Find the 2016th power of a complex number Calculate $\left( \frac{-1 + i\sqrt 3}{1 + i} \right)^{2016}$.
Here is what I did so far:
I'm trying to transform $z$ into its trigonometric form, so I can use De Moivre's formula for calculating $z^{2016}$.
Let $z = \frac{-1 + i\sqrt 3}{1 + i}$. This can be rewritten as $\frac{\sqrt 3 - 1}{2} + i\frac{\sqrt 3 + 1}{2}$.
$$z = r(\cos \phi + i \sin \phi)$$
$$r = |z| = \sqrt 2$$
$$\phi = \arctan {\sqrt 3 + 1}$$
Now, I don't know what to do with that $\sqrt 3 + 1$. How do I calculate $\phi$ ?
Thank you in advance!

$$\left( \frac { -1+i\sqrt { 3 }  }{ 1+i }  \right) ^{ 2016 }=\frac { { { 2 }^{ 2016 }\left( { \cos { \left( \frac { 2\pi  }{ 3 }  \right) +i\sin { \left( \frac { 2\pi  }{ 3 }  \right)  }  }  } \right)  }^{ 1013 } }{ { \left( { \left( 1+i \right)  }^{ 2 } \right) ^{ 1013 } } } =\frac { { { 2 }^{ 2016 }\left( { \cos { \left( \frac { 2016\pi  }{ 3 }  \right) +i\sin { \left( \frac { 2016\pi  }{ 3 }  \right)  }  }  } \right)  } }{ { 2 }^{ 1008 }{ i }^{ 1008 } } =\\ ={ 2 }^{ 1008 }{ e }^{ 2016i\pi /3 }={ 2 }^{ 1008 }$$
