Tính giá trị của biểu thức sau D = ( 1+i căn 3 ) ^5 / ( 1-i)^4 + (1 - i căn 3 )^5/(1+i)^4

* Hướng dẫn giải

Chọn A.

$=\frac{2^5{\left(\cos {\displaystyle \frac{\pi }{3}}+i\sin {\displaystyle \frac{\pi }{3}}\right)}^5}{2^2{\left(\cos \left(-{\displaystyle \frac{\pi }{4}}\right)+i\sin \left(-{\displaystyle \frac{\pi }{4}}\right)\right)}^4}+\frac{2^5{\left(\cos {\displaystyle \left(-\frac{\pi }{3}\right)}+i\sin \left(-\frac{\pi }{3}\right)\right)}^5}{2^2{\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)}^4}$

$$\begin{gathered} =8\frac{\cos {\displaystyle \frac{5\pi }{3}}+i\sin {\displaystyle \frac{5\pi }{3}}}{\cos \left(-\pi \right)+i\sin \left(-\pi \right)}+8\frac{\cos {\displaystyle \left(-\frac{5\pi }{3}\right)}+i\sin {\displaystyle \left(-\frac{5\pi }{3}\right)}}{\cos \pi +i\sin \pi } \\ = \frac{8\left({\displaystyle \frac{1}{2}}-{\displaystyle \frac{i\sqrt{3}}{2}}\right)}{-1}+\frac{8\left({\displaystyle \frac{1}{2}}+{\displaystyle \frac{i\sqrt{3}}{2}}\right)}{-1}=-8 \end{gathered}$$