# 4.4 Self-assessment questions and problems

## SAQ 26

Find the distance between the numbers 2 − *i* and 1 + *3i*.

### Answer

## SAQ 26

In each of the following cases, classify the statement as True or False.

(i) The set , where *a*, *b*, *c*, *d* **R**, is a rectangle.

(ii) The set in (i) is a closed rectangle

(iii) The set in (i) is an open rectangle.

(iv) The setis a disc

(v) The set in (iv) is a closed disc.

(vi) The sets *S*_{n}, *n* = 1, 2, 3 …, where

form a nested sequence.

(vii) The sets *S*_{n}, *n* = 1, 2, 3…, where

form a nested sequence.

(viii) The sets *S*_{n}, *n* = 1, 2, 3…, where

form a nested sequence.

(ix) The sets *S*_{n}, *n* = 1, 2, 3…, where

form a nested sequence.

## SAQ 27

Sketch the curves corresponding to the following equations.

(i) |*z* + 1 | = 1.

(ii) |*z* + 3−*i*| = |*z*−2 |.

## SAQ 28

Find the Cartesian equation (that is an equation connecting *x* and *y* directly) of each of the curves in SAQ 27.

### Answer

(i) (*x* + 1)^{2} + *y*^{2} = 1.

(ii) 5*x*−*y* + 3 = 0.

## SAQ 29

If you have ever tried to cut a flower bed by tying the ends of a piece of string to two canes so that the string is slack and then fiddling about with another cane keeping the string taut, as in Figure 33, then you will know that the equation

is the equation of an ellipse. Find the Cartesian form of the equation

### Answer

You have to tread with caution to avoid as much arithmetic as possible. One way is to write

| *z*−1 | = 3−| *z*−2 |

and so

| *z*−1 |^{2} = 9−6 | *z*−2 | + | *z* + 2 |^{2}

giving

36 | *z*−2 |^{2} = ( 9 + | *z*−2 |^{2}−| *z*−1 |^{2})^{2}.

The right-hand side simplifies considerably and (after a few days’ work) you should arrive at

(It is possible to go straight to *x* and *y* in the first equation, but in initial manipulations it is often simpler to work in terms of *z*.)

## SAQ 30

Sketch each of the following sets of points, where *a*, *b,*, *c*, *d* **C**:

### Answer

(i)

(ii)

(iii)

(Note that is the set of all points whose distance from *a* is less than or equal to their distance from *b*.)

## SAQ 31

(i) We sometimes want an algebraic description of a **line segment** in the complex plane, rather than a complete line. One way of doing this is to use a real parameter. Describe the set where *t* is real and *z*_{1} and *z*_{2} are fixed complex numbers. Identify the points corresponding to and 1 respectively.

(ii) Describe the set

### Answer

(i) The set is the set of points on the line segment joining *z*_{1} and *z*_{2}, which is denoted by [*z*_{1}, *z*_{2}] by analogy with an interval on the real line. The points corresponding to correspond to the points: *z*_{2}, the point of trisection of [*z*_{1}, *z*_{2}] nearer to *z*_{2}, the midpoint, and *z*_{1}, respectively.

(ii) Since

is real and . Hence *A* is the line which includes the line segment of part (i).

## SAQ 32

(i) Prove the following inequalities.

(ii) Show that

(iii) Use the formula

together with part (ii) and a variation of (a) of part (i) to prove the triangle inequality:

(iv) By writing , prove that

Prove, similarly, that

and hence that

Interpret this result geometrically.

### Answer

(i) (a)

(b)

(c) Since, for any real numbers *x* and *y*, we have

we can deduce that

and so the result follows.

(d) We have

Also

So

The result follows because moduli are nonnegative.

(ii)For any complex number α,

( see Problem 3(i) of Section 1.4)

Put and note that ; then

(iii)We have

Also

Thus

since both sides are nonnegative.

(iv)From (iii), we have . Thus . Interchanging *z* and *w*, gives and combining the two results gives

The result corresponds to the fact that the difference between the lengths of any two sides of a triangle does not exceed the length of the third side.

## SAQ 33

Show that the function where *d*(*z*,*w*) = |*z*−*w*| is a distance function: that is to say *d* satisfies:

### Answer

(i)

(ii) If *z* = *x* + *iy* and *w* = *u* + *iv* then

Thus *d*(*z, w*) = 0 if and only if *x*= *u* and *y* = *v*; that is if *z* = *w*, *d*(*z*, *w*) > 0 otherwise, because |*z* − *w*| is nonnegative.

(iii) We have

using the result of SAQ 32 (iii), with *z* replaced by *z* − *a* and *w* replaced by *a* − *w*. Thus

(Note that the set **C** and the function *d* with domain **C** × **C** constitutes a *metric space*.)

## SAQ 34

Prove that

### Answer

(i) by part (a) of SAQ 32 (i).

(ii) Since

we know that

and so

from which

## SAQ 35

Find an upper bound for each of ihe following sets.

### Answer

(i) We know that |*z*| = 2 and we want to replace by something larger. This means replacing by something smaller. We can write and use the result of SAQ 32 (iv) to give So

(ii) A similar argument to that in (i) gives

The next problem deals with the Bolzano–Weierstrass Theorem. **Bernard Bolzano** (1781–1848) was professor of the Philosophy of Religion, in Prague. He wrote a treatise entitled *Paradoxes of the Infinite*, but this did not appear until after his death. **Karl Weierstrass** (1815–1897) was, together with Cauchy and Riemann, one of the founders of complex analysis. He led an uneventful life, much of it spent in Berlin where he was the professor of Mathematics. He is remembered both for his work on elliptic functions, and also for his attempts to put analysis on a firm logical foundation. You will come across his name in connection with the Bolzano–Weierstrass Theorem, the Casorati–Weierstrass Theorem, and the Weierstrass *M*-test for uniform convergence.

To pose this problem we require the following definition.

**Definition**

Let *S* be any set, and *z*_{0} any point in C. Then *z*_{0} is called a **cluster point** of *S* if for all > 0 there is *z* *S* such that . (Note that z_{0} need not belong to *S*.)

## SAQ 36

Prove the Bolzano–Weierstrass Theorem: if *S* is a bounded set and contains infinitely many complex numbers then *S* has at least one cluster point.

### Answer

Since *S* is bounded there is a closed disc with centre the origin and radius *K*, say, which contains *S*. Therefore there is a closed square, *S*_{0}, with vertices at ±*K* ±*iK* which encloses *S*. The imaginary axis divides this square into two closed rectangles. At least one of these must contain infinitely many points. Choose this rectangle. If both rectangles contain infinitely many points, choose the right-hand one. The real axis divides this rectangle into two closed squares. At least one of these contains infinitely many points of *S*. Choose this one, or the upper one if they both do. Call this square *S*_{1}. If we continue in this way, dividing each square by a vertical line and then by a horizontal line, we get a nested sequence of closed squares {*S*_{n}} where *S*_{n} has side of length .

Also . By the Nested Rectangles Theorem, there is one and only one point *z*_{0} in every square. This is a cluster point, because given > 0, the disc |*z* − *z*_{0}| < contains the squares *S*_{n} where , that is, (Figure 37). (Notice that *z*_{0} could be at the corner of *S*_{n}: this position giving the maximum size of disc required.) Thus the disc contains an infinite number of points of the set.

Notice that there may be more than one cluster point. The cluster point will be unique if at each stage of the process only one of the two rectangles contains an infinite number of points of *S*.