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IAL 2023 Jan Q6

A Level / Edexcel / FP2

IAL 2023 Jan Paper · Question 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|\\[2mm] \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}\\[2mm] \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\\[2mm] \end{align*}

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

(6)
解答

(a)

The equation ∣z−2i∣=∣z−3∣|z - 2\mathrm{i}| = |z - 3| represents the perpendicular bisector of the line segment joining the points representing 2i2\mathrm{i} and 33 (i.e., (0,2)(0, 2) and (3,0)(3, 0)). The midpoint is (32,1)\left( \frac{3}{2}, 1 \right). The gradient of the line joining them is −23-\frac{2}{3}, so the gradient of the perpendicular bisector is 32\frac{3}{2}. The locus is a straight line passing through (32,1)\left( \frac{3}{2}, 1 \right) with a positive gradient, mainly lying in the 1st, 3rd, and 4th quadrants.

(Sketch instruction: Draw an Argand diagram with real and imaginary axes. Draw a straight solid line with a positive slope passing through the first, fourth and third quadrants, not passing through the origin.)

(b)

The transformation is given by w=izz−2iw = \frac{\mathrm{i}z}{z - 2\mathrm{i}}. Rearrange to make zz the subject:

w(z−2i)= izwz−2iw= izz(w−i)= 2iwz= 2iww−i\begin{align*} w(z - 2\mathrm{i}) = &\,\mathrm{i}z\\[4mm] wz - 2\mathrm{i}w = &\,\mathrm{i}z\\[4mm] z(w - \mathrm{i}) = &\,2\mathrm{i}w\\[4mm] z = &\,\frac{2\mathrm{i}w}{w - \mathrm{i}} \end{align*}

Substitute this expression for zz into the locus equation ∣z−2i∣=∣z−3∣|z - 2\mathrm{i}| = |z - 3|:

∣2iww−i−2i∣= ∣2iww−i−3∣∣2iw−2i(w−i)w−i∣= ∣2iw−3(w−i)w−i∣\begin{align*} \left| \frac{2\mathrm{i}w}{w - \mathrm{i}} - 2\mathrm{i} \right| = &\,\left| \frac{2\mathrm{i}w}{w - \mathrm{i}} - 3 \right|\\[4mm] \left| \frac{2\mathrm{i}w - 2\mathrm{i}(w - \mathrm{i})}{w - \mathrm{i}} \right| = &\,\left| \frac{2\mathrm{i}w - 3(w - \mathrm{i})}{w - \mathrm{i}} \right| \end{align*}

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

∣2iw−2iw−2∣= ∣2iw−3w+3i∣∣−2∣= ∣w(2i−3)+3i∣2= ∣−3+2i∣∣w+3i−3+2i∣\begin{align*} | 2\mathrm{i}w - 2\mathrm{i}w - 2 | = &\,| 2\mathrm{i}w - 3w + 3\mathrm{i} |\\[4mm] |-2| = &\,| w(2\mathrm{i} - 3) + 3\mathrm{i} |\\[4mm] 2 = &\,| -3 + 2\mathrm{i} | \left| w + \frac{3\mathrm{i}}{-3 + 2\mathrm{i}} \right| \end{align*}

Calculate the modulus ∣−3+2i∣=(−3)2+22=13|-3 + 2\mathrm{i}| = \sqrt{(-3)^2 + 2^2} = \sqrt{13}. Now, simplify the complex fraction:

3i−3+2i= 3i(−3−2i)(−3+2i)(−3−2i)= −9i−6i2(−3)2+22= 6−9i13= 613−913i\begin{align*} \frac{3\mathrm{i}}{-3 + 2\mathrm{i}} = &\,\frac{3\mathrm{i}(-3 - 2\mathrm{i})}{(-3 + 2\mathrm{i})(-3 - 2\mathrm{i})}\\[4mm] = &\,\frac{-9\mathrm{i} - 6\mathrm{i}^2}{(-3)^2 + 2^2}\\[4mm] = &\,\frac{6 - 9\mathrm{i}}{13}\\[4mm] = &\,\frac{6}{13} - \frac{9}{13}\mathrm{i} \end{align*}

Substitute back into the equation:

2= 13∣w+613−913i∣∣w−(−613+913i)∣= 213\begin{align*} 2 = &\,\sqrt{13} \left| w + \frac{6}{13} - \frac{9}{13}\mathrm{i} \right|\\[4mm] \left| w - \left( -\frac{6}{13} + \frac{9}{13}\mathrm{i} \right) \right| = &\,\frac{2}{\sqrt{13}} \end{align*}

This is in the required form ∣w−(p+qi)∣=r|w - (p + q\mathrm{i})| = r, where p=−613p = -\frac{6}{13}, q=913q = \frac{9}{13}, and r=213r = \frac{2}{\sqrt{13}}.