
原帖由 36661124 於 11-4-27 01:19 PM 發表
=.=
有 i 果d 係complex no.
所以將real no. group 埋一齊
complex no group 埋一齊 就係ans . 注意i²= -1
4b-4ai-3bi+3ai^2=7+5i
4b + 3a(i²) - 3bi -4ai=7 + 5i
4b + 3a(-1) -(3b +4a)i =7 + ...

原帖由 36661124 於 11-4-27 01:19 PM 發表
=.=
有 i 果d 係complex no.
所以將real no. group 埋一齊
complex no group 埋一齊 就係ans . 注意i²= -1
4b-4ai-3bi+3ai^2=7+5i
4b + 3a(i²) - 3bi -4ai=7 + 5i
4b + 3a(-1) -(3b +4a)i =7 + ...

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