题目

请在以下答案中选出((-16))^dfrac (1{4)}的根A((-16))^dfrac (1{4)}B ((-16))^dfrac (1{4)}C((-16))^dfrac (1{4)}D ((-16))^dfrac (1{4)}E ((-16))^dfrac (1{4)}F ((-16))^dfrac (1{4)}

请在以下答案中选出 $(-16)^{\frac{1}{4}}$ 的根

A $-1+i$

B $\sqrt{2}(-1+i)$

C $i$

D $\sqrt{2}(1-i)$

E $\sqrt{2}(-1-i)$

F $\sqrt{2}(1+i)$

题目解答

$(-16)^{\frac{1}{4}}$ 化为极坐标形式

$-16=16(\cos\pi+i\sin\pi)$

根据复数的n次方根公式

其4次方根为 $\sqrt[4]{16}(\cos(\frac{\pi+2k\pi}{4})+i\sin(\frac{\pi+2k\pi}{4}))$ $k=0,1,2,3$

当 $k=0$ 时， $2(\cos(\frac{\pi}{4})+i\sin(\frac{\pi}{4}))=\sqrt{2}(1+i)$

当 $k=1$ 时， $2(\cos(\frac{3\pi}{4})+i\sin(\frac{3\pi}{4}))=\sqrt{2}(-1+i)$

当 $k=2$ 时， $2(\cos(\frac{5\pi}{4})+i\sin(\frac{5\pi}{4}))=\sqrt{2}(-1-i)$

当 $k=3$ 时， $2(\cos(\frac{7\pi}{4})+i\sin(\frac{7\pi}{4}))=\sqrt{2}(1-i)$

答案为BDEF