PHYS20672 Complex Variables and Vector Spaces:
Solutions 1
1. .
a) −(7/10) + (12/5)i; r = 5/2,
θ = 2 arctan(4/3) = 0.59π
√
b) −(8/25) + (31/25)i; r = 41/5,
θ = arctan(−31/8)
= 0.58π
√
c) 3/2 + (3 3/2)i; r = 3,
θ =√
5π/3 ≡ −π/3
d) e/2 + √
( 3e/2)i; √
r = e, θ = π/3
e) 12 (3 − 3i); r = 3, θ = 11π/6 ≡ −π/6
3
æ
a
ò
2
à
-3
-2
d
b1
-1
1
-1
2
ô
e
ì
c
3
-2
-3
p
2. In terms of x and y, we have a) (x−1)2 +y 2 = 4; b) y = x+1 for x ≥ 0; c) x = ± y 2 + 3,
d) x = − ln(cos y) for −π/2 < y < π/2. [Had (c) been for x > 0, only the RH branch
would have appeared, symmetric under reflection in the x-axis.]
Abs@z-1D=2
5
2
2
1
-4
1
-1
0.5
2
2
3
-2
1
2
3
-2
2
-1
4
-2
4
Re@ãz D=1
1.0
3
3
1
1.5
4
4
1
-1
Re@z2 D=3, y>0
arg@z-iD=А4
-0.5
1
2
3
4
-1.0
-1.5
3. In (b) the boundaries are part of the region. For (d), we have
p
p
(x − 1)2 − y 2 < (x + 1)2 − y 2 ⇒ (x − 1)2 < (x + 1)2 ⇒ 0 < x.
So the region is Re z > 0. More intuitively, we are looking for all points that are closer
to 1 + 0i than to −1 + 0i, which is clearly all points to the right of the y-axis!
Re@zDr-3
1<Èz-1-äÈ£2
ÈArg@z-äDÈ<þ4
Èz-1È<Èz+1È
6
4
4
3
4
3
3
2
2
2
2
1
0
1
1
0
0
-2
0
-1
-1
-4
-1
-2
-2
-6
-4 -3 -2 -1
-2
-3
0
1
2
-2 -1
0
1
2
3
4
-3
-1
0
1
2
3
4
-1
0
1
2
3
4. Since a complex number z can be represented by the vector (x, y) in the plane, the
inequality will hold for two complex numbers z1 and z2 if it holds for the corresponding
vectors. When two vectors add the magnitude of the resultant vector will depend on the
angle between them, but it cannot be greater than the sum of the two magnitudes (which
is the result when they are parallel) or less than the magnitude of the longer minus that
of the shorter (antiparallel). Hence the inequality holds for vectors, as required.
Maximum
Arbitrary angle
Minimum
a) In this case the two complex mumbers are z1 = R2 ei2θ and z2 = 1, so |z1 | = R2 and
|z2 | = 1 respectively.
b) Note that for any two complex numbers z1 and z2 , |z1 /z2 | = |z1 |/|z2 |. The maximum
value of the ratio will come from the maximum value of the numerator and the minimum
value of the denominator.
5. a) ±21/4 eiπ/8 = ±(1.10 + 0.46i)
b) 21 ln 2 + iπ/4 = 0.347 + 0.785i
√
c) cos(π/4 + i) = cos(π/4) cos i − sin(π/4) sin i = (1/ 2)(cosh 1 − i sinh 1) = 1.09 − 0.83i
d) If z = arcsin i, then i = sin z = −i sinh(iz). Hence sinh(iz) = −1 and z = i arcsinh 1 =
0.881i is a solution. There are many possible answers because the identities sin z =
sin(π − z) and sin z = sin(z + 2πk) still hold for complex z.
6. a) sinh(iz) = 12 (eiz − e−iz ) = i sin z
b) sin(iz) = 2i1 (e−z − ez ) = i sinh z
c) Let iw = arcsin(iz). Then z = −i sin(iw) = sinh w and so w = arcsinh z. Hence
arcsin(iz) = i arcsinh z.
√
d) Let w = arcsinh z. Then √
2z = ew − e−w ⇒ e2w − 2zew − 1 = 0. Hence ew = z ± 1 + z 2
and w = arcsinh z = ln z + 1 + z 2 . (This is another example where there is more than
one answer: however, taking the positive root and the principal value of the logarithm
gives the branch for which arcsinh of a real number is real.)
e) Let w = arctanh z. Then z =(ew − e−w )/(ew + e−w ) ⇒ e2w = (1 + z)/(1 − z). Hence
arctanh z = 21 ln (1 + z)/(1 − z) .
cos2 z + sin2 z =
1
4
(eiz + e−iz )2 − (eiz − e−iz )2 = eiz e−iz = 1 whether z is real or complex.
7. a) The domain of 1/(z 2 + 1) is the whole complex plane excluding z = ±i
b) The domain of z/(z + z) is the complex plane excluding the imaginary axis x = 0
(where the denominator vanishes). c) The domain of 1/(|z|2 − 1) is the complex plane
excluding the points for which |z| = 1. d) The domain of Ln(z) is the complex plane
excluding the origin, and with the restriction −π < θ ≤ π.) In (a) and (d) the domain is
open and connected.
8. In the plots below, the lines change colour as θ increases to help trace the path. For
f (z) = z 3 + 5z 2 + 2 the total phase change as θ increases from 0 to 2π is ∆φ = 4π and
so there must be two zeros inside the unit circle. (The roots are actually at −5.077 and
0.039 ± 0.626i). For f (z) = z 3 + 5z 2 + 8 the path in the w plane doesn’t encircle the origin
at all, so ∆φ = 0 and there are no zeros inside the unit circle. (The roots are at −5.286
and 0.143 ± 1.222i).
z3 +5z2 +2 vs Θ for z=ãäΘ
6
6
12
æ
4
æ -2
2
4
æ
6
8
10
æ
12
æ
14
1
2
3
4
5
6
1
-0.2
2
3
4
5
6
-0.4
-4
æ
æ
0.2
6
-2
2
-4
0.4
2 æ 4
4
-2
0.6
æ
6
æ
8
0.8
æ
2
8
æ
Arg@z3 +5z2 +8D vs Θ for z=ãäΘ
æ
4
10
2
-6
z3 +5z2 +8 vs Θ for z=ãäΘ
Arg@z3 +5z2 +2D vs Θ for z=ãäΘ
æ
-0.6
æ
æ
-6
9. Warning: The question and solution don’t match the corrected version of Handout 1,
where lines of constant x and y are shown in the complex w plane. Here we’re looking
for lines of constant u and v in the z plane, and these are not illustrated in the current
version of the handout!
w = z 2 : u = x2 − y 2 and v p
= 2xy. So in the z plane, lines of constant u = a are
hyperbolae of the form x = ± a + y 2 and lines of constant v = b are hyperbolae of the
form y = b/(2x).
w = Ln z: u = 21 ln(x2 + y 2 ) and v = arctan(y/x). Then in the z plane, curves of constant
u = a are circles of radius e2a and lines of constant v = b (where −π < b ≤ π) are straight
lines which make an angle tan b with the real axis.
The two plots on the left show lines of constant u (blue) and v (red) in the complex z
plane. The heavy lines are u = 1 and v = 1 and the black dot is the point w = 0.
w=z2
w=ln z
w-plane, lines of u,v=n2
1.5
4
1.0
5
2
0.5
0
0.0
0
-0.5
-2
-5
-1.0
-1.5
-1.5 -1.0 -0.5
-4
0.0
0.5
1.0
1.5
-5
0
5
-4
-2
0
2
4