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標題:

Imaginary number problem (help fast)?

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發問:

( 1 - i ) ^40 how do i do dis??????????????????????????????/////////////

最佳解答:

Method 1 : (1 - i)^40 = [(1 - i)^2]^20 = [1^2 - 2i + i^2]^20 = (-2i)^20 = (-2)^20 × (i)^20 = 2^20 × {[(i)^2]^2}^5 = 1048576 × 1 = 1048576 ...... (Ans) Method 2 : (1 - i)^40 = {(√2)[(1/√2) - (1/√2)i]}^40 = (√2)^40 × [cos(315°) + i sin(315°)]^40 = (√2)^40 × [cos(315°) + i sin(315°)]^40 = 2^20 × [cos(315° × 40) + i sin(315° × 40)] ...... (By De Moivre's Theorem) = 1048576 × [cos(12600°) + i sin(12600°)] = 1048576 × [cos(360° × 35) + i sin(360° × 35)] = 1048576 × [cos(0°) + i sin(0°)] = 1048576 × [1 + 0] = 1048576 ...... (Ans)

其他解答:A215E4A2B88AAE64