Complex Exercise asked by Mohamad mathe in the math group

Exercise:

Consider a complex polynomial $p\left( z \right)=1+z+{{z}^{2}}+{{z}^{3}}$ with $z={{e}^{i\theta }}$ , $\theta \in \left[ 0,2\pi  \right]$

Determine the $\left| p\left( z \right) \right|$ and $\arg \left( p\left( z \right) \right)$ in terms of $\theta $

Solution: We have $1+z+{{z}^{2}}+{{z}^{3}}+....+{{z}^{n}}=\frac{1-{{z}^{n+1}}}{1-z}$ which is geometric series of common ratio is $z$

 so here $n=3$ so $1+z+{{z}^{2}}+{{z}^{3}}=\frac{1-{{z}^{4}}}{1-z}=\frac{-\left( -1+z \right)\left( z+1 \right)\left( {{z}^{2}}+1 \right)}{1-z}=\left( z+1 \right)\left( {{z}^{2}}+1 \right)$

$\Rightarrow p\left( z \right)=\left( z+1 \right)\left( {{z}^{2}}+1 \right)=\left( \cos \theta +i\sin \theta +1 \right)\left( \cos 2\theta +i\sin 2\theta +1 \right)$

          $=\left( 2{{\cos }^{2}}\frac{\theta }{2}+i\sin \theta  \right)\left( 2{{\cos }^{2}}\theta +i\sin 2\theta  \right)$

          $=\left( 2{{\cos }^{2}}\frac{\theta }{2}+i2\sin \frac{\theta }{2}\cos \frac{\theta }{2} \right)\left( 2{{\cos }^{2}}\theta +i2\sin \theta \cos \theta  \right)$

         $=2\cos \frac{\theta }{2}\left( \cos \frac{\theta }{2}+i\sin \frac{\theta }{2} \right)2\cos \theta \left( \cos \theta +i\sin \theta  \right)=4\cos \frac{\theta }{2}\cos \theta {{e}^{i\left( \frac{\theta }{2}+\theta  \right)}}$

            $=4\cos \theta \cos \frac{\theta }{2}{{e}^{i\left( 3\theta /2 \right)}}$

So the $\arg \left( p\left( z \right) \right)=\frac{3\theta }{2}+2k\pi $ and $\left| p\left( z \right) \right|=4\cos \theta \cos \frac{\theta }{2}$