數學一問come help me?

(a) (a-bi)(b-ai)=ab -a^2 i -b^2i -ab= -(a^2 + b^2)i ---------- (*)
which is a imaginary no. ∴(a-bi)(b-ai) is a imaginary no.

(b) (i) m/n
=[(a-bi)(b-ai)] / [(a+bi)(b+ai)]
= -(a^2 + b^2)i / [(ab-ab +(a^2 + b^2)i]
= -(a^2 + b^2)i / (a^2 +b^2)i
= -1

(ii) From the result of (b)(i) ,
get [(a-bi)(b-ai)] / [(a+bi)(b+ai)]= -1 ---------- (**)
If let a=1, b=2, then (**) becomes :
[(1-2i)(2-i)i] / [(1+2i)(2+i)] = -1
∴ {[(1-2i)(2-i)i]/[(1+2i)(2+i)]}^3002 = (-1)^3002 = 1

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Some notes for you :
Q(a): a complex no. 的格式 : x + yi,
where x- real no. (實數), yi - imaginary part(虛數部份)

∴ 如 計到 (a-bi)(b-ai) = ( ??? )i, 則 可說: 這是個 purely imaginary no.

Q(b ii): "hence" 這個字是一個「貼士」:- 即叫你 由(i)部的結果 砌出(ii)部 !
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有不明白的步驟嗎？ 歡迎提出！