 An introduction to complex numbers

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# 1.2 Audio files

The following files accompany the exercise in Section 4.2

Clicking on the link below opens an extract from Section 4.2 of the course (PDF, 1.7 MB) which accompanies the audio clips, also below. Listen to each of them in turn with the extracted pages open (you may like to print them out). Work on the problems at the appropriate places – you'll find the answers at the foot of this page.

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#### Transcript: Track 2

PHIL RIPPON:
In this audio-tape, we'll explain how to use complex numbers and the curly bracket set notation to
describe various subsets of the complex plane Look at Frame 1
In Frame 1 we've shown two subsets of the complex plane -. both of them ate straight lines On the left of the Frame is the vertical straight line through the point z = 1, and that has equation X = 1 On the right of the Frame is the straight line with equation X + 2y = 3, which crosses the axes at the points shown
Below the diagrams, we indicate two ways that these straight line sets can be described using curly b~ackent otation The straight line with equation x = 1 can be described as: the set of complex numbers z = x + iy such that X is equal to 1
Next, we've given the same description, but without referring explicitly to x and y Since xis the real part of z, we have: the set of complex numbers z such that the real part of z is equal to 1
Either description is acceptable Similarly, the straight line on the right can be described using curly bracket notation in two ways
Either: the set of complex numbers z =X + iy such that X + 2y equals 3, or, equivalently, the set of all z such that the real part of z plus twice the imaginary part of z equals 3
For this example, the second version may seem a little more cumbersome, but you'll meet examples later where it's more concise
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PHIL RIPPON:
In Frame 2, we show two more subsets of the complex plane These are both half-planes, and they're indicated by grey shading, or "tone" They consist of all points lying on one side of a straight line. On the left is the half-plane consisting of all points lying to the right of the line x = 1, not including the line itself In curly bracket notation, this is: the set of z such that the real part of z is strictly greater than 1 On the right of Frame 2 is: the set of z such that the imaginary part of z is less than or equal to 2 This consists of all z which lie on or below the line y = 2 This time, the boundary line is included in the set
Now an important difference between these two half-planes is the role played by the boundary line, We've indicated this difference in the Frame by using a broken line in the first case, and an unbroken one in the second Also, when we refer to specific points on the boundary, we indicate them by empty cirdes or solid circles, depending on whether they're excluded or included from the set in question So z = 1 is excluded from the first set, but z = 2i is included in the second set,
We also use the words "open" and "closed" to indicate whether or not the set includes its boundary The half-plane is called open if .the boundary is excluded, and closed if the boundary is included This use of the words "open" and "closed" should remind you of the way we describe open and closed intervals of the real line
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PHIL RIPPON:
Frame 3 shows two more half-planes - the first one open, and the second one closed. In each case, the equation of the boundary line is closely connected with the curly bracket notation for the set. On the left, the line X + 2y = 3 divides the plane into two halves - in the shaded half, x + 2y is greater than 3, whereas in the other half, it's less than 3. For example, the point z = 0 doesn't lie in the shaded set because X + 2y is less than 3 at this point. On the right of the Frame, the equation of the boundary is x - y = -1, and here the point z = 0 does lie in the shaded set because X - y is greater than -1 at that point.
In the box at the bottom of the Frame we've given the general form of an open or closed half-plane. Here a, b and c represent fixed real numbers, with a and b not both zero. The type of inequality determines whether the half-plane is open or closed - strict inequality gives an open half-plane, weak inequality gives a closed half-plane. Notice that, although we've shown the direction of the inequality as "greater than", you can also describe a half-plane with a "less than" inequality For example, in Frame 2 the set of z such that the real part of z is greater than 1 wuld also be described as the set of z such that minus the real part of z is less than -1. If you're asked to sketch a half-plane, all you have to do is to plot the straight lime ax + by = c, using either~a broken line or unkoken line depending on the inequality. Then you choose which side to shade by checking just one pojnt which is not on the line - for example, the point z = 0 if that's not on the lime
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PHIL RIPPON:
The sets in the next four frames are circles and discs Frame 5 shows two circles. The one on the left is centred at 0 and has radius 1. Its equation is x2 + y2 = 1, which can be written in complex notation as lzl2 = 1: that is, lzl = 1 or for shore mod z = 1. Thus the circle is: the set of z such that IzI= 1
The circle on the right is centred at L and has radius lI2. Each point on the circle is at a distance 112 from i; that is lz - il = 112 In curly bracket notation this is: the set of'z such that lz - il = 'lz.
In this case the equation of the circle in its complex form, namely lz - il = 112, is more concise than the (x,y)-version, xz + (y - 1)2 = 114
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PHIL RIPPON:
Frame 6 shows two discs, both centred at z = 0 A disc is just the set of points inside a circle, and it may also include the circle itself. The disc on the left is the set ofz such that lzl is strictly < 1. This consists of all points inside the circle centred at 0 and with radius 1, excluding the circle itself, which is the boundaxy of the disc. Since the boundary is excluded, this is called an open disc. The disc on the light of Frame 6 includes its boundary, the circle izl = 2, so it's a closed disc.
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PHIL RIPPON:
Frame 7 shows two more discs. Here, the centres of the discs are different from z = 0. On the left is the set of points whpse distance to 1 + i is strictly less than 1. Since the distance from z to 1 + i can be written as iz - 1 - il, as in the cloud, thus disc is: the set of z such that lz - 1 - il is strictly less than 1.
It's open because the boundary is excluded: for instance, the points z = 1 and z = i are not in the set. The example on the right of the Frame is: the set of z such that iz - il, the distance from z to i, is less than or equal to 1/2
This is closed because the boundary is included: for instance, the points z = 1/2i and z = 3/2i are both in the set.
The general form of a disc is shown in the box at the bottom of the Frame. Here, a denotes the centre of the disc, and r is its radius The type of inequality, strict or weak, determines whether the disc is open or closed.
Now practise this technique by attempting the problems in Frame 8.
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PHIL RIPPON:
The next four frames show various other sets which are related to circles and discs Frame 9 shows two sets; each one is the outside of a disc. The set on the left is described in curly bracket notation as:
the set of z such that lzl is strictly > 1
notice that the inequality here is "greater than", whereas the inequality for the inside of a disc is "less than" Also the boundary of the set, the circle lzl = 1, is excluded, because the inequality is strict.
The set on the right is the outside of a disc centred at z = 1 + i. This time, the inequality is weak, and so the boundary, the circle fz - 1 - il = 1, is included. Notice that, in these figures, we've indicated z = 0 and z = 1 + i to show that they're the centres of the boundary circles - they're not included in these sets
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PHIL RIPPON:
Frame 10 we introduce a new type of set called an annulus. An annulus is the set which lies between two concentric circles, which form its boundary. On the left, the annulus consists of those points which lie outside a circle of radius 1, but inside a circle of radius 2, with both cucles centred at z = 0. The boundary is excluded, because both inequalities are strict, and so this is an open annulus. On the right, the annulus consists of the points outside a circle of radius 113 and Inside a circle of radius 213, where both circles are centred on z = i. This time, both circles are included in the set, so that it's a closed annulus. Notice that, sometimes, one of the boundary circles may be included and the other excluded: in that case the annulus is neither open nor closed.
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PHIL RIPPON:
Frame 11 shows two examples of punctured discs. A punctured disc is obtained by removing the centre point from a disc. On the diagrams, we've indicated the puncture by drawing a small empty circle - in the curly bracket notation, the missing centre point corresponds to the restriction" 0 strictly < Izl" for the left-hand disc, and "0 strictly < lz + il" for the one on the right.
The problems in Frame 12 give several examples of sets related to circles and discs - try sketching them now.
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PHIL RIPPON:
Frame 13 shows two half-lines we shall call these "rays". In each case, the ray consists of half of a straight line, with the end point missing, indicated by the empty circle On the left we have:
the set of all z such that the principal argument of z (that's Arg Z with a capital A) is
On the right is the same set translated so that 0 moves to 1 + i. The culy bracket notation for the set on the right is the same as the curly bracket notation for the set on the left, except that z is replaced by z-(1+i).
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PHIL RIPPON:
Frame 14 shows examples of sets which we call sectors A sector is a set whose boundary is two rays. The sector on the left consists of all points z whose principal argument lies strictly between and 3x/4. The boundary of this set is excluded and so we call this an open sector. Notice the empty circle at z = 0 to emphasize that this point is excluded.
The sector on the right of Frame 14 consists of all points whose principal argument is less than or equal to x/6 in modulus: that is, the principal argument of z lies between -x/6 and x/6 inclusive. At first sight, you might guess that this set is closed because it includes its boundary, but this isn't so. The boundary point z = 0 is excluded because Arg 0 is not defined, and so this sector is neither open nor closed.
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PHIL RIPPON:
Frame 15 shows two sectors with their vertices at non-zero points. To plot each of these it may help to visualize the set obtained with only k g z inside the curly brackets and then to translate the vertex from z = 0 to the appropriate point. For instance, the left-hand sector can be obtained from the set of z with the modulus of Arg z less than 3x/4 by translating a distance of 1 to the right. For the definition of a general open sector, at the bottom of Frame 15, we need to specify the location of the vertex a of the sector, and the values a and b associated with the two rays which form the boundary of the sector.
Now try the problems in Frame 16.
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PHIL RIPPON:
Frame 17 defines three basic operations on sets, which can be used to generate new sets from old. They're each illustrated using an appropriate Venn diagram. The union of A and B, read as A union B , consists of those points which lie in either A or B, or in both. On the otha hand A intersection B consists of those points which lie in both A and B. Finally the difference "A minus B", consists of those points which lie in A but not in B. Notice that, as shown below, A - B, A n B and B - A are always mutually disjoint sets; no two of them have any points in common, also their union is equal toA U B.
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PHIL RIPPON:
defined by one condition on z, namely Re z > 1 for A, and lz - 11 <= 1 for B Thus A u B is the set of z such that either Re z > 1 or lz - 11 <= 1, or both On the other hand, A n B is the set of z such that both the Re z > 1 and lz-lI<=1. Finally, A-B is the set of z such that Re z >1 but lz - lI is not less than or equal to 1. This condition can be written more concisely as Re z > 1 and lz - 11 > 1, giving the form for A -B shown at the bottom of the Frame. Some examples to illustrate these operations are given in Frame 18 Each of the sets A and B is
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PHIL RIPPON:
Frame 19 describes an important special case of the difference operation. If the fust set is the entire complex plane C, then the difference is the complement of the second set. Thus C - A consists of all points z which do not lie in A. You've already seen examples of complements of discs in Frame 9, and two further examples are shown in this Frame On the left is the complex plane with the single point 1 + i deleted - this is the complement of the single point set [l + i). On the right is the complex plane with the negative real axis and zero deleted.
This set will be important later in the course, and it's called a cut plane. We've given two alternative definitions of these sets at the bottom of'the frame, to indicate that there's often more than one way to describe a given set. Finally, try the problems in Frame 20, and also those which come after the tape You'll find that these problems give you good practice in describing many of the sets that you'll meet later in the couIse
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Click on the link below to open the answers to the audio section (PDF, 0.3 MB).

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