$\displaystyle u=\left(r+\frac{a^2}{r}\right)\cos{\theta{}}=\left(r+a^2r^{-1}\right)\cos{\theta{}} \\[2ex] \displaystyle Differentiating\ partially\ w.r.t.\ ‘r’ \\[2ex] \displaystyle \frac{\partial{}u}{\partial{}r}=\left(1+a^2\bullet{}-1r^{-2}\right)\cos{\theta{}} \\[2ex] \displaystyle =\left(1-\frac{a^2}{r^2}\right)\cos{\theta{}}\ \ \ \ \rightarrow{}\left(i\right) \\[2ex] $

$\displaystyle Using\ Cauchy’s\ Reimann’s\ equation\ in\ polar\ form, \\[2ex] \displaystyle \ \frac{\partial{}u}{\partial{}r}=\frac{1}{r}\frac{\partial{}v}{\partial{}\theta{}} \\[2ex] \displaystyle \\[2ex] \displaystyle \therefore{}\left(1-\frac{a^2}{r^2}\right)\cos{\theta{}}=\frac{1}{r}\frac{\partial{}v}{\partial{}\theta{}}\ \ \ from\ \left(i\right) \\[2ex] \displaystyle \therefore{}\partial{}v=r\left(1-\frac{a^2}{r^2}\right)\cos{\theta{}\partial{}\theta{}} \\[2ex] $

$\displaystyle Integrating\ on\ both\ sides\ \\[2ex] \displaystyle \\[2ex] \displaystyle \therefore{}v=\left(r-\frac{a^2}{r}\right)\sin{\theta{}} \\[2ex] \displaystyle \\[2ex] $

$\displaystyle Hence\ f(z)=u+iv \\[2ex] \displaystyle =\left(r+\frac{a^2}{r}\right)\cos{\theta{}}+i\bullet{}\ \left(r-\frac{a^2}{r}\right)\sin{\theta{}} \\[2ex] \displaystyle =\ r\cos{\theta{}}+\frac{a^2}{r}\cos{\theta{}}+i\bullet{}r\sin{\theta{}}-i\bullet{}\frac{a^2}{r}\sin{\theta{}} \\[2ex] \displaystyle =r\left(\cos{\theta{}}+i\bullet{}\sin{\theta{}}\right)+\frac{a^2}{r}\left(\cos{\theta{}}-i\bullet{}\sin{\theta{}}\right) \\[2ex] \displaystyle =re^{i\theta{}}+\frac{a^2}{r}e^{-i\theta{}} \\[2ex] \displaystyle =\ re^{i\theta{}}+\frac{a^2}{re^{i\theta{}}} \\[2ex] \displaystyle =z+\frac{a^2}{z}\ \ \ (since\ z=\ re^{i\theta{}}) \\[2ex] \displaystyle \therefore{}Analytic\ function\ f\left(z\right)=\ z+\frac{a^2}{z} \\[2ex]$