(a)
(1+i)^i = exp[i log(1+i)] = exp[i(ln |1+i| + i arg(1+i))]
= exp[i ln sqrt(2) - arg(1+i)]
= exp[(i ln 2)/2 - (pi)/4 + 2npi]
= exp[-(pi)/4 + 2npi] exp[(i ln 2)/2]
where n is an arbitrary integer.
(b)
(-1)^1/pi = exp[ 1/(pi) log(-1)]
= exp[1/(pi) (ln |-1| + i arg(-1))]
= exp[1/(pi)(0 + i pi + i 2npi)]
= exp[(2n+1)i]
where n is an arbitrary integer.
Page 83 #2
(b) The principal value is
exp[3pi i Log( (e/2)(-1 - sqrt(3)i) )]
= exp[3pi i (ln|(e/2)(-1 - sqrt(3)i)| + i Arg((e/2)(-1 - sqrt(3)i)) ]
= exp[3pi i (1 - i 2pi/3)] = exp(2 pi^2 + 3 pi i)
= exp(2pi^2)exp(3 pi i) = exp(2pi^2)(-1) = - exp(2pi^2).
(c) The principal value is
exp[ 4i Log(1-i) ] = exp[ 4i(ln|1-i| + i Arg(1-i) )]
= exp( 2i ln 2 - 4(-pi/4) ) = exp( pi+ 2 i ln 2)
= e^pi exp(i 2 ln 2)
= e^pi(cos(2 ln 2) + i sin(2 ln 2)).
Page 83 #3
(-1 + sqrt(3)i)^3/2
= exp[3/2 log(-1 + sqrt(3)i)]
= exp[ 3/2 (ln 2 + i (2pi)/3 + i 2n pi)]
= exp[ 3/2 ln 2] exp[i pi(1+3n)]
= +/- exp[3/2] = +/- 2sqrt(2)
(since exp[i pi(1+3n)] = 1 if n is odd, -1 if n is even).
Page 84 #4
(a) (-1 + sqrt(3)i)^(1/2) = [2 exp(i 2pi/3)]^(1/2) = sqrt(2) exp(i pi/3 + i 2npi/2) = sqrt(2) exp(i pi/3) exp(n pi i) = +/- sqrt(2) exp(i pi/3) since exp(n pi i) is 1 if n is even, -1 if n is odd.
Thus [(-1 + sqrt(3)i)^(1/2)]^3 = { (- sqrt(2) exp(i pi/3))^3, (sqrt(2) exp(i pi/3))^3 } = { - 2 sqrt(2) exp(i pi), 2 sqrt(2) exp(i pi) } = { + 2 sqrt(2), - 2 sqrt(2) } = +/- 2 sqrt(2), the same answer as in question 3.
(b) (-1 + sqrt(3)i)^3 = [2 exp(i 2pi/3)]^3 = 8 exp(2pi i) = 8 so [(-1 + sqrt(3)i)^3]^(1/2) = 8^(1/2) = +/- sqrt(8) = +/- 2 sqrt(2), the same answer as in question 3.
Page 84 #6
|z^a| = |exp(a log z)| = |exp[a(ln |z| + i arg z)]| = exp(Re(a ln |z| + i a arg z)] = exp(a ln |z|) (using the fact that |exp(w)| = exp(Re w) for any complex number w).
Now you could simply observe that this equals the usual real-variable power |z|^a (which is the principal value of the complex-variable interpretation of |z|^a). Or, you could write it as exp(a Log |z|) (since Log t = ln t when t is a positive real number), and exp(a Log |z|) is by definition the principal value of |z|^a.
Page 84 #7
i^c = exp(c log i)
= exp[c(ln |i| + i arg i)]
= exp[ c(ln 1 + i (pi)/2 + i 2 n pi)]
= exp[ c i pi (2n + 1/2)]
= exp[(a + bi) i pi (2n + 1/2)]
= exp[-b pi (2n + 1/2)] exp[a i pi(2n + 1/2)].
Therefore, |i^c| = exp(-b pi(2n + 1/2))
= exp(-bpi/2) exp(-2 n b pi).
The only way these moduli can all be have the same value (regardless of the value of n) is if exp(-2 n b pi) always has the same value. When n=0 it has the value exp(0)=1, so that means we need to have exp(-2 n b pi) = exp(0) for all n, which means -2 n b pi must be zero for all n, which means b must be zero.
Therefore, the condition on c in order for |i^c| to have just one single value is that b=0, in other words, that c be real.
Page 91 #1
/ 2 2 / 2
(a) | ( (1/t) - i ) dt = | (1/t^2) - (2i/t) + i^2 dt
/ 1 / 1
/ 2 |2
= | (1/t^2) - (2i/t) - 1 dt = -1/t - 2 i ln t - t |
/ 1 |1
= -1/2 - 2 i ln 2 - 2 + 1 + 0 + 1 = -1/2 - i 2 ln 2.
/ pi/6 i2t 1 i2t | pi/6
(b) | e dt = -- e |
/ 0 2i | 0
which equals (-i/2)(exp(ipi/3)-exp(i0))
= (-i/2)(exp(ipi/3)-1) = (-i/2)(cos(pi/3) + i sin(pi/3) - 1)
= (-i/2)(-1/2 + i sqrt(3)/2) = sqrt(3)/4 + i/4.
/ infinity -zt / T -zt
(c) | e dt = lim | e dt
/ 0 T->infinity / 0
-zt | T -zT
= lim -(1/z) e | = lim (1/z - 1/z e ).
T->infinity | 0 T->infinity
Write z=x+iy; then e^(-zT) = e^(-xT)e^(-iyT). The modulus of this is e^(-xT). If x > 0, this goes to zero as T -> infinity , and hence e^(-zT) -> 0 (because if the modulus of a complex number goes to zero, so does that number). Therefore, in the case Re(z) > 0, we have
/ infinity -zt
| e dt = 1/z - (1/z)(0) = 1/z.
/ 0
If Re(z) <= 0, the limit of e^(-zT) (as T goes to infinity) does not exist. To see this, note that if Re(z) < 0, |e^(-zT)| = e^(-xT) goes to infinity. ("x" and "Re(z)" are just different names for the same thing). So clearly the limit does not exist in this case. In the remaining case, where Re(z)=0, e^(-zT) = e^0 e^(-iyT) which has constant modulus 1 but continues to go round and round the unit circle has T -> infinity , so the limit fails to exist in this case as well.
Therefore, if Re(z) <= 0,
/ infinity -zt
| e dt
/ 0
does not exist.Page 91 #2
If m != n,
(using t for theta)
/2 pi imt -int /2 pi i(m-n)t 1 i(m-n)t |2 pi
| e e dt = | e dt = ------ e |
/0 /0 i(m-n) |0
1 2 pi i (m-n) 0
= ------ ( e - e )
i(m-n)
1
= ------ ( 1 - 1) = 0.
i(m-n)
If m=n,
/2 pi imt -int /2 pi |2 pi | e e dt = | 1 dt = t | = 2 pi - 0 = 2 pi. /0 /0 |0
Page 99 #1
(a) Here z(theta) = 2 e^(itheta) and z'(theta) = 2ie^(itheta). Therefore,
(using t for theta)
/ / pi 2e^(it) + 2 / pi
| f(z) dz = | ----------- 2i e^(it) dt = | (2e^(it)+2) i dt
/C / 0 2e^(it) / 0
|pi |pi
= (2/i e^(it) + 2t) i | = 2e^(it) + 2it |
|0 |0
= 2e^(i pi) + 2i pi - 2e^0 - 0
= -2 + 2i pi - 2 = -4 + 2 i pi.
(b) Again, z(theta) = 2 e^(itheta) and z'(theta) = 2ie^(itheta). Therefore,
(using t for theta)
/ / 2pi 2e^(it) + 2 / 2pi
| f(z) dz = | ----------- 2i e^(it) dt = | (2e^(it)+2) i dt
/C / pi 2e^(it) / pi
|2pi |2pi
= (2/i e^(it) + 2t) i | = 2e^(it) + 2it |
|pi |pi
= 2e^(i 2pi) + 4 pi i - 2e^(i pi) - 2 pi i
= 2 + 4 pi i - (-2) - 2 pi i = 4 + 2 pi i.
(c) We know that the answer must be the sum of the answers in (a) and (b), i.e., (-4 + 2 pi i) + (4 + 2 pi i) = 4 pi i. But we're asked to verify this by direct calculation. Again, z(theta) = 2 e^(itheta) and z'(theta) = 2ie^(itheta). Therefore,
(using t for theta)
/ / 2pi 2e^(it) + 2 / 2pi
| f(z) dz = | ----------- 2i e^(it) dt = | (2e^(it)+2) i dt
/C / 0 2e^(it) / 0
|2pi |2pi
= (2/i e^(it) + 2t) i | = 2e^(it) + 2it |
|0 |0
= 2e^(i 2pi) + 4 i pi - 2 e^0 - 0
= 2 + 4 pi i - 2 = 4 pi i.
Page 99 #2
(a) Here, z(theta) = 1 + e^(itheta) and z'(theta) = ie^(itheta). Therefore,
(using t for theta)
/ / 2pi / 2pi
| f(z) dz = | [(1+e^(it)) - 1] i e^(it) dt = | i e^(2it) dt
/C / pi / pi
|2pi
= i/(2i) e^(2it) | = (1/2) [ e^(4 pi i) - e^(2 pi i) ]
|pi
= (1/2) (1-1)
= 0.
(b) This can be parametrized by x(t)=t, y(t)=0, 0 <= t <= 2. Then z(t) = t and z'(t) = 1, so
/ / 2 2 | 2
| f(z) dz = | (t-1) dt = t /2 - t | = 4/2 - 2 - (0-0) = 0.
/ C / 0 | 0
Page 100 #3
Let C_1, C_2, C_3, C_4 be the legs shown.
y
| C2
i|-----<---+ 1+i
| |
|C3 |C1
| |
--+---->----+- x
0| C4 1
These can be parametrized by:
C_1: x(t) = 1, y(t) = t, 0 <= t <= 1. Here z(t) = 1+ti and z'(t) = i.
C_2: x(t) = t, y(t) = 1, t from 1 down to 0. Here z(t) = t+i and z'(t) = 1.
C_3: x(t) = 0, y(t) = t, t from 1 down to 0. Here z(t) = ti and z'(t) = i.
C_4: x(t) = t, y(t) = 0, 0 <= t <= 1. Here z(t) = t and z'(t) = 1.
/ / / / /
| f(z) dz = | f(z) dz + | f(z) dz + | f(z) dz + | f(z) dz
/C /C1 /C2 /C3 /C4
/1 ____ /0 ___
= | pi exp[pi(1+ti)] i dt + | pi exp[pi(t+i)] dt
/0 /1
/0 __ /1 _
+ | pi exp[pi(ti)] i dt + | pi exp[pi t] dt
/1 /0
/1 /0
= | pi exp(pi - pi t i) i dt + | pi exp(pi t - pi i) dt
/0 /1
/0 /1
+ | pi exp(-pi t i) i dt + | pi exp(pi t) dt
/1 /0
/1 pi - pi t i /1 - pi i pi t
= | pi i e e dt - | pi e e dt
/0 /0
/1 - pi t i /1 pi t
- | pi i e dt + | pi e dt
/0 /0
pi /1 -pi t i -pi i /1 pi t
= (pi i e - pi i) | e dt + pi (- e + 1) | e dt
/0 /0
pi /1 -pi t i /1 pi t
= pi i (e - 1) | e dt + 2 pi | e dt
/0 /0
pi -1 -pi t i |1 pi t |1
= pi i (e - 1) ---- e | + (2 pi/pi) e |
pi i |0 |0
pi pi
= -(e - 1) (-1 - 1) + 2 (e - 1)
pi
= 4(e - 1).
Page 100 #6
C can be parametrized by z(t) = e^(it), 0 <= t <= 2pi. With t in this interval and using the branch of log given in the question, log(e^(it)) = it. (This would not be true if, for instance, we were using the principal branch Log. Then we'd need to let t range over the interval -pi< t <= pi in order to have Log(e^(it)) = it.)
Then z'(t) = ie^(it), and
/ /2 pi it /2 pi -it -t it
| f(z) dz = | exp[ (-1+i)(it) ] i e dt = | e e i e dt
/C /0 /0
/2 pi -t -0 -2 pi -2 pi
= i | e dt = i(e - e ) = i (1 - e ).
/0
Page 100 #11
Recall that if M is a number such that |f(z)| <= M everywhere on C, then | INT f(z) dz| <= ML where L is the length of C.
In our case, L = 3+4+5 = 12 (C is a triangle with side lengths 3, 4, and 5).
Also,
z _ z _ Re z
|e - z| <= |e | + |z| = e + |z|.
The point on C where |z| is greatest is the point on C farthest from the origin, which is -4. Therefore, |z| <= 4 for all z on C.
Also, Re z ranges between -4 and 0 on C, so e^(Re z) ranges between e^(-4) and e^0. e^0 = 1 is the greater of these, so e^(Re z) <= 1 for all z on C.
Therefore, everywhere on C, we have
z _ Re z
|e - z| <= e + |z| <= 1 + 4 = 5,
so
| / |
| | f(z) dz | <= M x L = 5 x 12 = 60.
| /C |