Find the real constants a,b,c such that the function is analytic $f(z)=r^2a\cos2\theta+ br\sin\theta+i(cr^2 \sin\theta\cos\theta+ r\cos\theta)$ The cauchy-Rieman equations for polar coordinates are $\frac{\partial{u}}{\partial{r}}=\frac{1}{r}\times\frac{\partial{v}}{\partial{\theta}}$ $\frac{\partial{v}}{\partial{r}}=\frac{-1}{r}\times\frac{\partial{u}}{\partial{\theta}}$ The derivatives are, $\frac{\partial{v}}{\partial{\theta}}=cr^2\cos2\theta-r\sin\theta\\$ $\frac{\partial{u}}{\partial{r}}=2ra\cos2\theta+b\sin\theta\\$
So we get,$2ra\cos2\theta+b\sin\theta=cr\cos2\theta-\sin\theta$
And, $\frac{\partial{v}}{\partial{r}}=2rc\sin\theta\cos\theta+\cos\theta$ $\frac{\partial{u}}{\partial{\theta}}=-4r^2a\sin\theta\cos\theta+br\cos\theta\\$ We get, $2rc\sin\theta\cos\theta+\cos\theta=4ra\sin\theta\cos\theta-b\cos\theta\\$
I don't know what to do after this or how to solve. I have tried my best but the the equating part is confusing.Here after equating I got $a=\frac{c}{2}$ and $b=-1$. Since the question is to find the real constants we should get a real numbers for a and c right? So what to do now? My previous question was wrong and sorry for that.
Your two conditions $$2ra\cos2\theta+b\sin\theta=cr\cos2\theta-\sin\theta$$ $$2rc\sin\theta\cos\theta+\cos\theta=4ra\sin\theta\cos\theta-b\cos\theta$$ must hold for every $r,\theta$, so you must just "equate the coefficients", like in your previous post: $$c=2a,\quad b=-1,$$ as you wrote. There is no other constraint and $a$ may even be any complex number. Any function $$f(z)=r^2a\cos2\theta-r\sin\theta+i(2ar^2 \sin\theta\cos\theta+ r\cos\theta)$$ is analytic and even polynomial, as could be found directly: $$ar^2(\cos2\theta+2i\sin\theta\cos\theta)+ r(i\cos\theta-\sin\theta)=az^2+iz.$$