{2i(1+i)}{(1-i)(1+i)} \\ ?$
$\verb?&= \frac{-2+2i}{2}\\ &=-1+2i \\?$
$\verb?\end{aligned} \\?$
$\verb?\hspace{2cm}\\?$
$\verb?\begin{aligned}\\?$
$\verb?(2) &\frac{1-\sqrt{3}i} {1+ \sqrt{3}i} \\ ?$
$\verb?&= \frac{(1-\sqrt{3}i)^2} {(1+ \sqrt{3}i)(1-\sqrt{3}i)}\\ ?$
$\verb?&= \frac{1-2\sqrt{3}i+(\sqrt{3}i)^2}{1-(\sqrt{3}i)^2}\\ ?$
$\verb?\end{aligned} \\?$
$\verb?\quad \quad \\?$
$\verb?\begin{aligned}\\?$
$\verb?&=\frac{-2-2\sqrt{3}i}{1+3} \\ &= \frac{-2-2\sqrt{3}i}{4} \\ ?$
$\verb?&= \frac{-1-\sqrt{3}i} {2} \\?$
$\verb?\end{aligned} \\?$
$\verb?\hspace{7cm} \\?$
$\verb+\end{equation*}+$