While solving the Area theorem , i'm facing trouble in understanding the equation in these two black boxes, i know how they write $\displaystyle A=\frac{1}{2}\int_{c} R^2 d\phi$, but how they got its right hand term? and how they can use $\phi$ and $\theta$ in same integal , while they are polar coordinates in $w$ and $z$ planes respectively.
Please Help.
This question relies on the idea of writing $f(z)$ in polar coordinates in both the $z$ and $w$ planes. That is, if $z = r e^{i\theta}$, then there exist real-valued functions of two variables $R$ and $\phi$ such that $$ f(r e^{i\theta}) =R(r ,\theta)\exp[i\phi(r,\theta)] $$ Now, the integral in the $w$ plane is around the image of the contour $r = 1$ under $f$. Parameterizing this as $z = e^{i\theta}$, we get $$ \oint_C R^2d\phi = \oint_{|z| = 1} R(1,\theta)^2d\phi(1,\theta) = \int_0^{2\pi}R(1,\theta)^2 \frac{\partial\phi}{\partial\theta}d\theta $$ where they have dropped the argument of $R$ for brevity.
The second boxed equation comes from the Cauchy-Riemann equations that express the fact that $f$ is holomorphic. You're probably more familiar with them in Cartesian form where you write $f(x+iy) = u(x,y) + i v(x,y)$. Then they are $$ \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}\;\;\;\;;\;\;\;\;\frac{\partial u}{\partial y} =-\frac{\partial v}{\partial x}. $$ But no matter what representation you use for $z$ and $f$, $f$ still has to satisfy $\partial f/\partial\bar{z} = 0$ to be holomorphic. So the partial derivatives of $R$ and $\phi$ will have some constraint on them. I encourage you to work them out for yourself, but you will end up with $$ \frac{R}{r}\frac{\partial \phi}{\partial\theta} = \frac{\partial R}{\partial r}\;\;\;\;;\;\;\;\; R\frac{\partial \phi}{\partial r} = -\frac{1}{r}\frac{\partial R}{\partial \theta}. $$ And that's where the box comes from: it's just a restatement of one of the Cauchy-Riemann equations in this coordinate system, and is a consequence of $f(z)$ being holomorphic.