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!
 A: $$\left( \frac { -1+i\sqrt { 3 }  }{ 1+i }  \right) ^{ 2016 }=\frac { { { 2 }^{ 2016 }\left( { \cos { \left( \frac { 2\pi  }{ 3 }  \right) +i\sin { \left( \frac { 2\pi  }{ 3 }  \right)  }  }  } \right)  }^{ 1013 } }{ { \left( { \left( 1+i \right)  }^{ 2 } \right) ^{ 1013 } } } =\frac { { { 2 }^{ 2016 }\left( { \cos { \left( \frac { 2016\pi  }{ 3 }  \right) +i\sin { \left( \frac { 2016\pi  }{ 3 }  \right)  }  }  } \right)  } }{ { 2 }^{ 1008 }{ i }^{ 1008 } } =\\ ={ 2 }^{ 1008 }{ e }^{ 2016i\pi /3 }={ 2 }^{ 1008 }$$
A: Note that
$$
z=
\frac{-1 + i\sqrt 3}{1 + i} = 
\frac{-1 + i\sqrt 3}{2}
\frac{2}{1 + i}
=\sqrt2 \alpha \beta
$$
where
$$
\alpha=\frac{-1 + i\sqrt 3}{2},
\qquad
\beta = \frac{\sqrt2}{1 + i}
$$
Note that $\alpha^3 = 1 = \beta^8 $.
Therefore $z^{24}=2^{12}$ and so $z^{2016}=2^{1008}$.
A: The main trick that makes this easy is to immediately recognize by sight parts of the third and eighth roots of unity.
The two primitive third roots of unity are
$$ \frac{-1 \pm \mathbf{i} \sqrt{3}}{2} $$
and the four primitive eighth roots of unity are
$$ \frac{\pm 1 \pm \mathbf{i}}{\sqrt{2}} $$
Sometimes people like to write $\frac{\sqrt{2}}{2}$ rather than $\frac{1}{\sqrt{2}}$.
And since they're similar, I'll mention the two primitive sixth roots of unity are
$$ \frac{1 \pm \mathbf{i} \sqrt{3}}{2} $$
In each case, the root with smallest positive complex argument is the one where you take the positive signs.
With this in mind, we rewrite the base as
$$\frac{-1 + i\sqrt 3}{1 + i} 
= \frac{-1 + \mathbf{i} \sqrt{3}}{2} \frac{\sqrt{2}}{1 + \mathbf{i}} \sqrt{2} 
= \sqrt{2} \zeta_3 \zeta_8^{-1} $$
where I've used $\zeta_n$ to denote the principal $n$-th root of unity.
Now, the 2016-th power is easy to compute!
A: Here is another approach: Why don't you "distribute" that exponent on the numerator and denominator? Then raise both numerator and denominator to the power 2016. The thing is that both $\arctan(-\sqrt{3})$ as well as $\arctan 1$ are well known angles. From there you can apply your DeMoivre. Once you have those new numerators and denominators, you can simply divide. I will do the denominator for you: $r=\sqrt{2}$ and $\theta=45°$, so to the power 2016 is $2^{1008}(\cos(2016(45°))+i\sin(2016(45°)))$ which is $2^{1008}(1+0i)$. Can you do the numerator? 
A: $$\dfrac{\sqrt3-1}{2\sqrt2}=\cos\dfrac\pi6\cos\dfrac\pi4-\sin\dfrac\pi6\sin\dfrac\pi4=\cos\left(\dfrac\pi6+\dfrac\pi4\right)$$
$$\dfrac{\sqrt3+1}{2\sqrt2}=\cos\dfrac\pi6\sin\dfrac\pi4+\sin\dfrac\pi6\cos\dfrac\pi4=\sin\left(\dfrac\pi6+\dfrac\pi4\right)$$

OR
$$-1+\sqrt3i=2\left(\cos\dfrac{2\pi}3+i\sin\dfrac{2\pi}3\right)=2e^{2i\pi/3}$$
$$1+i=\sqrt2\left(\cos\dfrac\pi4+i\sin\dfrac\pi4\right)=\sqrt2e^{i\pi/4}$$
A: $\left( \frac{-1 + i\sqrt 3}{1 + i} \right)^{2016}$
Lets simplify $\frac{-1 + i\sqrt 3}{1 + i}$
$\frac {1}{1+i}(-1 + i\sqrt 3)\\
\frac {1-i}{2}(-1 + i\sqrt 3)\\
 (1-i)(-\frac12 + i\frac{\sqrt 3}2)$
Convert to polar:
$\sqrt 2 (\cos \frac {-\pi}{4} +\sin \frac {-\pi}{4} )(\cos \frac{2\pi}3 + i \sin \frac{2\pi}3)$
Now we have a choice...we could raise to the 2016 power right now, or we could mulitiply those two complex numbers first then raise to the 2016 power.
$\left( \frac{-1 + i\sqrt 3}{1 + i} \right)^{2016} =
\sqrt 2^{2016} (\cos \frac {-\pi}{4} +\sin \frac {-\pi}{4} )^{2016}(\cos \frac{2\pi}3 + i \sin \frac{2\pi}3)^{2016}$
Applying deMoivres theorem:
$(\cos \frac {-\pi}{4} +\sin \frac {-\pi}{4} )^{2016} = (\cos \frac {-2016\pi}{4} +\sin \frac {-2016\pi}{4} )$
$8$ divides $2016$ 
$\frac {-2016\pi}{4} = 2n\pi$
$(\cos \frac {-\pi}{4} +\sin \frac {-\pi}{4} )^{2016} = 1$
$6$ divides $2016$, too.
$2^{1008}$
alternatively:
$\sqrt 2 (\cos \frac {-\pi}{4} +\sin \frac {-\pi}{4} )(\cos \frac{2\pi}3 + i \sin \frac{2\pi}3) = \sqrt 2 (\cos \frac {5\pi}{12} +\sin \frac {5\pi}{12})$
and we get to the same place.
A: Note that
$$\left( \frac{-1 + i\sqrt 3}{1 + i} \right)^{2016} $$can be expressed as$$ \left( \sqrt{2}(cos(\frac{5\pi}{12}) + i \cdot sin(\frac{5\pi}{12})) \right)^{2016}$$
which gives
$$\left (\sqrt{2}\right)^{2016} \left((cos(840\pi) + i \cdot sin(840\pi)) \right)\implies 2^{1008}$$
