Answer
$x^{-1+2i}=x^{-1}\cos (2\ln x)+i[x^{-1}\sin (2\ln x)]$
Work Step by Step
We will use the Euler's Formula
\[e^{i\theta}=\cos\theta+i\sin\theta\]
$x^{-1+2i}=x^{-1}x^{2i}$
$x^{-1+2i}=x^{-1}e^{2i\ln x}$
$x^{-1+2i}=x^{-1}e^{i(2\ln x)}$
By using Euler's Formula
$x^{-1+2i}=x^{-1}[\cos (2\ln x)+i\sin (2\ln x)]$
$x^{-1+2i}=x^{-1}\cos (2\ln x)+i[x^{-1}\sin (2\ln x)]$
Here,
$u(x)=x^{-1}\cos (2\ln x)$ and $v(x)=x^{-1}\sin (2\ln x)$
Hence ,
$x^{-1+2i}=x^{-1}\cos (2\ln x)+i[x^{-1}\sin (2\ln x)]$.
