Giá trị cuả biểu thức: $z=(\frac{-\sqrt{3}+i}{1+i})^{12000}$

Lời giải:

Ta có:

$(\frac{-\sqrt{3}+i}{1+i})^{2}=-\sqrt{3}-i=\frac{1}{2}.(-\frac{\sqrt{3}}{2}-\frac{1}{i}i)=\frac{1}{2}.(cos\frac{-5π}{6}+isin\frac{-5π}{6})$

$=>(\frac{-\sqrt{3}+i}{1+i})^{12}=[\frac{1}{2}.(cos\frac{-5π}{6}+isin\frac{-5π}{2})]^6=\frac{1}{2^6}.(cos\frac{-6.5π}{6}+isin\frac{-6.5π}{2})$

$\frac{1}{2^6}.(cos(-5π)+isin(-5π))=\frac{1}{2^6}.(-1)=\frac{-1}{2^6}$

$z=(\frac{-\sqrt{3}+i}{1+i})^{12000}=[(\frac{-\sqrt{3+i}}{1+i})^{12}]^{1000}=(\frac{-1}{2^6})^{1000}=\frac{1}{2^{60000}}$