Further Maths Further complex numbers Practice Questions

Free Further Maths Further complex numbers practice questions with full step-by-step worked solutions. Covers further-complex, transformations, argand-plane, image-of-a-point. Practise exam-style problems and check your method.

further-complextransformationsargand-planeimage-of-a-pointinverse-imageimage-of-a-locus
Further Maths70 questionsStep-by-step solutions
Question 1
2 markseasy
The transformation TT from the zz-plane to the ww-plane is given by w=3z+2−iw=3 z+2 - i. Find the image of the point z=1+2iz=1 + 2 i under TT, giving your answer in the form a+bia+bi.
Show worked solution

Worked solution

  1. Substitute the given value of zz into the transformation

    w=3(1+2i)+2−iw=3 \left(1 + 2 i\right)+2 - i

    The image of a point is found by evaluating f(z)f\left(z\right) at that point.

  2. State the modulus of the image

    ∣w∣=52\left|w\right|=5 \sqrt{2}

    A quick modulus check confirms the arithmetic.

  3. State the image of the point

    w=5+5iw=5 + 5 i

    This is the image of z=1+2iz=1 + 2 i under TT.

Answer
5+5i5 + 5 i
Question 2
2 markseasy
The transformation TT from the zz-plane to the ww-plane is given by w=4zw=4 z. Which of the following best describes the image under TT of the locus ∣z∣=2\left|z\right|=2?
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Worked solution

  1. Make zz the subject of the transformation

    z=w4z=\frac{w}{4}

    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.

  2. Multiply through by ∣4∣\left|4\right|

    ∣w∣=2∣4∣\left|w\right|=2\left|4\right|

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

  3. Square both sides

    ∣w∣2=4∣4∣2\left|w\right|^{2}=4\left|4\right|^{2}

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

  4. Select the correct description

    u2+v2=64u^{2}+v^{2}=64

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

Answer
u2+v2=64u^{2}+v^{2}=64
Question 3
4 marksintermediate
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 Re⁡(z)=1\operatorname{Re}\left(z\right)=1?
Show worked solution

Worked solution

  1. 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.

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

    Re⁡(1w)=1\operatorname{Re}\left(\frac{1}{w}\right)=1

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

  3. Multiply through by the positive real denominator

    Re⁡((1)(w)‾)=1∣w∣2\operatorname{Re}\left(\left(1\right)\overline{\left(w\right)}\right)=1\left|w\right|^{2}

    ∣w∣2\left|w\right|^{2} is real and positive, so it can be moved to the other side.

  4. Put w=u+ivw=u+iv and take the required part

    u=u2+v2u=u^{2} + v^{2}

    Both sides are now real polynomials in uu and vv.

  5. Expand and collect every term on one side

    u2−u+v2=0u^{2} - u + v^{2}=0

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

  6. Write down the equation of the locus LL

    Re⁡(z)=1\operatorname{Re}\left(z\right)=1

    This is the line x=1x=1 in the zz-plane.

  7. Select the correct description

    (u−12)2+v2=14\left(u - \frac{1}{2}\right)^{2}+v^{2}=\frac{1}{4}

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

Answer
(u−12)2+v2=14\left(u - \frac{1}{2}\right)^{2}+v^{2}=\frac{1}{4}
Question 4
6 markshard
Which of the following best describes the locus of points zz in the Argand diagram satisfying ∣z−(1+2i)∣=2∣z−(4+2i)∣\left|z-\left(1 + 2 i\right)\right|=2\left|z-\left(4 + 2 i\right)\right|?
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Worked solution

  1. Note that the multiplier is not 11

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

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

  3. Square both sides

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

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

  4. Expand the left-hand side

    LHS=x2−2x+y2−4y+5\text{LHS}=x^{2} - 2 x + y^{2} - 4 y + 5

    Multiply out each bracket in turn.

  5. Expand the right-hand side

    RHS=4x2−32x+4y2−16y+80\text{RHS}=4 x^{2} - 32 x + 4 y^{2} - 16 y + 80

    Remember to multiply every term inside the bracket by 44.

  6. Collect all the terms on one side

    −3x2+30x−3y2+12y−75=0-3 x^{2} + 30 x - 3 y^{2} + 12 y - 75=0

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

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

    x2−10x+y2−4y+25=0x^{2} - 10 x + y^{2} - 4 y + 25=0

    Dividing by −3-3 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

    x2−10x=(x−5)2−25x^{2}-10 x=\left(x - 5\right)^{2}-25

    Half the coefficient of xx is −5-5.

  9. Complete the square in yy

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

    Half the coefficient of yy is −2-2.

  10. State the centre and the radius

    centre (5, 2),radius 2\text{centre }\left(5,\ 2\right),\qquad\text{radius }2

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

  11. Select the correct description

    (x−5)2+(y−2)2=4\left(x - 5\right)^{2}+\left(y - 2\right)^{2}=4

    The locus is the circle with centre (5, 2)\left(5,\ 2\right) and radius 22.

Answer
(x−5)2+(y−2)2=4\left(x - 5\right)^{2}+\left(y - 2\right)^{2}=4
Question 5
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?
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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}

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