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题:若复数z满足z~10=-1,则z~4-z~6-z~2+1的值是( )。0(A)-1;(B)1;(C)0;(D)以上都不对。此题见于一些书刊,其答案为(A) 解法一:(方程法或设值法) 设(1) (1)式两边同乘以z~4得 z~8-z~6-z~2+1=z~4t(2) 因为z~10=-1,(1)式分子,分母同乘以-z~6。并整理得 z~8-z~6+z~4-z~2=t(3) 由(2)得z~8-z~6-z~2=z~4t-1(4) (4)式代入(3)得 z~4t-1+z~4=t ∴(z~4-1)t=1-z~4 (5) 两边同除以z~4-1得:t=-1,故选(A)。解法二:(用数列求和解) ∵ z~10=-1, ∴原式的分子、分母同乘以-z~6,得: =-z~2+z~4-z~6+z~8 (看成首项,公比皆为-z~2的等比数列) 故选(A)。以上两种解法及原书刊给出的答案都是(A),可是若本题用满足z~10=-1的一值z=i(或-i)代入,得上面结果都错了,这就奇怪了,问题何在?
Question: If the complex z satisfies z~10=-1, then the value of z~4-z~6-z~2+1 is (). 0 (A) - 1; (B) 1; (C) 0; (D) None of the above. This question is found in some publications. The answer is (A) Solution One: (Equation or Set Value) Let (1) (1) Both sides multiply by z~4 z~8-z~6-z~2 +1=z~4t(2) Since z~10=-1, (1) numerator, the denominator is multiplied by -z~6. And put it together z~8-z~6+z~4-z~2=t(3) From (2) got z~8-z~6-z~2=z~4t-1(4) (4 The formula is substituted into (3) to get z~4t-1+z~4=t ∴(z~4-1)t=1-z~4 (5) divided by z~4-1 on both sides: t=- 1, so choose (A). Solution two: (with the series sum solution) ∵ z~10=-1, ∴ original numerator, denominator multiplied by -z~6, got: =-z~2+z~4-z~6+z~ 8 (For the first item, the ratio of the public figures is -z~2.) So choose (A). The above two solutions and the answers given in the original book are (A), but if this question is substituted with a value z=i (or -i) that satisfies z~10=-1, the above results are all wrong. Strange, what’s the problem?