题目
Problem
Find the complex numbers z z z z + 5 i z − 5 \frac{z+5\mathrm{i}}{z-5} z − 5 z + 5 i ∣ z ∣ = 17 |z|=\sqrt{17} ∣ z ∣ = 17 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 + 5 i z − 5 \frac{z+5\mathrm{i}}{z-5} z − 5 z + 5 i ∣ z ∣ = 17 |z|=\sqrt{17} ∣ z ∣ = 17 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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设实部与虚部，分母乘共轭。商为实数意味着所得分子的虚部为零，再与模长条件联立。
答题过程
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Put z = x + y i z=x+y\mathrm{i} z = x + y i
z + 5 i z − 5 = [ x + ( y + 5 ) i ] [ ( x − 5 ) − y i ] ( x − 5 ) 2 + y 2 . \frac{z+5\mathrm{i}}{z-5}
=\frac{[x+(y+5)\mathrm{i}][(x-5)-y\mathrm{i}]}
{(x-5)^2+y^2}. z − 5 z + 5 i = ( x − 5 ) 2 + y 2 [ x + ( y + 5 ) i ] [( x − 5 ) − y i ] . The numerator is
x ( x − 5 ) + y ( y + 5 ) + i [ ( y + 5 ) ( x − 5 ) − x y ] . x(x-5)+y(y+5)
+\mathrm{i}\big[(y+5)(x-5)-xy\big]. x ( x − 5 ) + y ( y + 5 ) + i [ ( y + 5 ) ( x − 5 ) − x y ] . Since ∣ z ∣ = 17 ≠ 5 |z|=\sqrt{17}\ne5 ∣ z ∣ = 17 = 5 z ≠ 5 z\ne5 z = 5
( y + 5 ) ( x − 5 ) − x y = 0 ⟹ x − y = 5. (y+5)(x-5)-xy=0
\quad\Longrightarrow\quad x-y=5. ( y + 5 ) ( x − 5 ) − x y = 0 ⟹ x − y = 5. Also x 2 + y 2 = 17 x^2+y^2=17 x 2 + y 2 = 17 y = x − 5 y=x-5 y = x − 5
x 2 + ( x − 5 ) 2 = 17 , x 2 − 5 x + 4 = 0 , ( x − 1 ) ( x − 4 ) = 0. \begin{align*}
x^2+(x-5)^2=&\,17,\\
x^2-5x+4=&\,0,\\
(x-1)(x-4)=&\,0.
\end{align*} x 2 + ( x − 5 ) 2 = x 2 − 5 x + 4 = ( x − 1 ) ( x − 4 ) = 17 , 0 , 0. The corresponding values of y y y − 4 -4 − 4 − 1 -1 − 1
z = 1 − 4 i or z = 4 − i . \boxed{z=1-4\mathrm{i}\quad\text{or}\quad z=4-\mathrm{i}}. z = 1 − 4 i or z = 4 − i .
解法二
思路
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把实数商设为实数 k k k k k k
答题过程
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Since ∣ z ∣ = 17 |z|=\sqrt{17} ∣ z ∣ = 17 z ≠ 5 z\ne5 z = 5
z + 5 i z − 5 = k , k ∈ R . \frac{z+5\mathrm{i}}{z-5}=k,\qquad k\in\mathbb R. z − 5 z + 5 i = k , k ∈ R . Then z + 5 i = k ( z − 5 ) z+5\mathrm{i}=k(z-5) z + 5 i = k ( z − 5 ) z = x + y i z=x+y\mathrm{i} z = x + y i
x = k ( x − 5 ) , y + 5 = k y . \begin{align*}
x=&\,k(x-5),\\[4mm]
y+5=&\,ky.
\end{align*} x = y + 5 = k ( x − 5 ) , k y . Rearranging,
( k − 1 ) x = 5 k , ( k − 1 ) y = 5. \begin{align*}
(k-1)x=&\,5k,\\[4mm]
(k-1)y=&\,5.
\end{align*} ( k − 1 ) x = ( k − 1 ) y = 5 k , 5. The second equation rules out k = 1 k=1 k = 1
( k − 1 ) ( x − y ) = 5 ( k − 1 ) , (k-1)(x-y)=5(k-1), ( k − 1 ) ( x − y ) = 5 ( k − 1 ) , so x − y = 5 x-y=5 x − y = 5 x 2 + y 2 = 17 x^2+y^2=17 x 2 + y 2 = 17
x 2 + ( x − 5 ) 2 = 17 , ( x − 1 ) ( x − 4 ) = 0. \begin{align*}
x^2+(x-5)^2=&\,17,\\[4mm]
(x-1)(x-4)=&\,0.
\end{align*} x 2 + ( x − 5 ) 2 = ( x − 1 ) ( x − 4 ) = 17 , 0. Thus the candidates are z = 1 − 4 i z=1-4\mathrm{i} z = 1 − 4 i z = 4 − i z=4-\mathrm{i} z = 4 − i 17 \sqrt{17} 17
( 1 − 4 i ) + 5 i ( 1 − 4 i ) − 5 = − 1 4 , ( 4 − i ) + 5 i ( 4 − i ) − 5 = − 4. \begin{align*}
\frac{(1-4\mathrm{i})+5\mathrm{i}}{(1-4\mathrm{i})-5}
=&\,-\frac14,\\[4mm]
\frac{(4-\mathrm{i})+5\mathrm{i}}{(4-\mathrm{i})-5}
=&\,-4.
\end{align*} ( 1 − 4 i ) − 5 ( 1 − 4 i ) + 5 i = ( 4 − i ) − 5 ( 4 − i ) + 5 i = − 4 1 , − 4. Both quotients are real, so
z = 1 − 4 i or z = 4 − i . \boxed{z=1-4\mathrm{i}\quad\text{or}\quad z=4-\mathrm{i}}. z = 1 − 4 i or z = 4 − i . 