麻煩大家幫手替我解答第四至九題既步驟同埋答案，係關於複數既四則運算的 4.(1+2i)(2+5i)-(3-i) 5.(2-3i)^2-(2+3i)^2 7.5-2i/3i + 3+i/2-7i 8.(2+i)^2/1 9.(2-開方3i)^4 唔該曬！?

4)(1+2i)(2+5i)-(3-i)
= 2+4i+5i +(2i)(5i) -3+i
= 2+9i +10i² -3+i
= 2+9i -10 -3+i
= -11+10i

5) (2-3i)² - (2+3i)²
= (2-3i)(2-3i) - (2+3i)(2+3i)
= (4-6i-6i+9i²) - (4+6i+6i+9i²)
= -24i

7) (5-2i) / 3i + (3+i) / (2-7i)
= [ (5-2i)(2-7i) + 3i (3+i)] / [(3i) (2-7i)]
= [(10-35i -4i +14i²) + (9i +3i²)] / (6i-21i²)
= [10-39i-14+9i+3(-1)] / [(6i-(-21)]
= (-7-30i) /(21+6i)

8)
. .1
---------
(2+i)²

. . . . 1
= --------------
. . (2+i) (2+i)

. . . . 1
= ----------------
. . 4+2i+2i+ i²

. . . . 1
= ------------
. . 4+4i+ -1

. . . 1
= --------
. . .3+4i

9) (2-√3i)⁴
= [(2-√3i)(2-√3i)] [(2-√3i)(2-√3i)]
= (4 - 2√3i - 2√3i + 3i²)(4 - 2√3i - 2√3i + 3i²)
= (4 - 4√3i -3)(4 - 4√3i -3) . . . . . . . . . . . . . . . { ∵ i² = -1
= (1- 4√3i)(1- 4√3i)
= 1-4√3i -4√3i + (-4√3i)²
= 1-8√3i + 16(3)(-1)
= 1-8√3i - 48
= -47 - 8√3i