题目
Problem
Find the complex numbers z z z z + 4 z + 4 i \frac{z+4}{z+4\mathrm{i}} z + 4 i z + 4 ∣ z ∣ = 10 |z|=\sqrt{10} ∣ z ∣ = 10 z = x + i y z=x+\mathrm{i}y z = x + i y x x x y y y
(6)
题目中文翻译
求使 z + 4 z + 4 i \frac{z+4}{z+4\mathrm{i}} z + 4 i z + 4 ∣ z ∣ = 10 |z|=\sqrt{10} ∣ z ∣ = 10 z z z z = x + i y z=x+\mathrm{i}y z = x + i y x x x y y y
(6)
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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分母乘共轭后令商的虚部为零，再与模长条件联立求点。
答题过程
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Put z = x + y i z=x+y\mathrm{i} z = x + y i ∣ z ∣ = 10 ≠ 4 |z|=\sqrt{10}\ne4 ∣ z ∣ = 10 = 4 z ≠ − 4 i z\ne-4\mathrm{i} z = − 4 i
Rationalising,
z + 4 z + 4 i = [ ( x + 4 ) + y i ] [ x − ( y + 4 ) i ] x 2 + ( y + 4 ) 2 . \frac{z+4}{z+4\mathrm{i}}
=\frac{[(x+4)+y\mathrm{i}][x-(y+4)\mathrm{i}]}
{x^2+(y+4)^2}. z + 4 i z + 4 = x 2 + ( y + 4 ) 2 [( x + 4 ) + y i ] [ x − ( y + 4 ) i ] . The imaginary part of the numerator is
x y − ( x + 4 ) ( y + 4 ) = − 4 x − 4 y − 16. xy-(x+4)(y+4)=-4x-4y-16. x y − ( x + 4 ) ( y + 4 ) = − 4 x − 4 y − 16. The quotient is real when this is zero, giving x + y = − 4 x+y=-4 x + y = − 4
Together with x 2 + y 2 = 10 x^2+y^2=10 x 2 + y 2 = 10 y = − x − 4 y=-x-4 y = − x − 4
x 2 + ( x + 4 ) 2 = 10 ⟹ x 2 + 4 x + 3 = 0. x^2+(x+4)^2=10
\quad\Longrightarrow\quad x^2+4x+3=0. x 2 + ( x + 4 ) 2 = 10 ⟹ x 2 + 4 x + 3 = 0. Thus x = − 1 x=-1 x = − 1 x = − 3 x=-3 x = − 3 y = − 3 y=-3 y = − 3 y = − 1 y=-1 y = − 1
z = − 1 − 3 i or z = − 3 − i . \boxed{z=-1-3\mathrm{i}\quad\text{or}\quad z=-3-\mathrm{i}}. z = − 1 − 3 i or z = − 3 − i .
解法二
思路
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把实商设为一个实参数，可以不用直接乘共轭。根据实部虚部表示 x , y x,y x , y
答题过程
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Since the quotient is real, write
z + 4 = c ( z + 4 i ) , c ∈ R . z+4=c(z+4\mathrm{i}),\qquad c\in\mathbb R. z + 4 = c ( z + 4 i ) , c ∈ R . With z = x + y i z=x+y\mathrm{i} z = x + y i
x + 4 = c x , y = c ( y + 4 ) . \begin{align*}
x+4=&\,cx,\\[4mm]
y=&\,c(y+4).
\end{align*} x + 4 = y = c x , c ( y + 4 ) . The value c = 1 c=1 c = 1
x = 4 c − 1 , y = − 4 c c − 1 . \begin{align*}
x=&\,\frac4{c-1},\\[4mm]
y=&\,-\frac{4c}{c-1}.
\end{align*} x = y = c − 1 4 , − c − 1 4 c . Using x 2 + y 2 = 10 x^2+y^2=10 x 2 + y 2 = 10
16 ( 1 + c 2 ) ( c − 1 ) 2 = 10. \frac{16(1+c^2)}{(c-1)^2}=10. ( c − 1 ) 2 16 ( 1 + c 2 ) = 10. Therefore
16 ( 1 + c 2 ) = 10 ( c − 1 ) 2 , 6 c 2 + 20 c + 6 = 0 , ( 3 c + 1 ) ( c + 3 ) = 0. \begin{align*}
16(1+c^2)=&\,10(c-1)^2,\\[4mm]
6c^2+20c+6=&\,0,\\[4mm]
(3c+1)(c+3)=&\,0.
\end{align*} 16 ( 1 + c 2 ) = 6 c 2 + 20 c + 6 = ( 3 c + 1 ) ( c + 3 ) = 10 ( c − 1 ) 2 , 0 , 0. For c = − 1 / 3 c=-1/3 c = − 1/3 ( x , y ) = ( − 3 , − 1 ) (x,y)=(-3,-1) ( x , y ) = ( − 3 , − 1 ) c = − 3 c=-3 c = − 3 ( x , y ) = ( − 1 , − 3 ) (x,y)=(-1,-3) ( x , y ) = ( − 1 , − 3 )
z = − 3 − i or z = − 1 − 3 i . \boxed{z=-3-\mathrm{i}\quad\text{or}\quad z=-1-3\mathrm{i}}. z = − 3 − i or z = − 1 − 3 i . 