Given, $u = r^2 cos2\theta - rcos\theta + 2$

$u_{r} = \frac{\partial u}{\partial r} = 2rcos\theta - cos\theta$

$u_{\theta} = -2r^2 sin2\theta + rsin\theta $

$u_{\theta} = \frac{\partial u}{\partial \theta} = r^2(-sin2\theta)2 - r(-sin\theta)$

Now by Cauchy Riemann equations

$\frac{\partial u}{\partial r} = \frac{1}{r} \frac{\partial v}{\partial \theta}$ and $\frac{\partial v}{\partial r} = \frac{-1}{r} \frac{\partial u}{\partial \theta} \\ \frac{\partial v}{\partial r} = \frac{-1}{r} \frac{\partial u}{\partial \theta} = \frac{-1}{r}(-2r^2 sin2\theta + rsin\theta)$

$v_{r} = \frac{\partial v}{\partial r} = (2r sin 2\theta - sin\theta)$

Now

$\frac{\partial v}{\partial \theta} = r\frac{\partial u}{\partial r} = r(2rcos2\theta - cos\theta)$

$v_{\theta} = \frac{\partial v}{\partial \theta} = 2r^2cos2\theta - rcos\theta$

Now integrating either $v_{\theta}$ w.r.t. $\theta$ or v, w.r.t. r.

Now integrating $v_{\theta}$ w.r.t. $\theta$

$V = \int 2r^2 sin2\theta - rcos\theta$

Now f(z) $= u + iv \\ = r^2 cos2\theta - rcos\theta + 2 + i(r^2 sin2\theta - rsin\theta) \\ = r^2 (cos2\theta + isin2\theta) - r(cos\theta + isin\theta) + 2 \\ = r^2 e^{i2\theta} - re^{i2\theta} + 2 \\ f(z) = (re^{i\theta})^2 - re^{i\theta} + 2 \\ \therefore f(z) = z^2 - z + 2.......(\because z = re^{i\theta}, polarform)$

Analytic function $f(z) = z^2 - z + 2$