题目
Problem
Solve the equation
1+2i5z−zz∗+30+10i=0,
giving your answers in the form x+iy, where x and y are real.
(5)
题目中文翻译
解方程
1+2i5z−zz∗+30+10i=0,
将答案表示为 x+iy 的形式,其中 x 和 y 为实数。
(5)
原卷通用作答要求
Answer all questions. You should use a calculator where appropriate.
You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.
Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.
请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程;仅给出计算器答案而没有过程,不予评分。除非题目另有精度要求,非精确数值答案保留 3 位有效数字,以度为单位的角度保留 1 位小数。
解答
解法一
思路
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先用共轭消去常数分母,再设 z=x+yi,利用 zz∗=x2+y2 比较实部、虚部。
答题过程
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Since
1+2i5=1+45(1−2i)=1−2i,
write z=x+yi and zz∗=x2+y2. The equation becomes
(x+yi)(1−2i)−(x2+y2)+30+10i=0.
Equating parts,
x+2y−x2−y2+30=−2x+y+10=0,0.
The second gives y=2x−10. Substituting into the first,
x+2(2x−10)−x2−(2x−10)2+30=0.
Hence
−5x2+45x−90=(x−3)(x−6)=0,0.
If x=3, y=−4; if x=6, y=2. Thus
z=3−4iorz=6+2i.