Find the 2016th power of a complex number

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?

$$\dfrac{\sqrt3-1}{2\sqrt2}=\cos\dfrac\pi6\cos\dfrac\pi4-\sin\dfrac\pi6\sin\dfrac\pi4=\cos\left(\dfrac\pi6+\dfrac\pi4\right)$$

$$\dfrac{\sqrt3+1}{2\sqrt2}=\cos\dfrac\pi6\sin\dfrac\pi4+\sin\dfrac\pi6\cos\dfrac\pi4=\sin\left(\dfrac\pi6+\dfrac\pi4\right)$$

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OR

$$-1+\sqrt3i=2\left(\cos\dfrac{2\pi}3+i\sin\dfrac{2\pi}3\right)=2e^{2i\pi/3}$$

$$1+i=\sqrt2\left(\cos\dfrac\pi4+i\sin\dfrac\pi4\right)=\sqrt2e^{i\pi/4}$$

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