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FP2 2023 Jan Q6

A Level / Edexcel / FP2

IAL 2023 Jan Paper · Question 6

题目

Problem

6.

A complex number zz is represented by the point PP in an Argand diagram.

Given that

∣z−2i∣= ∣z−3∣\begin{align*} |z - 2\mathrm{i}| =&\, |z - 3| \end{align*}

(a) sketch the locus of PP. You do not need to find the coordinates of any intercepts.

(2)

The transformation TT from the zz-plane to the ww-plane is given by

w= izz−2iz≠2i\begin{align*} w =&\, \frac{\mathrm{i}z}{z - 2\mathrm{i}} \qquad z \neq 2\mathrm{i} \end{align*}

Given that TT maps ∣z−2i∣=∣z−3∣|z - 2\mathrm{i}| = |z - 3| to a circle CC in the ww-plane,

(b) find the equation of CC, giving your answer in the form

∣w−(p+qi)∣= r\begin{align*} |w - (p + q\mathrm{i})| =&\, r \end{align*}

where pp, qq and rr are real numbers to be determined.

(6)
题目中文翻译

复数 zz 在 Argand 图上由点 PP 表示。

已知

∣z−2i∣= ∣z−3∣. \begin{align*} |z-2\mathrm{i}|=&\,|z-3|. \end{align*}

(a) 画出点 PP 的轨迹示意图。不必求任何截点的坐标。

(2)

从 zz 平面到 ww 平面的变换 TT 定义为

w= izz−2i,z≠2i. \begin{align*} w=&\,\frac{\mathrm{i}z}{z-2\mathrm{i}}, \qquad z\ne2\mathrm{i}. \end{align*}

已知 TT 把 ∣z−2i∣=∣z−3∣|z-2\mathrm{i}|=|z-3| 映射为 ww 平面内的圆 CC。

(b) 求圆 CC 的方程，将答案写成

∣w−(p+qi)∣= r \begin{align*} |w-(p+q\mathrm{i})|=&\,r \end{align*}

的形式，其中 pp、qq、rr 为待求的实数。

(6)

解答

(a)

解法一

思路

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∣z−2i∣=∣z−3∣|z-2\mathrm{i}|=|z-3| 表示点 PP 到 2i2\mathrm{i} 和 33 的距离相等，所以轨迹是连接 (0,2)(0,2) 与 (3,0)(3,0) 线段的垂直平分线。

答题过程

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The equation

∣z−2i∣=∣z−3∣\begin{align*} |z-2\mathrm{i}|=|z-3| \end{align*}

means that PP is equidistant from the points representing 2i2\mathrm{i} and 33.

So the locus is the perpendicular bisector of the line segment joining

(0,2)and(3,0).\begin{align*} (0,2)\quad\text{and}\quad (3,0). \end{align*}

The sketch should be a straight line with positive gradient, not passing through the origin, lying through quadrants I, III and IV.

(b)

解法一

思路

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把 zz 用 ww 表示，再直接代入原来的模长关系。这样可以避免先求直线方程再代入，整体更短。

答题过程

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Given

w=izz−2i.\begin{align*} w=\frac{\mathrm{i}z}{z-2\mathrm{i}}. \end{align*}

Rearrange:

w(z−2i)= izwz−2iw= izz(w−i)= 2iwz= 2iww−i.\begin{align*} w(z-2\mathrm{i})=&\,\mathrm{i}z\\[2mm] wz-2\mathrm{i}w=&\,\mathrm{i}z\\[2mm] z(w-\mathrm{i})=&\,2\mathrm{i}w\\[2mm] z=&\,\frac{2\mathrm{i}w}{w-\mathrm{i}}. \end{align*}

The original locus is

∣z−2i∣=∣z−3∣.\begin{align*} |z-2\mathrm{i}|=|z-3|. \end{align*}

Substitute z=2iww−iz=\dfrac{2\mathrm{i}w}{w-\mathrm{i}}:

∣2iww−i−2i∣= ∣2iww−i−3∣.\begin{align*} \bigg|\frac{2\mathrm{i}w}{w-\mathrm{i}}-2\mathrm{i}\bigg| =&\,\bigg|\frac{2\mathrm{i}w}{w-\mathrm{i}}-3\bigg|. \end{align*}

Since w≠iw\neq \mathrm{i}, multiply both sides by ∣w−i∣|w-\mathrm{i}|:

∣2iw−2i(w−i)∣= ∣2iw−3(w−i)∣∣−2∣= ∣(2i−3)w+3i∣.\begin{align*} |2\mathrm{i}w-2\mathrm{i}(w-\mathrm{i})| =&\,|2\mathrm{i}w-3(w-\mathrm{i})|\\[2mm] |-2|=&\,|(2\mathrm{i}-3)w+3\mathrm{i}|. \end{align*}

So

2= ∣(2i−3)w+3i∣= ∣2i−3∣∣w+3i2i−3∣.\begin{align*} 2=&\,|(2\mathrm{i}-3)w+3\mathrm{i}|\\[2mm] =&\,|2\mathrm{i}-3| \bigg|w+\frac{3\mathrm{i}}{2\mathrm{i}-3}\bigg|. \end{align*}

Now

∣2i−3∣=13.\begin{align*} |2\mathrm{i}-3|=\sqrt{13}. \end{align*}

Also,

3i2i−3= 3i(−3−2i)(2i−3)(−3−2i)= −9i−6i213= 6−9i13.\begin{align*} \frac{3\mathrm{i}}{2\mathrm{i}-3} =&\,\frac{3\mathrm{i}(-3-2\mathrm{i})}{(2\mathrm{i}-3)(-3-2\mathrm{i})}\\[2mm] =&\,\frac{-9\mathrm{i}-6\mathrm{i}^2}{13}\\[2mm] =&\,\frac{6-9\mathrm{i}}{13}. \end{align*}

Therefore

2= 13∣w+613−913i∣.\begin{align*} 2=&\,\sqrt{13}\bigg|w+\frac{6}{13}-\frac{9}{13}\mathrm{i}\bigg|. \end{align*}

Hence

∣w−(−613+913i)∣= 213= 21313.\begin{align*} \bigg|w-\bigg(-\frac{6}{13}+\frac{9}{13}\mathrm{i}\bigg)\bigg| =&\,\frac{2}{\sqrt{13}}\\[2mm] =&\,\frac{2\sqrt{13}}{13}. \end{align*}

So

∣w−(−613+913i)∣=21313.\begin{align*} \boxed{ \bigg|w-\bigg(-\frac{6}{13}+\frac{9}{13}\mathrm{i}\bigg)\bigg| =\frac{2\sqrt{13}}{13} }. \end{align*}

解法二

思路

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先沿用逆变换，把模长关系化成 ∣(2i−3)w+3i∣=2|(2\mathrm{i}-3)w+3\mathrm{i}|=2。再令 w=u+viw=u+v\mathrm{i}，将实部和虚部分别平方，最后配方得到圆心和半径。

答题过程

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Rearranging the transformation gives

z= 2iww−i,z−2i= −2w−i,z−3= (2i−3)w+3iw−i.\begin{align*} z=&\,\frac{2\mathrm{i}w}{w-\mathrm{i}},\\[2mm] z-2\mathrm{i}=&\,\frac{-2}{w-\mathrm{i}},\\[2mm] z-3=&\,\frac{(2\mathrm{i}-3)w+3\mathrm{i}}{w-\mathrm{i}}. \end{align*}

Hence ∣z−2i∣=∣z−3∣|z-2\mathrm{i}|=|z-3| becomes

∣(2i−3)w+3i∣= 2.\begin{align*} |(2\mathrm{i}-3)w+3\mathrm{i}|=&\,2. \end{align*}

Let w=u+viw=u+v\mathrm{i}. Then

(2i−3)(u+vi)+3i= (−3u−2v)+(2u−3v+3)i.\begin{align*} (2\mathrm{i}-3)(u+v\mathrm{i})+3\mathrm{i} =&\,(-3u-2v)+(2u-3v+3)\mathrm{i}. \end{align*}

Taking the square of the modulus,

(3u+2v)2+(2u−3v+3)2= 4.\begin{align*} (3u+2v)^2+(2u-3v+3)^2=&\,4. \end{align*}

Expanding,

 9u2+12uv+4v2 +4u2+9v2+9−12uv+12u−18v=4,13u2+13v2+12u−18v+9= 4.\begin{align*} &\,9u^2+12uv+4v^2\\[2mm] &\,\hspace{2pt}+4u^2+9v^2+9-12uv+12u-18v =4,\\[2mm] 13u^2+13v^2+12u-18v+9=&\,4. \end{align*}

Divide by 1313 and complete the square:

u2+1213u+v2−1813v= −513,(u+613)2+(v−913)2= −513+36+81169= 413.\begin{align*} u^2+\frac{12}{13}u+v^2-\frac{18}{13}v =&\,-\frac{5}{13},\\[2mm] \bigg(u+\frac{6}{13}\bigg)^2 +\bigg(v-\frac{9}{13}\bigg)^2 =&\,-\frac{5}{13}+\frac{36+81}{169}\\[2mm] =&\,\frac{4}{13}. \end{align*}

The centre is −613+913i-\frac{6}{13}+\frac{9}{13}\mathrm{i} and the radius is 213\frac{2}{\sqrt{13}}. Thus

∣w−(−613+913i)∣=213.\begin{align*} \boxed{\bigg|w-\bigg(-\frac{6}{13}+\frac{9}{13}\mathrm{i}\bigg)\bigg| =\frac{2}{\sqrt{13}}}. \end{align*}

解法三

思路

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先把原轨迹写成直线的直角坐标方程。然后将逆变换的实部、虚部代入，得到关于 w=u+viw=u+v\mathrm{i} 的圆方程。

答题过程

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Let z=x+yiz=x+y\mathrm{i}. The locus satisfies

x2+(y−2)2= (x−3)2+y2−4y+4= −6x+96x−4y= 5.\begin{align*} x^2+(y-2)^2=&\,(x-3)^2+y^2\\[2mm] -4y+4=&\,-6x+9\\[2mm] 6x-4y=&\,5. \end{align*}

With w=u+viw=u+v\mathrm{i}, the inverse transformation is

z= 2i(u+vi)u+(v−1)i= (−2v+2ui)(u−(v−1)i)u2+(v−1)2.\begin{align*} z=&\,\frac{2\mathrm{i}(u+v\mathrm{i})}{u+(v-1)\mathrm{i}}\\[2mm] =&\,\frac{(-2v+2u\mathrm{i})\big(u-(v-1)\mathrm{i}\big)} {u^2+(v-1)^2}. \end{align*}

The numerator expands to

 −2uv+2u(v−1) +(2v(v−1)+2u2)i= −2u+(2u2+2v(v−1))i.\begin{align*} &\,-2uv+2u(v-1)\\[2mm] &\,\hspace{2pt}+\big(2v(v-1)+2u^2\big)\mathrm{i}\\[2mm] =&\,-2u+\big(2u^2+2v(v-1)\big)\mathrm{i}. \end{align*}

Thus, writing D=u2+(v−1)2D=u^2+(v-1)^2,

x= −2uD,y= 2u2+2v(v−1)D.\begin{align*} x=&\,\frac{-2u}{D},& y=&\,\frac{2u^2+2v(v-1)}{D}. \end{align*}

Substitute into 6x−4y=56x-4y=5:

−12u−8u2−8v(v−1)= 5(u2+(v−1)2)−12u−8u2−8v2+8v= 5u2+5v2−10v+513u2+13v2+12u−18v+5= 0.\begin{align*} -12u-8u^2-8v(v-1)=&\,5\big(u^2+(v-1)^2\big)\\[2mm] -12u-8u^2-8v^2+8v=&\,5u^2+5v^2-10v+5\\[2mm] 13u^2+13v^2+12u-18v+5=&\,0. \end{align*}

Complete the square:

(u+613)2+(v−913)2= −513+36+81169= 413.\begin{align*} \bigg(u+\frac{6}{13}\bigg)^2 +\bigg(v-\frac{9}{13}\bigg)^2 =&\,-\frac{5}{13}+\frac{36+81}{169}\\[2mm] =&\,\frac{4}{13}. \end{align*}

Therefore

∣w−(−613+913i)∣=213.\begin{align*} \boxed{\bigg|w-\bigg(-\frac{6}{13}+\frac{9}{13}\mathrm{i}\bigg)\bigg| =\frac{2}{\sqrt{13}}}. \end{align*}