Write the complex number in the form a + bi. square root of six (cos 315° + i sin 315°)?
回答 (3)
2019-01-30 2:19 pm
square root of six (cos 315° + i sin 315°)?
√6 (1/√2 - 1/√2 i)
√3 - √3 i
(In the given form where a = √3 and b = -√3)
√6 (1/√2 - 1/√2 i)
√3 - √3 i
(In the given form where a = √3 and b = -√3)
2019-01-29 10:58 pm
:-
√6 [ (1/√2) + i (1/√2) ]
( √6/√2 ) [ 1 + i ]
( √12/2 ) [ 1 + i ]
√3 [ 1 + i ]
√6 [ (1/√2) + i (1/√2) ]
( √6/√2 ) [ 1 + i ]
( √12/2 ) [ 1 + i ]
√3 [ 1 + i ]
2019-01-29 4:49 am
sqrt(6) * (cos(315) + i * sin(315)) =>
sqrt(6) * (sqrt(2)/2 - i * sqrt(2)/2) =>
sqrt(6) * (sqrt(2)/2) * (1 - i) =>
(sqrt(12)/2) * (1 - i) =>
(2 * sqrt(3)/2) * (1 - i) =>
sqrt(3) * (1 - i) =>
sqrt(3) - sqrt(3) * i
sqrt(6) * (sqrt(2)/2 - i * sqrt(2)/2) =>
sqrt(6) * (sqrt(2)/2) * (1 - i) =>
(sqrt(12)/2) * (1 - i) =>
(2 * sqrt(3)/2) * (1 - i) =>
sqrt(3) * (1 - i) =>
sqrt(3) - sqrt(3) * i
收錄日期: 2021-05-01 22:15:57
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20190128204316AAJut25