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), $3r^2cosp \theta= \dfrac{3}{r}r^p cos 3 \theta$ or, $3r^2cosp \theta= 3r^{p-1} cos 3 \theta \Rightarrow p=3 (\because r^2 =r^{p-1} \Rightarrow p-1=2 \Rightarrow p=3 , cos p \theta =cos 3 \theta \Rightarrow p=3)$

From (2), $-r^3 p sinp \theta=-rp r^{p-1}sin 3 \theta$ or $r^2sinp \theta=r^{p-1}sin 3 \theta$ $\Rightarrow p=3 (\because r^2=r^{p-1} \Rightarrow p=3, sin p \theta= sin 3 \theta \Rightarrow p=3)$

Thus $p=3.$                                  Answer.