\[{{i}^{-39}}~=\text{ }1/\text{ }{{i}^{39}}~=\text{ }1/\text{ }{{i}^{4\text{ }x\text{ }9\text{ }+\text{ }3}}~\]

\[=\text{ }1/\text{ }({{1}^{9}}~x\text{ }{{i}^{3}})\text{ }=\text{ }1/\text{ }{{i}^{3}}~=\text{ }1/\text{ }\left( -i \right)\text{ }\]

\[\left[ {{i}^{4}}~=\text{ }1,\text{ }{{i}^{3}}~=\text{ }-I\text{ }and\text{ }{{i}^{2}}~=\text{ }-1 \right]\]

Presently, increasing the numerator and denominator by\[~I\] we get

\[{{i}^{-39}}~=\text{ }1\text{ }x\text{ }i\text{ }/\text{ }\left( -i\text{ }x\text{ }i \right)\]

\[=\text{ }i/\text{ }1\text{ }=\text{ }i\]

Consequently,

${{I}^{-}}39=0+I$