Question

已知复数z满足 zi+z.(1+i)=1+4i.-|||-(1)求z;-|||-(2)若 z^2+k/(x^2)(kR) 为

Answer 1

问：已知复数z满足 zi+z.(1+i)=1+4i.-|||-(1)求z;-|||-(2)若 z^2+k/(x^2)(kR) 为

答：我们先将题目中的公式展开，可以得到：zi + z + zi·i + z·i = 1 + 4i化简后得：(1+i)z = 1 + 3i因此，(1) z = (1+3i)/(1+i) = (1+3i)(1-i)/(1+i)(1-i) = (1+2i)/2(2) z^2 + k/(x^2) = -kR将 (1) 中求得的 z 带入可得：z^2 + k/(x^2) = -kR(1+2i)^2/4 + k/(x^2) = -kR(1+4i-4)/4 + k/(x^2) = -kR(-3+4i)/4 + k/(x^2) = -kR因此，这个等式成立当且仅当：-3/4-kR = 0，4k/(3x^2) = 0，4kR/(3x^2) ≠ 0解得 k = -3x^2/4，R = -3x^2/4。因此，z^2 - 3x^2/(4x^2) = 3/4