（1+√3i）∧6/（1-i）∧5
答案:2 悬赏:50 手机版
解决时间 2021-02-02 08:07
- 提问者网友：聂風
- 2021-02-01 13:37
（1+√3i）∧6/（1-i）∧5
最佳答案
- 五星知识达人网友：佘樂
- 2021-02-01 15:11
把1+√3i和1-i转化为三角式再做乘方和除。
1+√3i=2(cosπ/3＋sinπ/3)
1-i=√2(cos(-π/4)+sin(-π/4)i)
（1+√3i）∧6/（1-i）∧5
=（2(cosπ/3＋sinπ/3i)）∧6/（√2(cos(-π/4)+sin(-π/4))i）∧5
=（2∧6(cos2π＋sin2π)i）/（2√2(cos(-5π/4)+sin(-5π/4))i）
=2∧4√2(cos(13π/4)+sin(13π/4)i)
=2∧4√2(cos(π/4)+sin(π/4)i)
=2⁴(1+i)
=16+16i
1+√3i=2(cosπ/3＋sinπ/3)
1-i=√2(cos(-π/4)+sin(-π/4)i)
（1+√3i）∧6/（1-i）∧5
=（2(cosπ/3＋sinπ/3i)）∧6/（√2(cos(-π/4)+sin(-π/4))i）∧5
=（2∧6(cos2π＋sin2π)i）/（2√2(cos(-5π/4)+sin(-5π/4))i）
=2∧4√2(cos(13π/4)+sin(13π/4)i)
=2∧4√2(cos(π/4)+sin(π/4)i)
=2⁴(1+i)
=16+16i
全部回答
- 1楼网友：撞了怀
- 2021-02-01 15:49
-1＋i
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