题目
.当 =dfrac (1+i)(1-i) 时, ^100+(7)^75+(7)^50 的值等于 ()-|||-(A)i (B) -(1)^circ . (C)1
题目解答
答案
:因为 $z=\dfrac {1+i}{1-i}=\dfrac {(1+i)^2}{(1-i)(1+i)}=i$ 所以 ${7}^{100}+{7}^{75}+{7}^{50}$=(100+7i)^50+(-1+5i)^50$=[(100+7i)(-1-5i)]^25+[(100-7i)(-1+5i)]^25$=(-501-341i)^25+(-501+341i)^25$=(-501)^25·(-1)+(-501)^25·1$=0+1=1 故选CC
C
C
解析
步骤 1：化简 $z$
$z=\dfrac {1+i}{1-i}=\dfrac {(1+i)^2}{(1-i)(1+i)}=\dfrac {1+2i+i^2}{1-i^2}=\dfrac {2i}{2}=i$
步骤 2：计算 ${7}^{100}+{7}^{75}+{7}^{50}$
${7}^{100}+{7}^{75}+{7}^{50}=(i^{100}+i^{75}+i^{50})$
步骤 3：化简指数
$i^{100}=(i^4)^{25}=1^{25}=1$
$i^{75}=i^{4*18+3}=(i^4)^{18}*i^3=1^{18}*(-i)=-i$
$i^{50}=(i^4)^{12}*i^2=1^{12}*(-1)=-1$
步骤 4：计算结果
${7}^{100}+{7}^{75}+{7}^{50}=1-i-1=-i$
$z=\dfrac {1+i}{1-i}=\dfrac {(1+i)^2}{(1-i)(1+i)}=\dfrac {1+2i+i^2}{1-i^2}=\dfrac {2i}{2}=i$
步骤 2：计算 ${7}^{100}+{7}^{75}+{7}^{50}$
${7}^{100}+{7}^{75}+{7}^{50}=(i^{100}+i^{75}+i^{50})$
步骤 3：化简指数
$i^{100}=(i^4)^{25}=1^{25}=1$
$i^{75}=i^{4*18+3}=(i^4)^{18}*i^3=1^{18}*(-i)=-i$
$i^{50}=(i^4)^{12}*i^2=1^{12}*(-1)=-1$
步骤 4：计算结果
${7}^{100}+{7}^{75}+{7}^{50}=1-i-1=-i$