Answer: \(z^n+z^{-n}=2\cos n\theta\) and \(z^n-z^{-n}=2i\sin n\theta\).

\(\displaystyle \sin^4\theta\cos^2\theta=\frac{1}{32}(\cos6\theta-2\cos4\theta-\cos2\theta+2)\).

\(\displaystyle \int_0^{\pi/4}\sin^4\theta\cos^2\theta\,d\theta=\frac{3\pi-4}{192}\).

Let \(z=\cos\theta+i\sin\theta\). Then by De Moivre’s theorem, \(z^n=\cos n\theta+i\sin n\theta\), and since \(z^{-1}=\cos\theta-i\sin\theta\), we also have \(z^{-n}=\cos n\theta-i\sin n\theta\). Hence

\(z^n+\frac{1}{z^n}=z^n+z^{-n}=2\cos n\theta\), and \(z^n-\frac{1}{z^n}=z^n-z^{-n}=2i\sin n\theta\).

Now consider \(\left(z-\frac{1}{z}\right)^4\left(z+\frac{1}{z}\right)^2\). First rewrite the factors:

\(\left(z-\frac{1}{z}\right)^2=z^2-2+\frac{1}{z^2}\),

\(\left(z+\frac{1}{z}\right)^2=z^2+2+\frac{1}{z^2}\),

so

\(\left(z-\frac{1}{z}\right)^4\left(z+\frac{1}{z}\right)^2=\left(z^2-2+\frac{1}{z^2}\right)^2\left(z^2+2+\frac{1}{z^2}\right).\)

A neater route is to use

\(\left(z-\frac{1}{z}\right)^2=z^2-2+\frac{1}{z^2}\)

and

\(\left(z-\frac{1}{z}\right)^4\left(z+\frac{1}{z}\right)^2=\left(z-\frac{1}{z}\right)^2\left(z^2-\frac{1}{z^2}\right)^2.\)

Now expand each square:

\(\left(z-\frac{1}{z}\right)^2=z^2-2+\frac{1}{z^2},\)

\(\left(z^2-\frac{1}{z^2}\right)^2=z^4-2+\frac{1}{z^4}.\)

Therefore

\(\left(z-\frac{1}{z}\right)^4\left(z+\frac{1}{z}\right)^2=\left(z^2-2+\frac{1}{z^2}\right)\left(z^4-2+\frac{1}{z^4}\right).\)

Multiplying out and collecting symmetric terms gives

\(=\left(z^6+\frac{1}{z^6}\right)-2\left(z^4+\frac{1}{z^4}\right)-\left(z^2+\frac{1}{z^2}\right)+4.\)

Using the identities from the first part, this becomes

\(=2\cos 6\theta-4\cos 4\theta-2\cos 2\theta+4.\)

On the other hand,

\(z-\frac{1}{z}=2i\sin\theta,\qquad z+\frac{1}{z}=2\cos\theta,\)

so the left-hand side is

\((2i\sin\theta)^4(2\cos\theta)^2=64\sin^4\theta\cos^2\theta.\)

Hence

\(64\sin^4\theta\cos^2\theta=2\cos 6\theta-4\cos 4\theta-2\cos 2\theta+4,\)

and dividing by 64 gives

\(\sin^4\theta\cos^2\theta=\frac{1}{32}(\cos 6\theta-2\cos 4\theta-\cos 2\theta+2).\)

Now integrate:

\(\int_0^{\pi/4}\sin^4\theta\cos^2\theta\,d\theta=\frac{1}{32}\int_0^{\pi/4}(\cos 6\theta-2\cos 4\theta-\cos 2\theta+2)\,d\theta.\)

So

\(=\frac{1}{32}\left[\frac{1}{6}\sin 6\theta-\frac{1}{2}\sin 4\theta-\frac{1}{2}\sin 2\theta+2\theta\right]_0^{\pi/4}.\)

At \(\theta=\pi/4\), \(\sin(6\theta)=\sin(3\pi/2)=-1\), \(\sin(4\theta)=\sin\pi=0\), and \(\sin(2\theta)=\sin(\pi/2)=1\). Therefore

\(=\frac{1}{32}\left(-\frac{1}{6}-0-\frac{1}{2}+\frac{\pi}{2}\right)=\frac{3\pi-4}{192}.\)