题目
复数除法 计算1+i/1-i,1/i,7+i/3+4i (-1+i)(2+i)/-i
提问时间:2020-10-25
答案
楼上最后一题解错了.
1、(1+i)/(1-i) = (1+i)²/(1-i)(1+i) = (1+2i-1)/(1+1) = 2i/2 = i
2、1/i = i/i² = i/(-1) = -i
3、(7+i)/(3+4i) = (7+i)(3-4i)/(3+4i)(3-4i) = (21-25i+4)/(9+16) = 1-i
4、(-1+i)(2+i)/(-i) = (-2-i+2i-1)/(-i) = (-3+i)/(-i) = (-3+i)i/(-i²) = (-3i -1)/1 = -1-3i
1、(1+i)/(1-i) = (1+i)²/(1-i)(1+i) = (1+2i-1)/(1+1) = 2i/2 = i
2、1/i = i/i² = i/(-1) = -i
3、(7+i)/(3+4i) = (7+i)(3-4i)/(3+4i)(3-4i) = (21-25i+4)/(9+16) = 1-i
4、(-1+i)(2+i)/(-i) = (-2-i+2i-1)/(-i) = (-3+i)/(-i) = (-3+i)i/(-i²) = (-3i -1)/1 = -1-3i
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