Prove isoperimetric inequality for a closed curve parametrized by a smooth $2\pi$-peirodic complex valued function Problem:
(The isoperimetric inequality) Suppose we have a smooth closed curve in the (complex) plane
which encloses an area A and has perimeter P. We wish to prove that
$$P^2 ≥ 4πA$$
(Hint) To do this, assume that the curve is parametrized by a smooth 2π-periodic complex valued function $f(x) = u(x) + iv(x)$ such that $(u')^2 + (v')^2 = c^2$ is constant. Using that $c((u')^2 +(v')^2)^{1/2} = |f'|^2$
, relate $P^2$ to $\int^{\pi}_{-\pi} |f'(x)|^2dx$.
Relate $A = \int udv$ to the $L^2$-inner product
$$
(f',f)=\int^{\pi}_{-\pi} \overline{f'(x)}f(x)dx
$$
Using Plancherel's identity you should be able to deduce.
Plancherel's identity:
$$
(f,g)=\sum_{k\in \mathrm{z}}\overline{\hat f(k)}\hat g(k)
$$
My attemp: 
According to the hint, 
$$P^2=[\int_{-\pi}^{\pi}|f(x)|dx]^2$$
$$
A=\int_{-\pi}^{\pi}udv=\int_{-\pi}^{\pi}u(x)v'(x)dx
$$
\begin{align}
(f',f)
&=(u'+iv',u+iv)\\
&=\int_{-\pi}^{\pi}(u'-iv')(u+iv)dx\\
&=\int_{-\pi}^{\pi}(u'u-iuv'+iu'v+v'v)dx\\
&=\frac{1}{2}(u(x)^2+v(x)^2)\Big|^{\pi}_{-\pi}-2iA\\
&=-2iA
\end{align}
Question:
1) I tried to make $P^2$ larger than a item containing $\int_{-\pi}^{\pi} |f|^2dx$. But I tried multiple inequalities and can only get $P^2 \leq$ something.
2) I don't know how to apply Plancherel's identity to build the connection.
 A: I will use the following notation: $L$ -- length of curve, $S$ -- area limited by curve,  $x=\varphi(s),\;y=\psi(s)$ -- parametric curve equations, where $s$ is the natural parameter (length from start of curve), $\varphi,\psi\in C^1[0,L]$. Since the curve is closed, then $\varphi(0)=\varphi(L)$, $\psi(0)=\psi(L)$. We pass from the parameter $s$ to the parameter $t=2\pi\frac{s}{L}-\pi$, then parametric equations becomes $x=x(t),\;y=y(t)$, $-\pi\leq t\leq \pi$, $x(-\pi)=x(\pi)$, $y(-\pi)=y(\pi)$. If $z(t)=x(t)+iy(t)$, then $z=z(t)$ -- is the parametric equation of our curve in complex form, and $z(-\pi)=z(\pi)$.
Notice that $|z'(t)|^2=(x'(t))^2+(y'(t))^2=\left(\dfrac{ds}{dt}\right)^2=\dfrac{L^2}{4\pi^2}$. Further, $\overline{z}z'=(xx'+yy')+i(xy'-x'y)$, and $S=\dfrac{1}{2}\displaystyle\int\limits_{-\pi}^{\pi}(xy'-x'y)(t)dt=\dfrac{1}{2i}\displaystyle\int\limits_{-\pi}^{\pi}\overline{z}(t)z'(t)dt$ (notice that $\displaystyle\int\limits_{-\pi}^{\pi}(xx'+yy')(t)dt=0$ due to the closedness conditions of the curve).
Expand $z$ and $z'$ in a Fourier series: $z(t)=\displaystyle\sum\limits_{k=-\infty}^\infty c_ke^{ikt}$, $z'(t)=\displaystyle\sum\limits_{k=-\infty}^\infty ikc_ke^{ikt}$. Due the equalities obtained above $\frac{1}{2\pi}\|z'\|^2=\frac{1}{2\pi}\displaystyle\int\limits_{-\pi}^\pi|z'(t)|^2dt=\frac{L^2}{4\pi^2}$ and $\frac{1}{2\pi}(z',z)=\frac{1}{2\pi}\displaystyle\int\limits_{-\pi}^{\pi}z'(t)\overline{z}(t)dt=\dfrac{iS}{\pi}$.
Now, applying the following Plancherel identities: $\frac{1}{2\pi}(f,g)=\displaystyle\sum\limits_{k=-\infty}^\infty c_k(f)\overline{c}_k(g)$, and $\frac{1}{2\pi}\|f\|^2=\displaystyle\sum\limits_{k=-\infty}^\infty|c_k(f)|^2$ for our case, we will get: $L^2=4\pi^2\displaystyle\sum\limits_{k=-\infty}^\infty|kc_k|^2$, $S=\pi\displaystyle\sum\limits_{k=-\infty}^\infty kc_k\overline{c}_k$. So $L^2-4\pi S=4\pi^2\displaystyle\sum\limits_{k=-\infty}^\infty(k^2-k)|c_k|^2\geq0$. Moreover, equality is possible only for $c_k=0$, $k\ne0,1$, i.e. a curve in which inequality turns into equality is a circle $z(t)=c_0+c_1e^{it}$, $-\pi\leq t\leq\pi$.
