\[\begin{array}{*{35}{l}}
Re\text{ }({{z}^{2}})\text{ }=\text{ }0,\text{ }\left| z \right|\text{ }=\text{ }2 \\
Let\text{ }z\text{ }=\text{ }x\text{ }+\text{ }iy. \\
Then,\text{ }\left| z \right|\text{ }=\text{ }\surd ({{x}^{2}}~+\text{ }{{y}^{2}}) \\
\end{array}\]
Given in the question,
$$ \[\begin{array}{*{35}{l}}
\surd ({{x}^{2}}~=\text{ }{{y}^{2}})\text{ }~=\text{ }2 \\
\Rightarrow ~{{x}^{2}}~+\text{ }{{y}^{2}}~=\text{ }4\text{ }\ldots \text{ }\left( i \right) \\
{{z}^{2}}~=\text{ }{{x}^{2}}~+\text{ }2ixy\text{ }\text{ }{{y}^{2}} \\
=\text{ }({{x}^{2}}~-\text{ }{{y}^{2}})\text{ }+\text{ }2ixy \\
\end{array}\]
Now, Re (z2) = 0
\[\Rightarrow ~{{x}^{2}=}~\text{ }{{y}^{2}}~=\text{ }0\text{ }\ldots \text{ }\left( ii \right)\]
Equating (i) and (ii), we get
\[\begin{array}{*{35}{l}}
\Rightarrow ~{{x}^{2}}~=\text{ }{{y}^{2}}~=\text{ }2 \\
\Rightarrow ~x\text{ }=\text{ }y\text{ }=\text{ }\pm \surd 2 \\
Hence,\text{ }z\text{ }=\text{ }x\text{ }+\text{ }iy \\
=\text{ }\pm \surd 2\text{ }\pm \text{ }i\surd 2 \\
=\text{ }\surd 2\text{ }+\text{ }i\surd 2,\text{ }\surd 2-\text{ }\text{ }i\surd 2,\text{ }\surd 2\text{ }+\text{ }i\surd 2\text{ }and \\
-\text{ }\surd 2-\text{ }\text{ }i\surd 2 \\
\end{array}\]
Hence, we have four complex numbers.