With z=a+bi, the conjugate, \(\displaystyle\overline{{z}}\), is a-bi. Hence, the expression \(\displaystyle{z}^{{{2}}}-\overline{{{z}}}^{{{2}}}\), is equivalent to

\(\displaystyle{\left({a}+{b}{i}\right)}^{{{2}}}-{\left({a}-{b}{i}\right)}^{{{2}}}={\left[{\left({a}\right)}^{{{2}}}+{2}{\left({a}\right)}{\left({b}{i}\right)}^{{{2}}}\right]}-{\left[{\left({a}\right)}^{{{2}}}-{2}{\left({a}\right)}{\left({b}{i}\right)}+{\left({b}{i}\right)}^{{{2}}}\right]}\)

\(\displaystyle{\left({u}{s}{e}{\left({a}+{b}\right)}^{{{2}}}={\left({a}\right)}^{{{2}}}+{2}{\left({a}\right)}{\left({b}\right)}+{\left({b}\right)}^{{{2}}}\right)}\)

\(\displaystyle={\left[{a}^{{{2}}}+{2}{a}{b}{i}+{b}^{{{2}}}{i}^{{{2}}}\right]}-{\left[{a}^{{{2}}}-{2}{a}{b}{i}+{b}^{{{2}}}{i}^{{{2}}}\right]}={a}^{{{2}}}+{2}{a}{b}{i}+{b}^{{{2}}}{i}^{{{2}}}-{a}^{{{2}}}+{2}{a}{b}{i}-{b}^{{{2}}}{i}^{{{2}}}={\left({a}^{{{2}}}-{a}^{{{2}}}\right)}+{\left({2}{a}{b}{i}+{2}{a}{b}{i}\right)}+{\left({b}^{{{2}}}{i}^{{{2}}}-{b}^{{{2}}}{i}^{{{2}}}\right)}={0}+{4}{a}{b}{i}+{0}={\left({4}{a}{b}{i}\right)}{i}\).

Hence, \(z^{2}-\overline{z}^{2}=(4abi)i\).