Show that for a complex number z

\begin{align} \left(z+ \frac{1}{z}\right)=2cos\theta \end{align}

\begin{align} We\ know\ that \quad z & = (cos \theta + isin \theta) \\ and\ that \quad \frac{1}{z} & = (cos \theta + isin \theta)^{-1} \\ Also,\ by\ deMoivre's\ Theorem \\ (cos \theta + isin \theta)^{-1} & =(cos (-\theta) + isin( -\theta))\\ \\ Therefore \\ (z+ \frac{1}{z})& = (cos \theta + isin \theta) + (cos (-\theta) + isin( -\theta))\\ And\ because \quad cos(- \theta)& = cos (\theta)\quad and\quad sin(-\theta)= - sin(\theta)) \\ (z+ \frac{1}{z}) & = (cos \theta + isin \theta) + (cos (\theta) - isin(\theta)) \quad \\ & = 2cos\theta \end{align}

Show that for a complex number z

\begin{align} \left(z-\frac{1}{z} \right)=2isin\theta \end{align}

\begin{align} We\ know\ that \quad z & = (cos \theta + isin \theta) \\ and\ that \quad \frac{1}{z} & = (cos \theta + isin \theta)^{-1} \\ Also,\ by\ deMoivre's\ Theorem \\ (cos \theta + isin \theta)^{-1} & =(cos (-\theta) + isin( -\theta))\\ \\ Therefore \\ (z- \frac{1}{z})& = (cos \theta + isin \theta) - (cos (-\theta) + isin( -\theta))\\ And\ because \quad cos(- \theta)& = cos (\theta)\quad and\quad sin(-\theta)= - sin(\theta)) \\ (z- \frac{1}{z}) & = (cos \theta + isin \theta) - (cos (\theta) - isin(\theta)) \quad \\ & = 2isin\theta \end{align}