Assignment 3 Solutions

The principal argument Arg is defined for all non-zero numbers. Division is defined as long as the denominator is non-zero. The other operations used in these formulas (addition, subtraction, multiplication, conjugation, and the modulus function) are defined for all complex numbers.

Therefore, the domain of each function is the set of complex numbers for which all denominators in the formula are non-zero and for which Arg is applied only to non-zero numbers.

For (a), this is all numbers other than . For (b), this is all non-zero numbers (note that 1/z is always non-zero so the restriction on inputs to the Arg function having to be non-zero is automatically satisfied). For (c), this is the set of all numbers z for which which is the same as saying For (d), this is the set of all numbers z for which

Page 32 #2 (6th edition: page 31 #2)

Page 32 #3 (6th edition: page 31 #3)

Just plug the given expressions for x and y into the formula for f(z) and simplify (and use the fact that 1/i = -i). It's straightforward algebra so I won't write it all out.

Page 32 #4 (6th edition: page 33 #13)

(a) The sector

(b) The sector

(c) The sector

Page 32 #6 (6th edition: page 32 #12)

This proves the boundary of the indicated region in the x,y plane maps to the boundary of the indicated region in the u,v plane. You could also construct a rigorous argument that points inside the boundary get mapped to points inside, and points outside get mapped to points outside, but it's a bit of a pain to do it precisely using the machinery at our disposal so far, so don't bother.

Page 33 #10 (6th edition: page 33 #14)

Page 42 #3 (6th edition: #2)

Recall the basic theorems on limits of sums, products, quotients, compositions, and that

From these we can prove by induction that

for all integers n (this was done in class).

From this, the theorem on the limits of quotients, and the assumption , it follows that the limit in part (a) equals

The only time when you would write out such full and pedantic justifications of these limit formulas is in an exercise such as this. Normally you would just write down the answer (which you can see by inspection). Other than an in an exercise like this, you would not need to show any of the intermediate steps, except that you do need to show the denominator is non-zero when using the theorem on quotients.

Page 42 #5 (6th edition: #7)

Page 42 #8 (6th edition: #9)

Let me write out the three key properties of limits involving infinity, so I can refer to them in a uniform way without worrying about the fact that they have different equation numbers in different editions of the textbook:

(a) Because of property (II) this follows by showing

This in turn follows from

using part (c) of exercise 3 (exercise 2 in the 6th edition).

(b) Because of property (I) this follows by showing

This in turn follows from

(c) Because of property (III) this follows by showing

This in turn follows from

Page 43 #9 (6th edition: #10)

(a) Because of property (III) this follows from

where the last line follows from the limits on quotients together with the fact that, when c=0, the assumption guarantees that a is non-zero. (Without this assumption the numerator and denominator would both be zero and the result would not hold. In fact, in this case T(z) would be the constant function b/d so its limit would be b/d not infinity.)

(b) Because of property (II) the first statement follows from

where the last line follows from the limits on quotients together with the assumption that c is non-zero.

Because of property (I), the second statement follows from

where the second line follows from part (c) of exercise 3 (exercise 2 in the 6th edition), together with the fact that the denominator equals (-1/c)(ad-bc) which is well-defined and non-zero because of our assumptions that c and ad-bc are each non-zero.

Page 43 #13 (6th edition: #14)

S is unbounded if and only if there is no real number M with the property that for every .

This is true if and only if, for each positive real number M, there is at least one for which .

Letting denote 1/M, we see that this is true if and only if, for each positive real number , there is at least one for which .

This is the same as saying that, for each -neighbourhood of infinity, there is at least one for which z belongs to that neighbourhood.

This is the same as saying that every neigbourhood of infinity contains at least one point of S.

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