题目
Problem
3.
A complex number z z z P P P ∣ z ∣ = 1 |z|=1 ∣ z ∣ = 1
(a) Sketch the locus of P P P z z z
(1)
The transformation T T T z z z z = x + i y z=x+\mathrm{i}y z = x + i y w w w w = u + i v w=u+\mathrm{i}v w = u + i v
w = 9 i z − i z + 1 z ≠ − 1 \begin{align*}
w=\frac{9\mathrm{i}z-\mathrm{i}}{z+1}
\qquad z\neq -1
\end{align*} w = z + 1 9 i z − i z = − 1 Given that the image under T T T P P P z z z z ≠ − 1 z\neq -1 z = − 1 l l l w w w
(b) determine, in simplest form, a Cartesian equation for l l l
(5)
题目中文翻译
复数 z z z P P P ∣ z ∣ = 1 |z|=1 ∣ z ∣ = 1
(a) 当 z z z P P P
(1)
从 z z z w w w T T T
w = 9 i z − i z + 1 , z ≠ − 1 , \begin{align*}
w=&\,\frac{9\mathrm{i}z-\mathrm{i}}{z+1},
\qquad z\ne-1,
\end{align*} w = z + 1 9 i z − i , z = − 1 , 其中 z = x + i y z=x+\mathrm{i}y z = x + i y w = u + i v w=u+\mathrm{i}v w = u + i v
已知 z z z P P P z ≠ − 1 z\ne-1 z = − 1 T T T w w w l l l
(b) 求直线 l l l
(5)
解答
(a)
解法一
思路
展开
∣ z ∣ = 1 |z|=1 ∣ z ∣ = 1 1 1 1 1 1 1
答题过程
展开
The locus ∣ z ∣ = 1 |z|=1 ∣ z ∣ = 1 O O O 1 1 1
(b)
解法一
思路
展开
先把 z z z w w w ∣ z ∣ = 1 |z|=1 ∣ z ∣ = 1
答题过程
展开
From
w = 9 i z − i z + 1 , \begin{align*}
w=\frac{9\mathrm{i}z-\mathrm{i}}{z+1},
\end{align*} w = z + 1 9 i z − i , make z z z
w ( z + 1 ) = 9 i z − i w z + w = 9 i z − i z ( w − 9 i ) = − w − i z = − w − i w − 9 i \begin{align*}
w(z+1)
=&\,
9\mathrm{i}z-\mathrm{i}\\[2mm]
wz+w
=&\,
9\mathrm{i}z-\mathrm{i}\\[2mm]
z(w-9\mathrm{i})
=&\,
-w-\mathrm{i}\\[2mm]
z
=&\,
\frac{-w-\mathrm{i}}{w-9\mathrm{i}}
\end{align*} w ( z + 1 ) = w z + w = z ( w − 9 i ) = z = 9 i z − i 9 i z − i − w − i w − 9 i − w − i Since ∣ z ∣ = 1 |z|=1 ∣ z ∣ = 1
∣ − w − i w − 9 i ∣ = 1 ∣ − w − i ∣ = ∣ w − 9 i ∣ ∣ w + i ∣ = ∣ w − 9 i ∣ \begin{align*}
\Big|
\frac{-w-\mathrm{i}}{w-9\mathrm{i}}
\Big|
=&\,
1\\[4mm]
|-w-\mathrm{i}|
=&\,
|w-9\mathrm{i}|\\[2mm]
|w+\mathrm{i}|
=&\,
|w-9\mathrm{i}|
\end{align*} w − 9 i − w − i = ∣ − w − i ∣ = ∣ w + i ∣ = 1 ∣ w − 9 i ∣ ∣ w − 9 i ∣ This means that w w w − i -\mathrm{i} − i 9 i 9\mathrm{i} 9 i 4 i 4\mathrm{i} 4 i
v = 4 \begin{align*}
\boxed{v=4}
\end{align*} v = 4
解法二
思路
展开
也可以完全代数化。令 w = u + i v w=u+\mathrm{i}v w = u + i v
∣ w + i ∣ = ∣ w − 9 i ∣ \begin{align*}
|w+\mathrm{i}|=|w-9\mathrm{i}|
\end{align*} ∣ w + i ∣ = ∣ w − 9 i ∣ 两边平方后展开。
答题过程
展开
From the first method,
∣ w + i ∣ = ∣ w − 9 i ∣ \begin{align*}
|w+\mathrm{i}|=|w-9\mathrm{i}|
\end{align*} ∣ w + i ∣ = ∣ w − 9 i ∣ Let w = u + i v w=u+\mathrm{i}v w = u + i v
∣ w + i ∣ 2 = u 2 + ( v + 1 ) 2 \begin{align*}
|w+\mathrm{i}|^2
=&\,
u^2+(v+1)^2
\end{align*} ∣ w + i ∣ 2 = u 2 + ( v + 1 ) 2 and
∣ w − 9 i ∣ 2 = u 2 + ( v − 9 ) 2 \begin{align*}
|w-9\mathrm{i}|^2
=&\,
u^2+(v-9)^2
\end{align*} ∣ w − 9 i ∣ 2 = u 2 + ( v − 9 ) 2 Hence
u 2 + ( v + 1 ) 2 = u 2 + ( v − 9 ) 2 v 2 + 2 v + 1 = v 2 − 18 v + 81 20 v = 80 v = 4 \begin{align*}
u^2+(v+1)^2
=&\,
u^2+(v-9)^2\\[3mm]
v^2+2v+1
=&\,
v^2-18v+81\\[3mm]
20v
=&\,
80\\[2mm]
v
=&\,
4
\end{align*} u 2 + ( v + 1 ) 2 = v 2 + 2 v + 1 = 20 v = v = u 2 + ( v − 9 ) 2 v 2 − 18 v + 81 80 4 Therefore
v = 4 \begin{align*}
\boxed{v=4}
\end{align*} v = 4
解法三
思路
展开
题目已经给定像是直线 l l l z = 1 z=1 z = 1 z = i z=\mathrm{i} z = i
答题过程
展开
The image is given to be a line. Choose two points on ∣ z ∣ = 1 |z|=1 ∣ z ∣ = 1
For z = 1 z=1 z = 1
w = 9 i − i 1 + 1 = 4 i . \begin{align*}
w=&\,\frac{9\mathrm{i}-\mathrm{i}}{1+1}
=4\mathrm{i}.
\end{align*} w = 1 + 1 9 i − i = 4 i . For z = i z=\mathrm{i} z = i
w = 9 i 2 − i 1 + i = − 9 − i 1 + i ⋅ 1 − i 1 − i = − 9 + 9 i − i + i 2 2 = − 5 + 4 i . \begin{align*}
w=&\,\frac{9\mathrm{i}^2-\mathrm{i}}{1+\mathrm{i}}\\[3mm]
=&\,\frac{-9-\mathrm{i}}{1+\mathrm{i}}
\cdot\frac{1-\mathrm{i}}{1-\mathrm{i}}\\[3mm]
=&\,\frac{-9+9\mathrm{i}-\mathrm{i}+\mathrm{i}^2}{2}\\[3mm]
=&\,-5+4\mathrm{i}.
\end{align*} w = = = = 1 + i 9 i 2 − i 1 + i − 9 − i ⋅ 1 − i 1 − i 2 − 9 + 9 i − i + i 2 − 5 + 4 i . The two distinct image points are ( 0 , 4 ) (0,4) ( 0 , 4 ) ( − 5 , 4 ) (-5,4) ( − 5 , 4 )
v = 4 . \begin{align*}
\boxed{v=4}.
\end{align*} v = 4 .
解法四
思路
展开
也可以直接将 z = x + i y z=x+\mathrm{i}y z = x + i y x 2 + y 2 = 1 x^2+y^2=1 x 2 + y 2 = 1 w w w
答题过程
展开
For z = x + i y z=x+\mathrm{i}y z = x + i y x 2 + y 2 = 1 x^2+y^2=1 x 2 + y 2 = 1
w = − 9 y + i ( 9 x − 1 ) x + 1 + i y = [ − 9 y + i ( 9 x − 1 ) ] ( x + 1 − i y ) ( x + 1 ) 2 + y 2 . \begin{align*}
w=&\,\frac{-9y+\mathrm{i}(9x-1)}{x+1+\mathrm{i}y}\\[3mm]
=&\,\frac{[-9y+\mathrm{i}(9x-1)](x+1-\mathrm{i}y)}
{(x+1)^2+y^2}.
\end{align*} w = = x + 1 + i y − 9 y + i ( 9 x − 1 ) ( x + 1 ) 2 + y 2 [ − 9 y + i ( 9 x − 1 )] ( x + 1 − i y ) . The real numerator is
− 9 y ( x + 1 ) + ( 9 x − 1 ) y = − 10 y . \begin{align*}
-9y(x+1)+(9x-1)y=&\,-10y.
\end{align*} − 9 y ( x + 1 ) + ( 9 x − 1 ) y = − 10 y . The imaginary numerator is
9 y 2 + ( 9 x − 1 ) ( x + 1 ) = 9 ( x 2 + y 2 ) + 8 x − 1 = 8 ( x + 1 ) . \begin{align*}
9y^2+(9x-1)(x+1)
=&\,9(x^2+y^2)+8x-1\\[3mm]
=&\,8(x+1).
\end{align*} 9 y 2 + ( 9 x − 1 ) ( x + 1 ) = = 9 ( x 2 + y 2 ) + 8 x − 1 8 ( x + 1 ) . Also,
( x + 1 ) 2 + y 2 = x 2 + y 2 + 2 x + 1 = 2 ( x + 1 ) . \begin{align*}
(x+1)^2+y^2=&\,x^2+y^2+2x+1\\[3mm]
=&\,2(x+1).
\end{align*} ( x + 1 ) 2 + y 2 = = x 2 + y 2 + 2 x + 1 2 ( x + 1 ) . On ∣ z ∣ = 1 |z|=1 ∣ z ∣ = 1 x = − 1 x=-1 x = − 1 z = − 1 z=-1 z = − 1
w = − 10 y + 8 ( x + 1 ) i 2 ( x + 1 ) = − 5 y x + 1 + 4 i . \begin{align*}
w=&\,\frac{-10y+8(x+1)\mathrm{i}}{2(x+1)}\\[3mm]
=&\,-\frac{5y}{x+1}+4\mathrm{i}.
\end{align*} w = = 2 ( x + 1 ) − 10 y + 8 ( x + 1 ) i − x + 1 5 y + 4 i . The imaginary part is constant, so
v = 4 . \begin{align*}
\boxed{v=4}.
\end{align*} v = 4 . 
