If f(x) is given by $f(x)=(\cos x+i \sin x)(\cos 3 x+i \sin 3 x) ....... \{\cos (2 n-1) x+i \sin (2 n-1) x\}$, then f''(x) is equal to

$-n^4 f(x)$

We have,

$f(x)=\cos \{x+3 x+...+(2 n-1) x\} + i \sin \{x+3 x+5 x+...+(2 n-1) x\}$

$\Rightarrow f'(x)=\cos n^2 x+i \sin n^2 x$

$\Rightarrow f'(x)=-n^2\left(\sin n^2 x\right)+n^2\left(i \cos n^2 x\right)$

$\Rightarrow f''(x)=-n^4 \cos n^2 x-n^4 i \sin n^2 x$

$\Rightarrow f''(x)=-n^4\left(\cos n^2 x+i \sin n^2 x\right)$

$\Rightarrow f''(x)=-n^4 f(x)$