Answer:Given function $f(z)=r^3 cosp \theta+ir^psin3 \theta.$ Comparing it with the standard complex function $f(z)=u+iv.$

Here, $u=r^3cosp \theta, v=r^psin3 \theta.$

If the function is analytic, it should satisfy Cauchy-Reimann equation, i.e.,

 $\dfrac{\partial u}{\partial r}=\dfrac{1}{r}\dfrac{\partial v}{\partial \theta}$.................(1)

$\dfrac{\partial u}{\partial \theta}=-r\dfrac{\partial v}{\partial r}$..................(2)

Now from (1), $3r^2cosp \theta= \dfrac{3}{r}r^p cos 3 \theta$ …