Giá trị cuả biểu thức: $z=(\frac{-\sqrt{3}+i}{1+i})^{12000}$ 11/07/2021 Bởi Josephine Giá trị cuả biểu thức: $z=(\frac{-\sqrt{3}+i}{1+i})^{12000}$
Lời giải: Ta có: $(\frac{-\sqrt{3}+i}{1+i})^{2}=-\sqrt{3}-i=\frac{1}{2}.(-\frac{\sqrt{3}}{2}-\frac{1}{i}i)=\frac{1}{2}.(cos\frac{-5π}{6}+isin\frac{-5π}{6})$ $=>(\frac{-\sqrt{3}+i}{1+i})^{12}=[\frac{1}{2}.(cos\frac{-5π}{6}+isin\frac{-5π}{2})]^6=\frac{1}{2^6}.(cos\frac{-6.5π}{6}+isin\frac{-6.5π}{2})$ $\frac{1}{2^6}.(cos(-5π)+isin(-5π))=\frac{1}{2^6}.(-1)=\frac{-1}{2^6}$ $z=(\frac{-\sqrt{3}+i}{1+i})^{12000}=[(\frac{-\sqrt{3+i}}{1+i})^{12}]^{1000}=(\frac{-1}{2^6})^{1000}=\frac{1}{2^{60000}}$ Bình luận
Đáp án:???
Lời giải:
Ta có:
$(\frac{-\sqrt{3}+i}{1+i})^{2}=-\sqrt{3}-i=\frac{1}{2}.(-\frac{\sqrt{3}}{2}-\frac{1}{i}i)=\frac{1}{2}.(cos\frac{-5π}{6}+isin\frac{-5π}{6})$
$=>(\frac{-\sqrt{3}+i}{1+i})^{12}=[\frac{1}{2}.(cos\frac{-5π}{6}+isin\frac{-5π}{2})]^6=\frac{1}{2^6}.(cos\frac{-6.5π}{6}+isin\frac{-6.5π}{2})$
$\frac{1}{2^6}.(cos(-5π)+isin(-5π))=\frac{1}{2^6}.(-1)=\frac{-1}{2^6}$
$z=(\frac{-\sqrt{3}+i}{1+i})^{12000}=[(\frac{-\sqrt{3+i}}{1+i})^{12}]^{1000}=(\frac{-1}{2^6})^{1000}=\frac{1}{2^{60000}}$