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IAL 2024 June Q6

A Level / Edexcel / FP2

IAL 2024 June Paper · Question 6

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

w=z−iz+1z≠−1\begin{align*} w = \frac{z - \mathrm{i}}{z + 1} \qquad z \neq -1 \end{align*}

Given that TT maps the imaginary axis in the zz-plane to the circle CC in the ww-plane, determine

(i) the coordinates of the centre of CC

(ii) the radius of CC

(7)
解答

The transformation is given by:

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

Rearrange to make zz the subject:

w(z+1)= z−iwz+w= z−iz(w−1)= −w−iz= w+i1−w\begin{align*} w(z + 1) = &\,z - \mathrm{i}\\[4mm] wz + w = &\,z - \mathrm{i}\\[4mm] z(w - 1) = &\,-w - \mathrm{i}\\[4mm] z = &\,\frac{w + \mathrm{i}}{1 - w} \end{align*}

Given that zz lies on the imaginary axis, its real part is zero, i.e., Re⁡(z)=0\operatorname{Re}(z) = 0. Let w=u+ivw = u + \mathrm{i}v. Substitute this into the expression for zz:

z= u+iv+i1−(u+iv)= u+i(v+1)(1−u)−iv\begin{align*} z = &\,\frac{u + \mathrm{i}v + \mathrm{i}}{1 - (u + \mathrm{i}v)}\\[4mm] = &\,\frac{u + \mathrm{i}(v + 1)}{(1 - u) - \mathrm{i}v} \end{align*}

To find the real part of zz, multiply the numerator and the denominator by the complex conjugate of the denominator, (1−u)+iv(1 - u) + \mathrm{i}v:

z= [u+i(v+1)][(1−u)+iv](1−u)2+v2\begin{align*} z = &\,\frac{\big[u + \mathrm{i}(v + 1)\big] \big[(1 - u) + \mathrm{i}v\big]}{(1 - u)^2 + v^2} \end{align*}

The real part of the numerator is formed by multiplying the real parts together and the imaginary parts together:

Re⁡(z)= u(1−u)−v(v+1)(1−u)2+v2\begin{align*} \operatorname{Re}(z) = &\,\frac{u(1 - u) - v(v + 1)}{(1 - u)^2 + v^2} \end{align*}

Since Re⁡(z)=0\operatorname{Re}(z) = 0, the numerator must be zero:

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

Complete the square for both uu and vv:

(u−12)2−14+(v+12)2−14= 0(u−12)2+(v+12)2= 12\begin{align*} \left( u - \frac{1}{2} \right)^2 - \frac{1}{4} + \left( v + \frac{1}{2} \right)^2 - \frac{1}{4} = &\,0\\[4mm] \left( u - \frac{1}{2} \right)^2 + \left( v + \frac{1}{2} \right)^2 = &\,\frac{1}{2} \end{align*}

This represents the equation of a circle in the ww-plane.

(i) The coordinates of the centre of CC are (12,−12)\left( \frac{1}{2}, -\frac{1}{2} \right). (ii) The radius of CC is 12=12\sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}.