Locus Problems With Complex Numbers

Locus Locus Problems With Complex Numbers We start with a definition concept of locus. Then, we will present some basic examples of locus problems, given as regions in the complex plane. A locus is a set of points whose members are determined by a specific rule. This set of points, when marked on a coordinate system, may look like a line, curve or a surface. Loci can be given in equation forms which involve complex numbers. Some Basic Examples Ex 1 Let z be a complex number. Then the set of points which is defined as is an example of a locus. L {( x, y) z 5Re( z) C}, For brevity, we can denote this locus by the equation z 5 Re( z). From now on, we will use equation forma to denote a loci. We need to see, how this set of points appear, when we plot them on a Cartesian plane. To explore this, let z x iy. Then, x iy x iy x y x y x y x 5Re( ) x y 5 x 5 4 (Eq.1) This restricts the values of x. Why and how? 1

Is the locus of Ex 1, really y x? Which one of the following two graphs depicts the locus of z 5 Re( z)? Graph I: y x Graph II: y x, x 0 To answer this question, think of the question raised in the red-comment in (Eq.1): x y 5 x This quantity is The condition in the LHS, demands that x 0. always 0. Why? Graph I admits values of x 0, and Graph II does not. This will help you to choose the correct picture of the locus.

Consider the loci of the following equations. Ex. What is the shape of the locus given by the equation, z 5 Re( z)? z 5 Re( z) x y 5 x. The reader can see that the restriction marked in the (Eq.1), is no longer valid. Therefore, the correct graph of the locus is given by Graph I. From now on, we will use equation form to denote a locus. 3 Ex 3 Sketch the locus given by the equation, z z Re( ) Im( ) 64. Let z x iy. Then 3 Re( z) Im( z) 64 Re( z) Im( z) 64. x y 3 Therefore, the equation yields 3 x y 64. (Eq.) i. Can you notice that (Eq.) limits the domain of y? Why is this? ii. The reason is based on Year 7 or Year 8 Mathematics. Namely, x, being a real number, x 0. Therefore, 3 iii y 64 y 4. iv However, there is no restriction for the values of x, since for any value of x, there is a matching value of y. v When y 4 x 0. 3

vi Now, start the graph, beginning from the point, (0,4). When the magnitude of the variable, x, is increasing then magnitude of the variable, y, has to decrease, so that (Eq. ) continues to be satisfied. vii When x 0, y remains closer to zero. This should happen for a long interval of x since 3 x x, for x 0. much much smaller (viii) Therefore, for a longer interval of x near zero, the change in y is small. Therefore top of the graph should look flat. (ix) Also, when x 8, y 0. (x) When x 8 or x 8, the value of y 0. This means that near x = 8 and x = -8. The graph should look almost vertical. 3 3 (xi) Since x y 64 y 64 x, for values of x with very large magnitude, 1 1 3 3. y ( x ) ( x ) This means that as x gets larger and larger in magnitude y is tending to be a larger and larger negative number. (xii) Moreover, for large values of x, the graph of y 1 3 3 -( x ) -x. This is because when the variable x has a large magnitude, 64 x x. To convince yourself about this, use your calculator to compare the values of 641064 and 1064 ; 64 0000 and 0000. 3 3 3 3 (There was no magic in my choice of the values for the variable, x. I just needed some reasonably large values.) Now we present the sketch of the locus. 4

Can you figure out how one can use the features, (i)-(xii), to produce the graph above, without using a graphing calculator? After you figure it out, you may check this graph using your calculator. Ex 4 Sketch the locus given by the equation, z z. z z z ( z 1) 0 z 0 or z 1 The Null This eqution So, we can factorise Z out. Factor Law resembles an equation such as z0 unit circle t t where tz. So the required locus is the unit circle and its centre. If one does not include the point (0,0) in the locus, the solution is incomplete. The answer requires that you mark the point, (0,0), along with the circle. 5

Ex 5 Sketch the locus given by the equation, Z Z. Let z x iy. Then, z z x y xyi x y. In this side, Im Part =0 and Re Part0 The imaginary part in the RHS (Right Hand Side) is zero. Therefore, the imaginary part of the LHS must also be zero. Hence, we have This gives the equation: xy 0 x 0, or y 0. x y x y. (Eq.*) This value 0 Since x y 0, if x = 0, (Eq. *) gives Similarly, if y = 0, (Eq. *) gives y y y. This is not y or y, but y. x x y x. Then, the condition, x = 0 gives that y y y z a negative number or zero. positive number or zero. 6 0 (0,0).

And, the condition, y = 0, gives that x x x when x is nonnegative x when x is non-positive x x 1, or 0 (0,0), or (1,0) z 0 or 1, x x 1, or 0 (0,0), or ( 1,0) z 0 or 1. Then, the locus of z z is { z 0, z 1, z 1}. The answer requires that you mark these points on the plane. Ex 6 Sketch the locus given by the equation, ( z1) z 1. There is an almost self-harming way to proceed with this. With z x iy, we can write 7

( x iy 1) x iy 1 ( x 1) y ( x 1) iy x y 1 xyi. ( x 1) y 4( x 1) x y 1 4x y To solve this, first, one has to simplify the contents inside the square roots, square both sides, and then solve the resulting equation. You can see how humongous the task is, but there is an easier way. ( z 1) z 1 z 1 ( z 1) ( z 1) z 1 ( z 1)( z 1) 0. z1 Then, ( z1)( z1) ( z1)( z1) z 1 z 1 z 1 0 ( z 1) z 1 z 1 0. At this point, the humble Null Factor Law comes to our rescue: ( z 1) z 1 z 1 ( z 1) 0 0 or z1 z1 Absolute value of a number is zero iff the number is zero. Hence, z1. Accordingly, one of the points in the locus is z 1. To get the other points, we need to solve the second equation, z1 z 1. With z x iy, we get z1 z1 ( x 1) y ( x 1) y ( x 1) y ( x 1) y x x x 0. Finally, the answer requires that you mark the point z 1, and the whole y-axis. 8

y (1,0) x Hyperbolas Hyperbolas can be described by the following principle: In the diagram above, A and B are two given points, P is a variable point on the Cartesian plane, and k is a positive number. Then the locus, AP BP k is a hyperbola provided the number k satisfies a necessary condition. To further investigate this condition on k, first we refer to the triangle inequality (in the geometric form): Length of any side of the triangle must be smaller or equal to the sum of the lengths of the other two sides. Therefore, we can write that AB BP AP, 9

where the equality holds if and only if the point B lies in the straight line segment AP between A and P. Now, we rewrite the inequality in the following form: AB BP AP AP BP AB. Should this condition hold, even if the quantity, AP-BP is positive? To explore this, we also note that we can write another triangle inequality for the same triangle: BAAP BP, where the equality holds if and only if the point A lies in the straight line segment BP between B and P. Then, BA AP BP BP AP AB. Therefore both conditions AP BP AB and BP AP AB must hold. This can be rewritten in a more compact form as AP BP AB. (Eq.3) Therefore, it is clear that one cannot find points P such that AP BP AB. And, if AP BP AB Then, either AP BP AB or AP BP AB. Now, observe that AP BP AB AP AB BP ABP is a straight line. On the other hand, if AP BP AB, then we have AP BP AB BA AP BP BAPis a straight line. 10

Examples The discussion above is applicable to the following locus problems. They could be hyperbolas, depending on quantity in the RHS. a) z 4i z 3 4i 4 b) z 4i z 3 4i 5, c) z 4i z 3 4i 6. Each of these locus problems involve 3 points in the Cartesian plane: P: z x iy A: v 4i B: w3 4i AP z v z 4i BP z3 4i AB vw 4i 34i 5 5 11

Note that the equation, z 4i z 34i 4, can be written as but AP BP 4 AP BP 4, 4 AP BP AB 5. (Eq.3) Therefore, the equation, z 4i z 34i 4 should result in a part of a hyperbola. If the Example (a) is changed to z 4i z 34i 4, then the result would be a complete hyperbola. Moreover, the locus, z 4i z 3 4i p, will be empty for any p 5; for p 5 the locus will be a part of straight line, and for p 5, the locus will be a hyperbola. Detailed Solution: Example (a) To find out the exact locus of z 4i z 34i 4, first we let z x iy, and then write: z 4i z 34i 4 x iy 4i x iy 34i 4. This will lead to 1

xiy 4i xiy 34i 4 x i( y4) ( x3) i( y4) 4 xiy 4i xiy 34i 4 ( x ) ( y 4) ( x 3) ( y 4) Students should learn to come to this step, directly from z4i z34 i. 4. Now, move the second term in the LHS of the equation, ( x) ( y4) ( x3) ( y4) 4, to its RHS, and then square both sides to get: ( x ) ( y 4) y 4 ( x3) ( 4) = 16 + 8 ( x3) ( y 4) ( x3) ( y 4). Now, we cancel out the red terms in the equation above. Next, we isolate the square root term, by obtaining the following equation: ( x) ( x3) - 16 = 8( x3) ( y4) By simplifying the LHS, we get, 10x 1 8 ( x 3) ( y 4). (Eq.4) must be nonnegative since the RHS is nonnegative: x1 10 Now square both sides to get, 10x 1 64( x 3) 64( y 4). 13

Then, by moving the right hand term to the LHS and then simplifying and completing square on x terms, we get, 36 1 x 64( y 4) 144. At this point, by dividing both sides by 144, and then writing the result in the standard form, we get: 1 x ( 4) y 1 x 1 10 3 From(Eq.4) and. The hyperbola exists only when (to see this apply the principle, if a b 1a 1. ) 1 1 1 x x x 1 1 or 1,. These inequalities yield x 3 or x 5. One of these two inequalities should be 1 satisfied together with our early inequality, x. That is, 10 x 1 x 1 and x 1 10 10 10 3 5 3 5 x or x and x and or x Impossible x 5 No points for x 3.. (Eq.5) The overall condition, x 5, gives the RHS of the hyperbola. 14

The following sketch depicts the sketch of the hyperbola, x 1 ( 4) y 3 1. If you provide the sketch above as your answer for the locus, you will loose marks. The following is the correct locus is given in brown in the sketch below. 15

Here is the graph of the solutions for our original complex equation. What would have happened, if we had z 4i z 34i 4? z 4i z 34i 4 ( x ) ( y 4) ( 4 ( x3) ( y 4) ) ( x) ( y4) 168 ( x3) ( y4) ( x3) ( y4) Finally, we obtain ( x) ( x 3) 168 ( x3) ( y4). 10x1 8 ( x3) ( y4). (Eq.6) We could have obtained (Eq. 6), just by observation of the work we did to obtain (Eq. 4), without repeating all the work. Also, the only difference of this equation and (Eq.4) and (Eq. 6) is the presence of the negative sign before the square root in the RHS of the latter equation. Now, accordingly, the inequalities in (Eq. 5) must be modified: x 1 x 1 x 1 10 10 10 x 3 or x 5 and x 3 and or x 5 and x 3 Impossible No points for x 5.. (Eq.7) Hence, this new locus gives the, same hyperbola for x 3, which is LHS of the hyperbola. Consequently, the locus of z 4i z 34i 4 is the complete hyperbola as both inequalities of (Eq.5) and (Eq. 7) are satisfied. 16

Example (b) We now work with the second equation, z 4i z 3 4i 5: Note that this can be written as AP - BP 5. As we have discussed earlier, if AP-BP=AB, this equation must contain points lying somewhere on AB. Since AB = 5 and AP BP AB. we have the following diagram. Can you guess the solution set from the diagram? The solution set must be {( x,4): x 3}. This is because AP > AB and the point, P, lies on AB. Let us see, whether we can reach the same solution, by further analysis, Now, z 4i z 34i 5 xiy 4i xiy 34i 5. This final equation above, is equivalent to the equation: ( x ) ( y 4) ( x 3) ( y 4) 5 Now, move the second term to the left hand side and square both sides of the resulting equation: As a result, we get ( x ) ( y 4) 5 ( x 3) ( y 4) = 5 + 10 ( x 3) ( y 4) ( x 3) ( y 4) 17.

( x ) ( x 3) - 5 = 10 ( x 3) ( y 4). By simplifying we get, 10( x 3) 10 ( x 3) ( y 4). Since the RHS of the equation above is nonnegative, we have x 3. Now, after dividing by 10, and then squaring both sides, we obtain: This leads to the equation: x3 ( x3) ( y 4). ( y 4) 0 y 4 Now, having established that always y 4, let us investigate further whether we can also verify that x 3, employing this kind of analysis. Since y = 4, now the original equation, x iy 4i x iy 34i 5 reduces to x4i 4i x4i 34i x x3 5. For x 3, the condition, x x3 5, further reduces to x ( x3) 5. The left hand side always simplified to 5. This indicates that the original equation, x iy 4i x iy 34i 5, is valid for any x satisfying the constraint x 3, given that y = 4. Note that this solution interval is on the right hand side of the complex number v on the straight line joining w and v. Next, we consider another possible interval of solutions for x. 18

For x 3, the condition, x x3 5, becomes x ( x3) x ( x3) x15 x 3. This indicates that the original equation, xiy 4i xiy 34i 5, is valid for x = 3, in the interval x [,3], but this solution is already covered in the case that x 3. Finally, consider the interval x(, ). Then the condition, x x3 5, reduces to x ( x3) 5 5. This absurdity indicates that the original equation (the second equation) has no solutions for x(, ). Therefore, the complex numbers, z, satisfying the Example (b), lie on the right hand side of the straight line joining w and v, including the complex number v. Thus the locus of Example (b) is given in the following sketch. Here AP-BP =AB, as long as the point P on the line AB, to the right of the point B. Now consider Example (c). Then xiy 4i xiy 34i 6 ( x ) ( y 4) ( x 3) ( y 4) 6 19

Now, move the second term to the LHS and square both sides of the resulting equation to get: So, we get ( x ) ( y 4) 6 ( x3) ( y 4) = 36 + 1 ( x 3) ( y 4) ( x 3) ( y 4) ( x ) ( x 3) - 36 = 1 ( x 3) ( y 4) By simplifying we get, 10x 41 1 ( x 3) ( y 4). Since the RHS is nonnegative we should have: Now square both sides to obtain: x 41. (Eq.8) 4 That is, 10x 41 144( x 3) 144( y 4). 100x 80x 1681 144( x 6x 9) 144( y 4). Now, by moving the first two terms of the LHS to the RHS, and then simplifying and completing square on x terms, and then finally flipping the sides, we get, 1 44 x 144( y 4) 396. Next, by dividing both sides by 396, and then by writing the result in the standard form, we get, 0

1 x ( y 4) 1. 3 11 Therefore each of the term in the LHS should be less than or equal to 1. Thus, from the first term of the LHS, we have: x 1 1 5 x 7. 3 41 Since it is required by (Eq. 8) that x, the original complex number equation does not 4 have any solution, and consequently the graph is empty. Also we may say that since the original equation indicates a hyperbola and its simplified form is an equation an ellipse there are no complex numbers z satisfying the equation, z 4i z 34i 6. Ellipses Examples (A) z 4i z 34i 4 (B) z 4i z 34i 5 (C) z 4i z 3 4i 6 From a well-known definition of an ellipse, it follows that the respective simplified equation of each of the equations must yield a possible equation of an ellipse. 1

Note that the distance between w and v is 5 units. So, triangle inequality imposes that AP BP AB. As we have already mentioned before, since complex numbers can be identified with appropriate points in the plane, this triangular inequality dictates nature of the loci of the examples above. Since, we know that AB = 5, the locus of the relationship, z 4i z 3 4 i p. AP is a null set if p < 5. Again, this is because, any three points on the plane should abide by the triangular inequality. PB Now, let us consider the equation in Example (A). z 4i z 34i 4. Since, this implies that AP BP AB, 5 there are no solutions. That is, the solution set is empty. However, as a mathematical expedition, we carry out further analysis.

First, note that Then, z 4i z 34i 4 xiy 4i xiy 34i 4 x iy 4i xiy 34i x i( y 4) ( x3) i( y 4) Thus, = ( x ) ( y 4) ( x3) ( y 4). ( x ) ( y 4) ( x 3) ( y 4) 4. (Eq.9) Each of the quantities in the LHS of the equation above is nonnegative. Hence, each of them must be less than or equal to 4. We can write this in the form: and and ( x ) ( y 4) 16 ( x 3) ( y 4) 16 ( x ) 16 ( x 3) 16. That is, ( y4) 0 ( y4) 0. and ( x) 16 ( x3) 16. Even though a similar inequality, ( y4) 16 0 y8, can be derived for y, it is not useful at the moment. The first of the two aforementioned inequalities on x, gives that 4 ( x) 46 x. And, then, the second inequality, indicates that: 4 ( x3) 41 x 7. Again, both of these equations are dictates of the initial equation. Therefore, the initial equation has solutions only for 3

1 x. (Eq.10) Now, move the second term of the LHS of (Eq. 9), to the RHS and square both sides of the resulting equation, to obtain: ( x ) ( y 4) 4 ( x3) ( y4) Consequently, we get: = 16-8 ( x3) ( y4) ( x 3) ( y 4) ( x ) ( x 3) - 16 = 8 ( x 3) ( y 4) By simplifying, we arrive at: 10x 1-8 ( x 3) ( y 4). (Eq.11) Since, the LHS is always non-positive, we have 1 x. (Eq.1) 10 Now, square both sides of (Eq. 11), to get, 10x 1 64( x 3) 64( y 4). Next, move the RHS to the LHS of the equation above, then simplify, and complete square on x terms, to obtain, 1 36 x 64( y 4) 144. Dividing both sides by 144, and then writing in the standard form we receive that, 4

From this equation, it is evident that 1 x y 4 3 3 1. This yields the inequalities, 1 x 1 1 1 x 3 and x 3. 3 x 5 and x 7. (Eq.13) For our original complex equation to have solutions, all the inequalities given by (Eq.10), (Eq.1), and (Eq.13) must be obeyed. That is, if explicitly said, all the inequalities, 1 x, x 7, x 1, 10 be obeyed. It is obvious that no values of x can satisfy all of these inequalities. 5 and x, must Therefore, the locus of Example (A) is an empty set. Also, since the original equation indicates an ellipse, and its worked out form is an equation of a hyperbola we may conclude that there are no complex numbers, z satisfying the equation, z 4i z 34i 4. 5

We now work with Example (B). The equation gives x iy 4i x iy 34i 5 AP PB 5 AP PB AB This can be pictured in the following diagram: It is clear that P must lie between A and B. This gives the inequality, - x 3. Therefore, the solution set of Example (B) must be {( x,4): x 3}. (#) For the moment, we forget this, and will conduct all the calculations, to see, how the mathematics is being played out. We start with the equation ( x ) ( y 4) ( x 3) ( y 4) 5 Now, move the second term of square both sides of the resulting equation, to get: 6 ( x ) ( y 4) ( x 3) ( y 4) 5 to the RHS and

( x ) ( y 4) 5 ( x 3) ( y 4) Thus, we get = 5-10 ( x 3) ( y 4) ( x 3) ( y 4) ( x ) ( x 3) - 5 = 10 ( x 3) ( y 4) By simplifying we obtain, 10( x 3) 10 ( x 3) ( y 4). Now, after dividing by 10 and squaring both sides of the resulting equation, we arrive at: This leads to the condition: x 3 ( x 3) ( y 4). ( y4) 0 y 4. This means that all the solutions for Example B, must have just single y value, y= 4. Since y = 4, the equation ( x ) ( y 4) ( x3) ( y 4) 5 indicates that (since each of the terms in the LHS must be less than or equal to 5) ( x ) ( y 4) 5 and ( x 3) ( y 4) 5 ( x ) 5 and ( x 3) 5. y4 ( y4) 0 y4 ( y4) 0 This equation yields two inequalities, 5 x 5 and -5 x3 5, and y 4. Consequently, we conclude that 7

7 x 3 and x8 and y 4 7 x and y 4. x3 and y 4. (Eq. 14) The equation, (Eq.14) gives the only feasible solution for Example B. Just to double check this, let us revisit the very original form of Example (B). Since the condition, y = 4 must be satisfied with any possible solutions for x we start by substituting, as follows: x yi 4i x yi 34i 5 x4i 4i x4i 34i 5 x x3 5. x the equation, x x3 5, can be written as: For 3, x ( x3) x15 x 3. This indicates that Example (B), in the interval of [3, ), has only one solution, x 3. In the interval of[, 3], the equation, x x3 5, becomes x ( x3) x ( x3) 5. This indicates that the original equation, Example (B), is satisfied for all the values of x in the interval, x3 and y 4. See again (#) and (Eq. 14). Finally, consider the interval of (, ). Here, the LHS of the equation, x x3 5, reduces to 8

Then, we have, x ( x3) 5. -5=5. This absurdity indicates that the original equation, Example (B), has no solutions in the interval of (, ). Therefore, the set of complex numbers, z, that satisfy the Example (B) lie on the straight line segment that joins v and w. The complex numbers v and w are also in the locus. This is another verification of the already obtained result; see (Eq. 14). Next, we consider the Example (C): xiy 4i xiy 34i 6. Then, by setting z x yi, we obtain that ( x ) ( y 4) ( x 3) ( y 4) 6. Now, move the second term to the LHS and square both sides of the resulting equation: ( x ) ( y 4) This leads to the equation, 6 ( x3) ( y4) = 36 + 1 ( x3) ( y4) ( ) x 3 y ( 4) ( x ) ( x 3) - 36 = 1 ( x 3 ) ( y 4). By simplifying we get, 10x 41-1 ( x 3) ( y 4). (Eq.15) Since the LHS is always non-positive, we have the condition that 9

41 x. (Eq.16) 10 Now square both sides of (Eq. 15), to get: That is, 10x 41 144( x 3) 144( y 4). 100x 80x 1681 144( x 6x 9) 144( y 4). Now move the first two terms of the LHS of the equation above, to the RHS, simplify, complete square on x terms, and finally flip the sides, to get, 1 44 x 144( y 4) 396. (Eq.17) Since, each one of the two terms in the LHS is nonnegative, none of those terms can be larger than 396. Therefore, we get the inequality: x 1 x 1 1 x 3. 44 396 9 3 This gives the inequality: 5 7 x. (Eq.18) Now, by dividing both sides of (Eq. 17) by 396, and writing the resulting equation in the standard form, we get, 30

x 3 1 ( 4) y 11 1. Eq.19 41 5 7 Now the inequality, x (Eq. 16) is satisfied by the inequality, x, (Eq. 18). 10 This means that the whole of the domain of (Eq. 19), (or equivalently, (Eq.17)), is allowed by (Eq. 16). Thus the locus is the whole ellipse given by (Eq. 19). The ellipse shown below, is the required locus. 11 11 The condition on y, 4 y 4, which can be seen on the diagram above, can be obtained from (Eq. 17), since the largest value possible for the second term in the LHS of (Eq. 17) is 396. You may also refer to the derivation of the inequality in (Eq. 18). 31

In this article, I have repeated some information, even several times, within the same problem. This shouldn t be necessary. However, I adopted this style because it is my experience that many students are unable to remember the first sentence after reading the second sentence, while following a mathematical argument. This defect of many students may be due to playing of video games in which what matters is only the current screen, not anything before or anything after. In contrast, mathematics consists of trains of thoughts. If a link is broken somewhere, the rest and the whole becomes incomprehensible. 3