(a) Let me summarize the general procedure since I think this is giving some people trouble.
One way to do this problem is to write z in the form x+iy by calculating -2/(1 + sqrt(3) i) = -1 + sqrt(3) i, so x = -1 and y = sqrt(3). Then you can calculate Arctan y/x = -pi/3.
This is not a correct value of the argument; remember that Arctan y/x will either give you a correct value for the argument or else one that's off by pi, depending on the quadrant. In this case we are looking for a second-quadrant angle but Arctan y/x is a fourth-quadrant angle (it's giving you the argument for the complex number 1 - sqrt(3) i, which has the same y/x ratio as our desired number -1 + sqrt(3) i).
So, we need to add or subtract pi to -pi/3 in order to get a correct value of the argument. Either adding or subtracting pi to -pi/3 will give a correct value for the argument. But, since we want the principal argument (in the range -pi < theta <= pi), we add, getting -pi/3 + pi = 2pi/3 as our answer.
A second way to do this problem is to first find that pi/3 is a value for the argument of 1+sqrt(3)i (which you can do either as above or by just drawing a 30-60-90 triangle), then using the fact that arg(-2/(1+sqrt(3) i) = arg(-2) - arg(1+sqrt(3)i) and the fact that pi is a value for arg(-2) to see that pi- pi/3 = 2pi/3 is a value for the argument of -2/(1 + sqrt(3)i). It's in the right range, so it's the principal argument.
(b) By either of the methods above, you get -3pi/4 as the principal argument.
(c) Here it's easiest to use the second method: Arg(sqrt(3)-i) = -pi/6 (get this either by method 1 above or just be drawing the appropriate 30-60-90 triangle), so 6 times this, namely (6)(-pi/6) = -pi, is a value for arg (sqrt(3)-i)^6. It's not the principal argument; adding 2pi brings it in to the right range so the principal argument is pi.
Page 20 #2 (6th edition: Page 17 #2)
(c)
(-1+i)^7 = [sqrt(2) exp(i(3pi)/4)]^7
= (sqrt(2))^7 exp(i(21pi)/4)
= (sqrt(2))^7(cos (21pi)/4 + i sin (21pi)/4)
= (sqrt(2))^7 ( - 1/(sqrt(2)) - i/(sqrt(2)) )
= -8(1+i)
(d)
(1+sqrt(3) i)^(-10) = [2 exp(i(pi)/3)]^(-10)
= 2^(-10) exp(i(-10pi)/3)
= 2^(-10)(cos (-10pi)/3 + i sin (-10pi)/3)
= 2^(-10)( -1/2+ i (sqrt(3))/2)
= 2^(-11)(-1 + sqrt(3) i)
Page 21 #3(c-f) (6th edition: Page 22 #2(a-d))
(c) -1 = exp(i pi) and pi is the principal argument, so the principal root for (-1)^(1/3) is
exp(i pi/3) = cos(pi/3) + i sin(pi/3) = (1 + sqrt(3) i)/2and the other two roots are
exp(i (pi/3 + 2pi/3) ) = exp(ipi) = -1and
exp(i (pi/3 + 4pi/3) ) = exp(i5pi/3) = (1 - sqrt(3) i)/2(d) -16 = 16 exp(i pi) and pi is the principal argument, so the principal root for (-16)^(1/4) is
2exp(i pi/4) = 2cos(pi/4) + 2i sin(pi/4) = sqrt(2)(1+i)and the other three roots are
2exp(i (pi/4 + 2pi/4) ) = 2exp(i3pi/4) = sqrt(2)(-1+i)
2exp(i (pi/4 + 4pi/4) ) = 2exp(i5pi/4) = sqrt(2)(-1-i)and
2exp(i (pi/4 + 6pi/4) ) = 2exp(i7pi/4) = sqrt(2)(1-i)(e) and (f) are done similarly and the answers are in the book, so I won't write them out. The principal root for (e) is sqrt(2) and the principal root for (f) is 1 - sqrt(3)i. This may be worth going over:
-8 - 8sqrt(3)i = 16 exp(-i2pi/3) and -2pi/3 is the principal argument, so the principal fourth root is 2 exp(-i2pi/12) = 2 exp(-ipi/6) = 1 - sqrt(3)i.
Page 21 #5 (6th edition: Page 17 #4)
The equation is equivalent to
i theta 2
| e - 1 | = 4
2
<==> | cos(theta) -1 + i sin(theta)| = 4
2 2
<==> [cos(theta) - 1] + [sin(theta)] = 4
2 2
<==> cos theta - 2 cos theta + 1 + sin theta = 4
<==> - 2 cos theta = 2
<==> cos theta = -1
<==> theta = pi (or pi + 2n pi where n is an integer, but the
question asked for theta between 0 and 2 pi).
This can also be seen geometrically: numbers of the form z = exp(i theta) are those that lie on the unit circle |z|=1, while those that satisfy the equation |z-1|=2 are those on the circle of radius 2 centred at 1. Therefore, the desired number z = exp(i theta) must be on both those circles. The circles intersect only at the complex number -1 (the point (-1,0)), where theta = pi.
Page 21 #8 (6th edition: page 17 #6)
Since arg(z1 z2) = arg z1 + arg z2, we know that Arg z1 + Arg z2 is always a correct value for arg(z1 z2). It's not always the principal argument, but when Re z1 > 0 and Re z2 > 0 we know that Arg z1 and Arg z2 lie between -pi/2 and pi/2, so their sum lies between -pi and pi, so in this case the sum is the principal argument Arg(z1 z2).
Page 22 #12 (6th edition: page 22 #4)
(a) Since a+i = A exp(i alpha) we know that its two square roots are sqrt(A) exp(i alpha/2) and sqrt(A) exp(i ((alpha)/2 + (2pi)/2)) = - sqrt(A) exp(i alpha/2).
(b) Note that a+i lies in the first or second quadrants, so alpha is between 0 and pi, so 0 < alpha/2 < pi/2 and hence cos alpha/2 and sin alpha/2 are both positive. Also note that because a+i = A exp(i alpha) = A cos alpha + i A sin alpha, we have A cos alpha = a.
We can therefore write
sqrt(A) exp (i alpha/2) = sqrt(A) (cos alpha/2 + i sin alpha/2)
2 2
= sqrt(A) sqrt( cos alpha/2) + i sqrt(A) sqrt( sin alpha/2)
1 + cos alpha 1 - cos alpha
= sqrt(A) sqrt(-------------) + i sqrt(A) sqrt(-------------)
2 2
sqrt(A + A cos alpha) sqrt(A - A cos alpha)
= --------------------- + i --------------------
sqrt(2) sqrt(2)
sqrt(A + a) sqrt(A - a)
= ----------- + i -----------
sqrt(2) sqrt(2)
Page 22 #13 (6th edition: page 22 #5)
|z0| = 8 and Arg z0= 3 pi/4, so the principal cube root is 2 exp(i pi/4) = 2 (cos pi/4 + i sin pi/4) = 2(1/sqrt(2) + i/sqrt(2)) = sqrt(2)(1+i). The other roots can be found not only by the technique used in problem 3 (1 in the 6th edition), but also just by multiplying by the given values of omega_3 and omega_3 squared.
Page 22 #14 (6th edition: page 22 #6)
The desired roots are the fourth roots of -4. These are found just as in problem 3c (2a in the 6th edition) to be 1+i, -1+i, -1-i, and 1-i. This means the polynomial factors as [z-(1+i)][z-(-1+i)][z-(-1-i)][z-(1-i)] = (z-1-i)(z+1-i)(z+1+i)(z-1+i). The product of the first and fourth factors is z^2-2z+2 while the product of the second and third factors is z^2+2z+2.
This always happens for polynomials with real coefficients: if z0 is a root, so is its conjugate w0 (I'm writing w0 in place of
__ z0in this text-only version so I can fit things on one line). Therefore, when you factor the polynomial, you can pair up the non-real factors into products of the form (z - z0)(z - w0) = z^2 - (z0+w0)z + z0w0= z^2 - 2Re(z0)z + |z0|^2, proving that any polynomial with real coefficients can be factored into linear (from the real roots) and quadratic (from pairs of non-real complex roots) factors with real coefficients.
Page 22 #15 (6th edition: page 17 #9)
Write z1= A exp(is) and z2= B exp(it). Then
__ is -it i(s-t)
Re(z1 z2) = Re( A e B e ) = Re (AB e ) = AB cos (s-t),
while |z1 z2| = AB. These two are equal if and only if
cos(s-t)=1, which is true if and only if s-t is an integral multiple
of 2pi. (I've written s and t for the book's theta_1 and theta_2
just to avoid a preponderance of subscripts in the notation when converting to
graphics to put this on the web).
Page 22 #16(a) (6th edition: page 17 #10)
As in the derivation of the triangle inequality,
2 2 2 __
|z1 + z2| = |z1| + |z2| + 2 Re(z1 z2)
while
2 2 2
(|z1| + |z2|) = |z1| + |z2| + 2 |z1| |z2|.
These are equal if and only if
__
Re(z1 z2) = |z1| |z2|,
which by the previous exercise happens if and only if arg z1 and
arg z2 differ by an integral multiple of 2pi.
Geometrically, this happens if and only if z1 and z2 are vectors that point in the same direction; in all other cases it is clear that the third side of the triangle they form will be shorter than the sum of the first two sides.
Page 23 #19 (6th edition: page 22 #7)
Just plug in to the first identity of the previous question (using c in place of z and n-1 in place of n, together with the fact that c is not equal to 1 so it makes sense to divide by c-1).
Page 23 #21 (6th edition: page 18 #12)
To simplify typesetting in this text-only version, I will write c2bar in place of
__ c2
If z1 = c1 c2 and z2 = c1 c2bar, then clearly |z2| = |c1| |c2bar| = |c1| |c2| = |z1| so z1 and z2 have the same modulus.
It remains to prove the converse: that if z1 and z2 have the same modulus (say r), then there are numbers c1 and c2 for which z1 = c1 c2 and z2 = c1 c2bar.
There are many ways to find such numbers. One way is to write z1 = r exp(is) and z2 = r exp(it) and let S = (s+t)/2, T=s-S (note that s = S+T and t=S-T). Then define c1 = r exp(i S) and c2 = exp(i T).
With these definitions, c1c2= r exp(i(S+T) = r exp(is) = z1 and c1c2bar= r exp(i(S-T)) = r exp(it) = z2.
Geometrically, c1 is a number with the same modulus as z1 and z2 but whose argument is halfway between the arguments of z1 and z2. Multiplying c1 by c2 means rotating it by an angle T counterclockwise (just enough to rotate the vector c1 until it points in the z1 direction), while multiplying it by c2bar means rotating by angle T clockwise (just enough to make it point in the z2 direction).
Other solutions are possible as well.
Page 25 #1
(a) This is a closed disk (circle together with the points inside the circle) of radius 1 centred at 2-i. There are boundary points (the points on the circle); these are all contained within the set, so the set is closed and is not open. Therefore, it is not a domain. It is, however, connected.
(b) This can be rewritten |z - (-3/2)| > 2. It is the set of points outside the circle of radius 2 centred at -3/2. There are boundary points (the points on the circle); none of these boundary points is included in the set, so the set is not closed and is open. It is also connected. Therefore, it is a domain.
(c) This is the set of points above the line y=1. There are boundary points (the points on the line); none of these boundary points is included in the set, so the set is not closed and is open. It is also connected. Therefore, it is a domain.
(d) This is the line y=1. There are boundary points; in fact, the set of boundary points is the set itself. (To understand why every point on the line is a boundary point, note that if you take any neighbourhood of a point on the line, that neighbourhood will stick out above and below the line so it will contain both points in the set and points not in the set). Since all the boundary points are included in the set, the set is closed and is not open. Therefore, it is not a domain. (However, it is connected).
(e) This is the wedge consisting of all first-quadrant points that lie on or below the line y=x and on or above the x-axis, except for the origin. There are boundary points; points where arg z is 0 or pi/4 are boundary points (and these are included in the set), and the origin is also a boundary point (which is not included in the set). Since the set contains some but not all of its boundary points, it is neither open nor closed. It is therefore not a domain. (It is, however, connected).
(f) This is the set of points which are closer to or the same distance from the origin as they are to 4; in other words, the set of points on or to the left of the line x=2. This set has boundary points (points on the line are boundary points); these are all included in the set, so the set is closed and not open. It is therefore not a domain. (It is, however, connected).
Page 25 #2
See the answers to question 1.
Page 25 #3
(a) is the only bounded set.
Page 25 #5
S is not connected because if you take a point z in the |z|<1 portion of the set and a point w in the point |w-2|<1 portion of the set, you cannot draw a path from z to w without going outside the set.
(However, note that if the point 1 were part of the set, then you could get from z to w by a path that stays inside the set).