
Re: [中學] 虛數

: 想請問第一題
: 我怎麼算答案都是(a^5)i
: 哪裡錯了?
a < 0
√a = √[-(-a)] = i√(-a)
√[-a^2] = -ai
√[a^3] = √[-((-a)^3)] = i(-a)^(3/2)
√[a^4] = a^2
所以最後答案 = a^3 * √(-a) * i (-a)^(3/2)
= a^3 * (-a)^2 * i
= ia^5
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