Hard Further Maths Further complex numbers Questions

Challenging, exam-style Further Maths Further complex numbers questions with worked solutions. Stretch yourself on the hardest further-complex, loci, apollonius-circle, arc-of-a-circle problems.

further-complexlociapollonius-circlearc-of-a-circleargument-of-a-quotienttransformations
Further Maths34 questionsStep-by-step solutions
Question 1
9 markschallenging
The transformation TT from the zz-plane to the ww-plane is given by w=1zw=\frac{1}{z}. Which of the following best describes the image under TT of the locus ∣z−3i∣=3\left|z - 3 i\right|=3?
Show worked solution

Worked solution

  1. Write down the equation of the locus LL

    ∣z−3i∣=3\left|z - 3 i\right|=3

    This is the circle with centre 3i3 i and radius 33 in the zz-plane.

  2. Make zz the subject of the transformation

    z=1wz=\frac{1}{w}

    A point ww is on the image exactly when z=f−1(w)z=f^{-1}\left(w\right) lies on LL, so this is the expression that must be substituted.

  3. Substitute z=f−1(w)z=f^{-1}\left(w\right) into the equation of LL

    ∣1w−3i∣=3\left|\frac{1}{w}-3 i\right|=3

    Every zz in the equation of LL is replaced by its expression in ww.

  4. Combine each modulus over the common denominator

    ∣−3iw+1w∣=3\left|\frac{-3 i w + 1}{w}\right|=3

    Writing each bracket as a single fraction is what makes the moduli easy to handle.

  5. Multiply through by ∣w∣\left|w\right|

    ∣−3iw+1∣=3∣w∣\left|-3 i w + 1\right|=3\left|w\right|

    Clearing the denominator turns the equation into a statement about two moduli.

  6. Square both sides

    ∣−3iw+1∣2=9∣w∣2\left|-3 i w + 1\right|^{2}=9\left|w\right|^{2}

    Both sides are non-negative, so squaring adds no new points.

  7. Put w=u+ivw=u+iv and use ∣α∣2=(Re⁡α)2+(Im⁡α)2\left|\alpha\right|^{2}=\left(\operatorname{Re}\alpha\right)^{2}+\left(\operatorname{Im}\alpha\right)^{2}

    (3v+1)2+(−3u)2=9[u2+v2]\left(3 v + 1\right)^{2}+\left(-3 u\right)^{2}=9\left[u^{2}+v^{2}\right]

    Each modulus squared becomes a sum of two real squares.

  8. Expand and collect every term on one side

    6v+1=06 v + 1=0

    This is the Cartesian equation of the image, before it is tidied up.

  9. Make the equation explicit

    v=−16v=-\frac{1}{6}

    The image is a straight line in the ww-plane.

  10. Note the type of the image

    a straight line in the w-plane\text{a straight line in the }w\text{-plane}

    A circle through the origin maps to a line under w=1zw=\frac{1}{z}, and a line maps to a line under a linear map.

  11. Note what the transformation does

    w=1zw=\frac{1}{z}

    This is a Mobius transformation.

  12. Check the direction of the substitution

    z=f−1(w), not w=f(z)z=f^{-1}\left(w\right),\ \text{not }w=f\left(z\right)

    The equation of LL is a statement about zz, so zz must be replaced.

  13. Sanity-check with one mapped point

    take a point of L, apply w=1z, test the equation\text{take a point of }L,\ \text{apply }w=\frac{1}{z},\ \text{test the equation}

    One point of the source locus, pushed through the map, must satisfy the answer.

  14. Recall the circle-line property

    circles and lines⟶circles and lines\text{circles and lines}\longrightarrow\text{circles and lines}

    A Mobius map (and w=1zw=\frac{1}{z} in particular) sends circles and lines to circles and lines.

  15. Eliminate options with the wrong type

    the image is a straight line\text{the image is a straight line}

    Only one option has the right type as well as the right numbers.

  16. Select the correct description

    v=−16v=-\frac{1}{6}

    This is the image of the locus in the ww-plane.

Answer
v=−16v=-\frac{1}{6}
Question 2
9 markschallenging
The region RR in the Argand diagram consists of the points zz satisfying 1≤∣z−(2+2i)∣≤31\le\left|z-\left(2 + 2 i\right)\right|\le3 and 0≤arg⁡(z−(2+2i))≤π20\le\arg\left(z-\left(2 + 2 i\right)\right)\le\frac{\pi}{2}. Which of the following points lies in RR?
Show worked solution

Worked solution

  1. Interpret the first inequality

    1≤∣z−(2+2i)∣≤31\le\left|z-\left(2 + 2 i\right)\right|\le3

    This inequality describes the annulus between radii 11 and 33 centred at 2+2i2 + 2 i.

  2. Interpret the second inequality

    0≤arg⁡(z−(2+2i))≤π20\le\arg\left(z-\left(2 + 2 i\right)\right)\le\frac{\pi}{2}

    This inequality describes the sector of directions from 00 to π2\frac{\pi}{2}.

  3. Write every candidate as a Cartesian point

    (4, 3),(2, 2),(5, 6),(1, 3),(4, 1)\left(4,\ 3\right),\quad \left(2,\ 2\right),\quad \left(5,\ 6\right),\quad \left(1,\ 3\right),\quad \left(4,\ 1\right)

    Testing an inequality is easiest with real coordinates.

  4. Test the point z=4+3iz=4 + 3 i

    z=4+3i:satisfies both inequalitiesz=4 + 3 i:\quad \text{satisfies both inequalities}

    This point lies in RR.

  5. Test the point z=2+2iz=2 + 2 i

    z=2+2i:fails inequality 1z=2 + 2 i:\quad \text{fails inequality 1}

    This point breaks one of the two conditions, so it is not in RR.

  6. Test the point z=5+6iz=5 + 6 i

    z=5+6i:fails inequality 1z=5 + 6 i:\quad \text{fails inequality 1}

    This point breaks one of the two conditions, so it is not in RR.

  7. Test the point z=1+3iz=1 + 3 i

    z=1+3i:fails inequality 2z=1 + 3 i:\quad \text{fails inequality 2}

    This point breaks one of the two conditions, so it is not in RR.

  8. Test the point z=4+iz=4 + i

    z=4+i:fails inequality 2z=4 + i:\quad \text{fails inequality 2}

    This point breaks one of the two conditions, so it is not in RR.

  9. Recall what the intersection means

    R={z: both inequalities hold}R=\left\{z:\ \text{both inequalities hold}\right\}

    RR is the intersection of the two sets, not their union.

  10. Note the boundary is included

    ≤ and ≥ are not strict\le\ \text{and}\ \ge\ \text{are not strict}

    A point exactly on a boundary curve still belongs to RR.

  11. Recall the Cartesian form of a complex number

    z=x+iy,w=u+ivz=x+iy,\qquad w=u+iv

    Every locus and every image is finally written in terms of real coordinates.

  12. Recall the modulus of a complex number

    ∣x+iy∣=x2+y2\left|x+iy\right|=\sqrt{x^{2}+y^{2}}

    The modulus is the distance of the point (x,y)\left(x,y\right) from the origin.

  13. Recall what a modulus of a difference measures

    ∣z−a∣=distance from z to a\left|z-a\right|=\text{distance from }z\text{ to }a

    This is why so many loci in the Argand diagram are circles and lines.

  14. Recall the standard circle locus

    ∣z−a∣=r\left|z-a\right|=r

    This is a circle of radius rr centred at the point representing aa.

  15. Select the point lying in RR

    z=4+3iz=4 + 3 i

    This point satisfies both defining inequalities.

Answer
z=4+3iz=4 + 3 i
Question 3
9 markschallenging
Which of the following best describes the locus of points zz in the Argand diagram satisfying ∣z−2∣=12∣z+4∣\left|z - 2\right|=\frac{1}{2}\left|z + 4\right|?
Show worked solution

Worked solution

  1. Note that the multiplier is not 11

    k=12≠1k=\frac{1}{2}\neq1

    With k≠1k\neq1 the locus is a circle (an Apollonius circle), not a perpendicular bisector.

  2. Write z=x+iyz=x+iy and put the locus in Cartesian form

    (x−2)2+(y)2=12(x+4)2+(y)2\sqrt{\left(x-2\right)^{2}+\left(y\right)^{2}}=\frac{1}{2}\sqrt{\left(x+4\right)^{2}+\left(y\right)^{2}}

    Each modulus is the distance from (x,y)\left(x,y\right) to a fixed point.

  3. Square both sides

    (x−2)2+(y)2=14[(x+4)2+(y)2]\left(x-2\right)^{2}+\left(y\right)^{2}=\frac{1}{4}\left[\left(x+4\right)^{2}+\left(y\right)^{2}\right]

    Both sides are non-negative, so squaring is reversible and adds no points.

  4. Expand the left-hand side

    LHS=x2−4x+y2+4\text{LHS}=x^{2} - 4 x + y^{2} + 4

    Multiply out each bracket in turn.

  5. Expand the right-hand side

    RHS=x24+2x+y24+4\text{RHS}=\frac{x^{2}}{4} + 2 x + \frac{y^{2}}{4} + 4

    Remember to multiply every term inside the bracket by 14\frac{1}{4}.

  6. Collect all the terms on one side

    3x24−6x+3y24=0\frac{3 x^{2}}{4} - 6 x + \frac{3 y^{2}}{4}=0

    The coefficients of x2x^{2} and of y2y^{2} are both 34\frac{3}{4} and there is no xyxy term, so this is a circle.

  7. Divide through by the coefficient of x2+y2x^{2}+y^{2}

    x2−8x+y2=0x^{2} - 8 x + y^{2}=0

    Dividing by 34\frac{3}{4} puts the equation into the standard form x2+y2+2gx+2fy+c=0x^{2}+y^{2}+2gx+2fy+c=0.

  8. Complete the square in xx

    x2−8x=(x−4)2−16x^{2}-8 x=\left(x - 4\right)^{2}-16

    Half the coefficient of xx is −4-4.

  9. Rearrange into the standard circle equation

    (x−4)2+y2=16\left(x - 4\right)^{2}+y^{2}=16

    The constant terms produced by completing the square move to the right.

  10. State the centre and the radius

    centre (4, 0),radius 4\text{centre }\left(4,\ 0\right),\qquad\text{radius }4

    This circle is called the Apollonius circle of the two points.

  11. Check the two points of the locus on the line ABAB

    0 and 8 divide AB in the ratio 12:10\ \text{and}\ 8\ \text{divide }AB\ \text{in the ratio }\frac{1}{2}:1

    These are the internal and external division points, and they are the ends of a diameter.

  12. Note why k≠1k\neq1 matters

    k=1 ⇒ the x2+y2 terms cancelk=1\ \Rightarrow\ \text{the }x^{2}+y^{2}\ \text{terms cancel}

    With k=1k=1 the locus degenerates to the perpendicular bisector of ABAB.

  13. Verify a point on the circle

    8:∣z−a∣=12∣z−b∣8:\quad \left|z-a\right|=\frac{1}{2}\left|z-b\right|

    Substituting the right-hand end of the horizontal diameter satisfies the original equation.

  14. Reject the perpendicular bisector

    only k=1 gives a straight line\text{only }k=1\ \text{gives a straight line}

    The straight-line option is the answer to ∣z−a∣=∣z−b∣\left|z-a\right|=\left|z-b\right|, which is a different locus.

  15. Reject a circle centred at 22

    the centre is (4, 0)\text{the centre is }\left(4,\ 0\right)

    The centre of an Apollonius circle is not one of the two fixed points.

  16. Check the radius

    r=4r=4

    The radius comes from the completed square, not from the distance ABAB.

  17. Select the correct description

    (x−4)2+y2=16\left(x - 4\right)^{2}+y^{2}=16

    The locus is the circle with centre (4, 0)\left(4,\ 0\right) and radius 44.

Answer
(x−4)2+y2=16\left(x - 4\right)^{2}+y^{2}=16
Question 4
9 markschallenging
The transformation TT from the zz-plane to the ww-plane is given by w=z+1z−1w=\frac{z + 1}{z - 1}. Which of the following is a Cartesian equation of the image of the locus ∣z∣=3\left|z\right|=3 under TT, where w=u+ivw=u+iv?
Show worked solution

Worked solution

  1. Write down the equation of the locus LL

    ∣z∣=3\left|z\right|=3

    This is the circle with centre 00 and radius 33 in the zz-plane.

  2. Make zz the subject of the transformation

    z=w+1w−1z=\frac{w + 1}{w - 1}

    A point ww is on the image exactly when z=f−1(w)z=f^{-1}\left(w\right) lies on LL, so this is the expression that must be substituted.

  3. Substitute z=f−1(w)z=f^{-1}\left(w\right) into the equation of LL

    ∣w+1w−1∣=3\left|\frac{w + 1}{w - 1}\right|=3

    Every zz in the equation of LL is replaced by its expression in ww.

  4. Multiply through by ∣w−1∣\left|w - 1\right|

    ∣w+1∣=3∣w−1∣\left|w + 1\right|=3\left|w - 1\right|

    Clearing the denominator turns the equation into a statement about two moduli.

  5. Square both sides

    ∣w+1∣2=9∣w−1∣2\left|w + 1\right|^{2}=9\left|w - 1\right|^{2}

    Both sides are non-negative, so squaring adds no new points.

  6. Put w=u+ivw=u+iv and use ∣α∣2=(Re⁡α)2+(Im⁡α)2\left|\alpha\right|^{2}=\left(\operatorname{Re}\alpha\right)^{2}+\left(\operatorname{Im}\alpha\right)^{2}

    (u+1)2+v2=9[(u−1)2+v2]\left(u + 1\right)^{2}+v^{2}=9\left[\left(u - 1\right)^{2}+v^{2}\right]

    Each modulus squared becomes a sum of two real squares.

  7. Expand and collect every term on one side

    2u2−5u+2v2+2=02 u^{2} - 5 u + 2 v^{2} + 2=0

    This is the Cartesian equation of the image, before it is tidied up.

  8. Divide through by the coefficient of u2+v2u^{2}+v^{2}

    u2−5u2+v2+1=0u^{2} - \frac{5 u}{2} + v^{2} + 1=0

    Dividing by 22 puts the equation in the standard form u2+v2+2gu+2fv+c=0u^{2}+v^{2}+2gu+2fv+c=0.

  9. Complete the square in uu

    u2−52u=(u−54)2−2516u^{2}- \frac{5}{2} u=\left(u - \frac{5}{4}\right)^{2}- \frac{25}{16}

    Half the coefficient of uu is −54- \frac{5}{4}.

  10. Rearrange into the standard circle equation

    (u−54)2+v2=916\left(u - \frac{5}{4}\right)^{2}+v^{2}=\frac{9}{16}

    This puts the image in the form (u−p)2+(v−q)2=r2\left(u-p\right)^{2}+\left(v-q\right)^{2}=r^{2}.

  11. State the centre and radius of the image circle

    centre (54, 0),radius 34\text{centre }\left(\frac{5}{4},\ 0\right),\qquad\text{radius }\frac{3}{4}

    The image is a circle, as a Mobius or linear map always predicts.

  12. Check one point of the image

    (2, 0) lies on the image circle\left(2,\ 0\right)\ \text{lies on the image circle}

    Substituting this point into the equation gives zero, as required.

  13. Note what the transformation does

    w=z+1z−1w=\frac{z + 1}{z - 1}

    This is a Mobius transformation.

  14. Check the direction of the substitution

    z=f−1(w), not w=f(z)z=f^{-1}\left(w\right),\ \text{not }w=f\left(z\right)

    The equation of LL is a statement about zz, so zz must be replaced.

  15. Sanity-check with one mapped point

    take a point of L, apply w=z+1z−1, test the equation\text{take a point of }L,\ \text{apply }w=\frac{z + 1}{z - 1},\ \text{test the equation}

    One point of the source locus, pushed through the map, must satisfy the answer.

  16. Select the matching equation

    (u−54)2+v2=916\left(u - \frac{5}{4}\right)^{2}+v^{2}=\frac{9}{16}

    This is the Cartesian equation of the image of LL.

Answer
(u−54)2+v2=916\left(u - \frac{5}{4}\right)^{2}+v^{2}=\frac{9}{16}
Question 5
9 markschallenging
The region RR in the Argand diagram consists of the points zz satisfying ∣z∣≤6\left|z\right|\le6 and Im⁡(z)≥3\operatorname{Im}\left(z\right)\ge3. Find the exact area of RR.
Show worked solution

Worked solution

  1. Interpret the first inequality

    ∣z∣≤6\left|z\right|\le6

    This is the closed disc of radius 66 centred at 00.

  2. Interpret the second inequality

    Im⁡(z)≥3\operatorname{Im}\left(z\right)\ge3

    This is the half-plane y≥3y\ge3.

  3. Recognise the region as a circular segment

    x2+y2≤36,y≥3x^{2}+y^{2}\le36,\qquad y\ge3

    The chord y=3y=3 cuts a segment off the disc of radius 66.

  4. Set up the area as an integral

    A=∫36236−y2 dyA=\int_{3}^{6}2\sqrt{36-y^{2}}\,dy

    For each yy the disc contributes a horizontal chord of length 236−y22\sqrt{36-y^{2}}.

  5. Find the half-angle subtended by the chord

    cos⁡α=36=12 ⇒ α=π3\cos\alpha=\frac{3}{6}=\frac{1}{2}\ \Rightarrow\ \alpha=\frac{\pi}{3}

    The chord is at distance 33 from the centre.

  6. Apply the segment formula

    A=r2α−dr2−d2=36×π3−327A=r^{2}\alpha-d\sqrt{r^{2}-d^{2}}=36\times\frac{\pi}{3}-3\sqrt{27}

    This is the standard exact area of a circular segment.

  7. Simplify the surd

    327=933\sqrt{27}=9\sqrt{3}

    Always simplify surds before stating an exact answer.

  8. Give a decimal check

    12π−93≈22.112\pi-9\sqrt{3}\approx22.1

    A sketch confirms the segment is a little under a quarter of the disc.

  9. Recall the antiderivative behind a circular cut

    ∫2r2−t2 dt=tr2−t2+r2arcsin⁡(tr)+c\int 2\sqrt{r^{2}-t^{2}}\,dt=t\sqrt{r^{2}-t^{2}}+r^{2}\arcsin\left(\frac{t}{r}\right)+c

    This standard integral is what produces every circular-segment area.

  10. Recall the exact values that keep the answer exact

    arcsin⁡12=π6,arcsin⁡1=π2,27=33\arcsin\frac{1}{2}=\frac{\pi}{6},\qquad\arcsin1=\frac{\pi}{2},\qquad\sqrt{27}=3\sqrt{3}

    Exact values are what turn the integral into an answer in π\pi and surds.

  11. Confirm the region is bounded

    R⊆{z: ∣z∣≤6}R\subseteq\left\{z:\ \left|z\right|\le6\right\}

    The first inequality already confines RR to a bounded set.

  12. State the area as a decimal check

    A≈22.1107A\approx 22.1107

    The decimal value is a quick sanity check on the exact answer.

  13. Recall the Cartesian form of a complex number

    z=x+iy,w=u+ivz=x+iy,\qquad w=u+iv

    Every locus and every image is finally written in terms of real coordinates.

  14. Recall the modulus of a complex number

    ∣x+iy∣=x2+y2\left|x+iy\right|=\sqrt{x^{2}+y^{2}}

    The modulus is the distance of the point (x,y)\left(x,y\right) from the origin.

  15. State the exact area of RR

    A=12π−93A=12 \pi-9 \sqrt{3}

    This is the exact area of the region RR.

Answer
A=12π−93A=12 \pi-9 \sqrt{3}

Unlock 29 more Further complex numbers questions

Create a free account to work through every Further Maths Further complex numbers question with instant step-by-step worked solutions, progress tracking and interactive lessons.

  • Full worked solutions for every question
  • Interactive lessons and instant feedback
  • Track your mastery across every topic
Create a Free Account

No card required · Free forever

More Further complex numbers practice

Practice QuestionsWorked Solutions

Related Further Pure topics

Conic sections 1Conic sections 2GroupsInequalitiesFurther integration techniquesFurther matrix algebra