Question

fie z apartine lui C,
$z = \sqrt{2} - i \sqrt{6}$
Sa se calculeze
${z}^{2016}$
​

Answer 1

$z = \sqrt{2}-i\sqrt{6}\\ \\ \Rightarrow z = \sqrt{2}\cdot(1-i\sqrt 3)\\ \\ \Rightarrow \dfrac{z\sqrt 2}{2} = 1-i\sqrt 3 = -2\cdot\left(\cos \dfrac{2\pi}{3}+i\sin \dfrac{2\pi}{3}\right) \\ \\ \Rightarrow \dfrac{z^{2016}\cdot 2^{1008}}{2^{2016}} =(-2)^{2016}\cdot \left(\cos\dfrac{2\cdot 2016\pi}{3}+i\sin\dfrac{2\cdot 2016\pi}{3}\right) =\\ \\ = 2^{2016}\cdot \Big(\cos(2\cdot 672\pi)+i\sin(2\cdot 672\pi)\Big)= \\ \\ =2^{2016}\cdot (1+i\cdot 0) \\ \\ \\\Rightarrow z^{2016}= 2^{2016+1008}$

$\Rightarrow \boxed{z^{2016}= 2^{3024}}$