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 +/- i. 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
_
z + z is non-zero,
which is the same as saying Re z !=
0. For (d), this is the set of all numbers z for which |z| !=
1.
Page 32 #2 (6th edition: page 31 #2)
f(x+iy) = (x+iy)^3 + x + iy + 1 = (x^3 - 3xy^2 + x + 1) + i(3x^2 y - y^3 + y) so u(x,y) = x^3 - 3xy^2 + x + 1 and v(x,y) = 3x^2 y - y^3 + y.
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 r <= 1, 0 <= theta <= pi/2
(b) The sector r <= 1, 0 <= theta <= 3pi/4
(c) The sector r <= 1, 0 <= theta <= pi
Page 32 #6 (6th edition: page 32 #12)
The line segment AB (the part of the line y=-x as x runs from 0 to 1) is mapped onto the line segment A'B' (the part of the line u=0 as v runs from 0 to -2) because u = x^2-y^2 = x^2-(-x)^2 = 0 (proving the image is part of the line u=0) and v = 2xy = -2x^2 which runs from 0 to -2 as x runs from 0 to -1.
The line segment BD (the part of the line x=1 as y runs from -1 to 1) is mapped onto the parabolic arc B'D' (the part of the parabola u = 1 - v^2/4 as v runs from -2 to 2) because v = 2xy = 2y (so y=v/2) and u = x^2 - y^2 = 1 - y^2 = 1 - v^2/4. This proves that the image is part of the parabola u = 1 - v^2/4, and as y runs from -1 to 1, v=2y runs from -2 to 2, proving that the image is the B' to D' part of the parabola.
(The question also asked you to verify that C maps to C'; this is easy to see because C is the point (x,y)=(1,0) and f(C) = (1+0i)^2 = 1+0i so f(C) is the point (u,v)=(1,0), i.e., C'.)
Finally, the line segment AD (the part of the line y=x as x runs from 0 to 1) is mapped onto the line segment A'D' (the part of the line u=0 as v runs from 0 to 2) because u = x^2-y^2 = x^2-x^2 = 0 (proving the image is part of the line u=0) and v = 2xy = 2x^2 which runs from 0 to 2 as x runs from 0 to 1.
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)
(a) The vector iz has the same length as the vector z but has been rotated by an angle of 90 degrees counterclockwise. In our graphical representation, we would draw at the point z = (x,y) the vector iz = (-y,x) which has the same length as the vector from the origin to (x,y) but has been rotated by 90 degrees counterclockwise.
Picture available in graphical version only
(b) The vector z/|z| has the same direction as the vector z but magnitude 1. In our graphical representation, we would draw at the point z=(x,y) the vector z/|z| = (-y,x)/sqrt(x^2+y^2) which has the same direction as the vector from the origin to (x,y), but has length 1.
Picture available in graphical version only
Page 42 #3 (6th edition: #2)
Recall the basic theorems on limits of sums, products, quotients, compositions, and that
(*) lim z = w and (**) lim c = c (c a constant).
z->w z->w
From these we can prove by induction that
n n
(***) lim z = w
z->w
for all integers n (this was done in class).
From this, the theorem on the limits of quotients, and the assumption z0 != 0, it follows that the limit in part (a) equals 1/z0^n.
(b) From (***) it follows that the limit of z^3 = i^3.
From this, (**) applied when c=i, and the theorem on the limit of products, it follows that the limit of iz^3 = i i^3 = i^4 = 1.
From this, (**) applied when c=-1, and the theorem on the limit of sums, it follows that the limit of iz^3 - 1 = 1 - 1 = 0.
From (*), (**) applied when c=i, and the theorem on the limit of sums, it follows that the limit of z+i = i+i = 2i.
This is non-zero, so we can apply the theorem on the limit of quotients to conclude that the limit of (iz^3 - i)/(z+i) = 0/2i = 0.
(c) From (**), (***), and the theorem on products, it follows that for any monomial cz^n we have the limit (as z goes to z0) of c z^n is c z0^n.
It now follows, using the theorem on the limit of sums and mathematical induction (you need not write out the details of the induction proof), that
lim a0 + a1 z + ... + an z^n = a0 + a1 z0 + ... + an z0^n
z->z0
so
lim P(z) = P(z0) and lim Q(z) = Q(z0).
z->z0 z->z0
Since Q(z0) is non-zero, we can apply the theorem on the limits of quotients to conclude that
lim P(z)/Q(z) = P(z0)/Q(z0).
z->z0
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)
We need to show that, for any neighbourhood of 0, there is a punctured neighbourhood of z0 which is mapped into it by the function f(z)g(z).
In other words: we need to show that, for any number e > 0, there is a number d with the property that |f(z)g(z)-0| < e whenever 0 < |z-z0| < d.
We know there is a neighbourhood of z0 on which |g(z)| <= M, so, as long as d is smaller than the radius r of this neighbourhood, |g(z)| <= M whenever 0 < |z-z0| < d.
We also know that the limit of f(z) as z -> z0 is 0. This means that, for any positive number, it is possible to make |f(z)| smaller than that number by taking 0 < |z-z0| < d with d small enough.
Because the preceding statement is true for any positive number it is, in particular, true for e/M. This means we can find a value of d with the property that |f(z)| < e/M whenever 0 < |z-z0| < d. Let s stand for this value of d.
We now know that, as long as d < r, |g(z)| <= M whenever 0 < |z-z0| < d; and, as long as d < s, |f(z)| <= e/M whenever 0 < |z-z0| < d.
Therefore, as long as d is less than the smaller of r and s, both inequalities |g(z)| <= M and |f(z)| <= e/M will be true whenever 0 < |z-z0| < d, and therefore
|f(z)g(z)-0| = |f(z)| |g(z) < (e/M)(M) = e whenever 0 < |z-z0| < d,which is exactly what we needed to show.
Note: you cannot assume that the limit of g(z) as z -> 0 exists. If it did exist (say it equalled L), the proof would be very easy: just appeal to the theorem on limits of products to say that the limit of f(z)g(z) equals (0)(L)=0. However, this exercise shows that even if
lim g(z)
z->z0
does not exist, the limit of f(z)g(z)
still exists and equals zero, provided g(z) is bounded and the limit of
f(z) is zero.
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:
1
(I) lim f(z) = infinity if and only if lim ---- = 0
z->w z->w f(z)
(II) lim f(z) = L if and only if lim f(1/z) = L
z->infinity z->0
1
(III) lim f(z) = infinity if and only if lim ------ = 0
z->infinity z->0 f(1/z)
(a) Because of property (II) this follows by showing
2
4(1/z)
lim ---------- = 4.
z->0 2
(1/z - 1)
This in turn follows from
2 2 2
4(1/z) 4(1/z) z
lim ---------- = lim --------------
z->0 2 z->0 2 2
(1/z - 1) (1/z - 1) z
4
= lim ------
z->0 2
(1-z)
= 4
using part (c) of exercise 3 (exercise 2 in the 6th edition).
(b) Because of property (I) this follows by showing
1
lim ---------- = 0.
z->1 1
------
3
(z-1)
This in turn follows from
1 3 3
lim ---------- = lim (z-1) = 0 = 0.
z->1 1 z->1
------
3
(z-1)
(c) Because of property (III) this follows by showing
1
lim --------------- = 0.
z->0 2
(1/z) + 1
----------
(1/z) - 1
This in turn follows from
1 (1/z) - 1
lim --------------- = lim ----------
z->0 2 z->0 2
(1/z) + 1 (1/z) + 1
----------
(1/z) - 1
2
(1/z) - 1 z
= lim ---------- --
z->0 2 2
(1/z) + 1 z
2
z - z
= lim ------
z->0 2
1 + z
0
= - = 0.
1
Page 43 #9 (6th edition: #10)
(a) Because of property (III) this follows from
1 1
lim ------ = lim --------------
z->0 T(1/z) z->0 a(1/z) + b
----------
d
d
= lim -------
z->0 a/z + b
d z
= lim ------- -
z->0 a/z + b z
dz
= lim ------
z->0 a + bz
0
= ----- = 0
a + 0
where the last line follows from the limits on quotients together with the fact that, when c=0, the assumption ad - bc != 0 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
a(1/z) + b
lim T(1/z) = lim --------------
z->0 z->0 c(1/z) + d
a/z + b z
= lim ------- -
z->0 c/z + d z
a + bz
= lim ------
z->0 c + dz
a + 0 a
= ----- = -
c + 0 c
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
1 cz + d
lim ---- = lim ------
z->-d/c T(z) z->-d/c az + b
c(-d/c) + d
= -----------
a(-d/c) + b
0
= -----------
-(ad)/c + b
= 0
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 |z| <= M for every z in S.
This is true if and only if, for each positive real number M, there is at least one z in S for which |z| > M.
Letting e denote 1/M, we see that this is true if and only if, for each positive real number e, there is at least one z in S for which |z| > 1/e.
This is the same as saying that, for each e-neighbourhood of infinity, there is at least one z in S 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.