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Complex Numbers : Further Mathematics (A2-Level) : Mathematics Content Knowledge


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Further Mathematics (A2-Level)
Complex Numbers
Question 1 of 9

1. How confident are you that

a. you can represent simple sets of complex numbers as Loci in the Argand diagram?


Example

A Circle
Consider |z| = 5 where z = x + iy|z| is found by working out \sqrt{x^2 + y^2  }  so |z| = 5 is equivalent to writing \sqrt{x^2 + y^2  }  and thus x^2 + y^2  = 25.  This is easily recognised as the formula for a circle, centre and origin and radius 5.

Consider the general equation of a circle as (x - a)^2 + (y - b)^2 = r^2 giving a circle with centre (a,b) and radius r. In complex numbers we rewrite z = x + iy as z = (x - a) + i(y - b) which can be rearranged to z = x + iy - (a + ib) and thus the initial equation is transformed into |z - (a + ib)| = r. So this also represents a circle with centre a + ib and radius r.


A half-line
Consider argz = \frac{\pi }{4} .

This represents the set of points (shown by the blue line) on the following diagram. The blue line continues infinitely away from the origin. Students may enjoy the challenge in picturing this line being only half-infinitely long.

A similar line of thinking as with the circle yields the general equation arg(z - (a + bi)) = \theta . This will give a line starting at the point a = bi and proceeding at an angle \theta  relative to the horizontal axis and in an anti-clockwise direction.


Perpendicular bisector

Consider |z - (a - bi)| = |z - (c = di)|.
|z - (a + bi)| measures the distance from z to the point a + bi and the equation says that this distance is equation to the distance from z to the point c + di. Thus the equation dictates that z has to be the same distance from a + bi as it is from c + di. This will be recognised as the perpendicular bisector of the line joining a + bi to c + di - and is best illustrated with a simple sketch.

Example: sketch the locus given by |z - (1 - 2i)| = |z - (2 + 3i)|. 

 

Note: the red dashed line is not necessary but is often helpful – so that the middle point can be found more easily. It is not necessary to draw this diagram accurately – a sketch is usually all that is required.


A harder equation

This example illustrates a technique for finding the equation of a locus – it is a little long winded and is best reserved for situations which do not appear obvious, even after some thought.

For example - the locus of points such that |z - 2| = 3|z - (1 + i)|.

This is interpreted as meaning that the distance between z and the point 2 must be three times the distance between z and the point 1 + i. This is not a shape that is easily recognised, and an alternative approach for finding the equation fo the locus in a recognisable form is required.

Let z=x+\iota y then the equation can be re-written as follows:

\mid z-2 \mid =3 \mid z-(1+\iota )\mid  now replacing z with x+\iota ygives \mid x+\iota y-2\mid =3\mid x+iy-(1+\iota )\mid .

Re-order the terms on each side so as to separate the real parts and the imaginary parts gives \mid (x-2) + \iota y \mid=3 \mid(x-1) + (\iota y-\iota ).

Calculating the modulus on each side, but also squaring to remove the square root:

(x-2)^2-y^2=9[(x-1)^2+(y-1)^2]

Now expand the brackets on each side:

x^2-4x+4+y^2=9x^2-18x+9y^2-18y+18

Collect terms:

0=8x^2-14x+8y^2-18y+14

Complete the square with respect to x and y:

(x-\frac{7} {8})^2+(y-\frac{9} {8})^2=\frac{49} {64}-\frac{81} {64}+\frac{14} {8}=0

Rearrange:

(x-\frac{7} {8})^2+(y-\frac{9} {8})^2=\frac{49} {64}+\frac{81} {64}-\frac{14} {8}

(x-\frac{7} {8})^2+(y-\frac{9} {8})^2=\frac{9} {32}

 

Now in this form the locus is recognisable as a circle, centre \frac{7} {8}+\frac{9} {8}\iota , with radius \frac{3\sqrt{2}} {8}.

What this might look like in the classroom

These loci divide into two groups, the first three are fairly straightforward and should help the students consolidate their previous knowledge of loci, and equations of circles, and definitions of angles and so on.

The final example is much harder. Students would benefit from a slow run through the explanation, perhaps with some pre-printed instructions with some parts missing – so that they can fill in the gaps, or explain what has been done to move from one line to the next.

Students should also be able to work backwards  - when presented with a some sketches of loci they ought to be able to write the equation that would produce that sketch, or perhaps match up equations to loci. The more time spent helping students become familiar with the four loci the easier they will find remembering the four types later in the course.

Taking this mathematics further

Students would do well to consider additional loci – in order to stretch their minds further. Can they picture what would be necessary in order to draw a spiral? Or perhaps they can start to think in three dimensions – a sphere?  A plane?

Making connections

The equation of a circle has been seen before, and students should find little difficulty in linking the circle drawn on the complex plane with their previous knowledge of coordinate geometry. The perpendicular bisector has similar links to coordinate geometry.

The half-line is a newer concept but the students might be drawn into a useful discussion about infinity. Examples of helpful aspects of infinity can be found in the external links section below.
 

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