题目
2.必答[单选题]设z_(1)=5-5i,z_(2)=-3+4i,问(z_(1))/(z_(2))等于?A. (7)/(5)-(1)/(5)iB. (7)/(5)+(1)/(5)iC. -(7)/(5)-(1)/(5)iD. -(7)/(5)+(1)/(5)i
2.必答[单选题]设$z_{1}=5-5i$,$z_{2}=-3+4i$,问$\frac{z_{1}}{z_{2}}$等于?
A. $\frac{7}{5}-\frac{1}{5}i$
B. $\frac{7}{5}+\frac{1}{5}i$
C. $-\frac{7}{5}-\frac{1}{5}i$
D. $-\frac{7}{5}+\frac{1}{5}i$
题目解答
答案
C. $-\frac{7}{5}-\frac{1}{5}i$
解析
复数除法运算的关键在于有理化分母,即通过乘以分母的共轭复数,将分母转化为实数。本题需计算$\frac{z_1}{z_2}$,其中$z_1=5-5i$,$z_2=-3+4i$。解题核心步骤为:
- 确定共轭复数:$z_2$的共轭复数为$-3-4i$;
- 分子分母同乘共轭复数,展开并化简;
- 分离实部与虚部,得到最终结果。
步骤1:有理化分母
将分子和分母同时乘以$z_2$的共轭复数$-3-4i$:
$\frac{z_1}{z_2} = \frac{(5-5i)(-3-4i)}{(-3+4i)(-3-4i)}$
步骤2:展开分子
计算分子:
$\begin{aligned}(5-5i)(-3-4i) &= 5 \cdot (-3) + 5 \cdot (-4i) -5i \cdot (-3) -5i \cdot (-4i) \\&= -15 -20i +15i +20i^2 \\&= -15 -5i +20(-1) \quad (\text{因} \ i^2=-1) \\&= -15 -5i -20 \\&= -35 -5i\end{aligned}$
步骤3:计算分母
利用平方差公式:
$(-3+4i)(-3-4i) = (-3)^2 - (4i)^2 = 9 - 16i^2 = 9 - 16(-1) = 25$
步骤4:化简结果
将分子和分母代入:
$\frac{-35 -5i}{25} = -\frac{35}{25} - \frac{5}{25}i = -\frac{7}{5} - \frac{1}{5}i$