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IAL 2023 June Q5

A Level / Edexcel / FP2

IAL 2023 June Paper · Question 5

The transformation TT from the zz-plane, where z=x+iyz = x + \mathrm{i}y , to the ww-plane, where w=u+ivw = u + \mathrm{i}v is given by

w= z+1z−3z≠3\begin{align*} w =\,& \frac{z + 1}{z - 3} \qquad z \neq 3\\[2mm] \end{align*}

The straight line in the zz-plane with equation y=4xy = 4x is mapped by TT onto the circle CC in the ww-plane.

(a) Show that CC has equation

3u2+3v2−2u+v+k= 0\begin{align*} 3u^2 + 3v^2 - 2u + v + k =\,& 0\\[2mm] \end{align*}

where kk is a constant to be determined.

(5)

(b) Hence determine

(i) the coordinates of the centre of CC

(ii) the radius of CC

(2)
解答

(a)

The transformation is given by:

w= z+1z−3\begin{align*} w = &\,\frac{z+1}{z-3} \end{align*}

Rearrange to make zz the subject:

w(z−3)= z+1wz−3w= z+1z(w−1)= 3w+1z= 3w+1w−1\begin{align*} w(z-3) = &\,z+1\\[4mm] wz - 3w = &\,z+1\\[4mm] z(w-1) = &\,3w+1\\[4mm] z = &\,\frac{3w+1}{w-1} \end{align*}

Let z=x+iyz = x + \mathrm{i}y and w=u+ivw = u + \mathrm{i}v:

x+iy= 3(u+iv)+1u+iv−1= (3u+1)+i(3v)(u−1)+iv\begin{align*} x + \mathrm{i}y = &\,\frac{3(u+\mathrm{i}v)+1}{u+\mathrm{i}v-1}\\[4mm] = &\,\frac{(3u+1) + \mathrm{i}(3v)}{(u-1) + \mathrm{i}v} \end{align*}

Multiply the numerator and denominator by the complex conjugate of the denominator, (u−1)−iv(u-1) - \mathrm{i}v:

x+iy= [(3u+1)+i(3v)][(u−1)−iv](u−1)2+v2= (3u+1)(u−1)+3v2+i[3v(u−1)−v(3u+1)](u−1)2+v2\begin{align*} x + \mathrm{i}y = &\,\frac{[(3u+1) + \mathrm{i}(3v)][(u-1) - \mathrm{i}v]}{(u-1)^2 + v^2}\\[4mm] = &\,\frac{(3u+1)(u-1) + 3v^2 + \mathrm{i}[3v(u-1) - v(3u+1)]}{(u-1)^2 + v^2} \end{align*}

Equate the real parts (xx) and the imaginary parts (yy):

x= 3u2−3u+u−1+3v2(u−1)2+v2=3u2−2u+3v2−1(u−1)2+v2y= 3uv−3v−3uv−v(u−1)2+v2=−4v(u−1)2+v2\begin{align*} x = &\,\frac{3u^2 - 3u + u - 1 + 3v^2}{(u-1)^2 + v^2} = \frac{3u^2 - 2u + 3v^2 - 1}{(u-1)^2 + v^2}\\[4mm] y = &\,\frac{3uv - 3v - 3uv - v}{(u-1)^2 + v^2} = \frac{-4v}{(u-1)^2 + v^2} \end{align*}

We are given the line equation y=4xy = 4x. Substituting our expressions for xx and yy:

−4v(u−1)2+v2= 4(3u2−2u+3v2−1(u−1)2+v2)\begin{align*} \frac{-4v}{(u-1)^2 + v^2} = &\,4\left( \frac{3u^2 - 2u + 3v^2 - 1}{(u-1)^2 + v^2} \right) \end{align*}

Since (u−1)2+v2≠0(u-1)^2 + v^2 \neq 0:

−4v= 12u2−8u+12v2−4−v= 3u2−2u+3v2−10= 3u2+3v2−2u+v−1\begin{align*} -4v = &\,12u^2 - 8u + 12v^2 - 4\\[4mm] -v = &\,3u^2 - 2u + 3v^2 - 1\\[4mm] 0 = &\,3u^2 + 3v^2 - 2u + v - 1 \end{align*}

This matches the form 3u2+3v2−2u+v+k=03u^2 + 3v^2 - 2u + v + k = 0, where k=−1k = -1.

(b)

Divide the equation of the circle by 3 to complete the square:

u2−23u+v2+13v−13= 0(u−13)2−19+(v+16)2−136−13= 0(u−13)2+(v+16)2= 19+136+13(u−13)2+(v+16)2= 4+1+1236(u−13)2+(v+16)2= 1736\begin{align*} u^2 - \frac{2}{3}u + v^2 + \frac{1}{3}v - \frac{1}{3} = &\,0\\[4mm] \left(u - \frac{1}{3}\right)^2 - \frac{1}{9} + \left(v + \frac{1}{6}\right)^2 - \frac{1}{36} - \frac{1}{3} = &\,0\\[4mm] \left(u - \frac{1}{3}\right)^2 + \left(v + \frac{1}{6}\right)^2 = &\,\frac{1}{9} + \frac{1}{36} + \frac{1}{3}\\[4mm] \left(u - \frac{1}{3}\right)^2 + \left(v + \frac{1}{6}\right)^2 = &\,\frac{4 + 1 + 12}{36}\\[4mm] \left(u - \frac{1}{3}\right)^2 + \left(v + \frac{1}{6}\right)^2 = &\,\frac{17}{36} \end{align*}

(i) The coordinates of the centre of CC are (13,−16)\left(\frac{1}{3}, -\frac{1}{6}\right). (ii) The radius of CC is 1736=176\sqrt{\frac{17}{36}} = \frac{\sqrt{17}}{6}.