3F1-solutions-2 - Isaac Castillo Soto

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MSM3F1/4F1: Complex Variable Theory for Physicists
Solutions to Examples Sheet 2
1. If w(z) = u(x, y) + iv(x, y), where z = x + iy, is analytic, the
functions u and v satisfy the Cauchy-Riemann equations, namely
ux = vy, uy = −vx, where the subscripts denote differentiation
with regard to x and y. Derive the equivalent of these equations
for the case where u and v are the functions of r and θ, where
z = reiθ.
Solution. From the definition of the differential we have that,
depending on what the function is the function of,
du = ux dx+ uy dy,(1)
dv = vx dx+ vy dy,(2)
du = ur dr + uθ dθ,(3)
dv = vr dr + vθ dθ,(4)
dr = rx dx+ ry dy,(5)
dθ = θx dx+ θy dy.(6)
Then, we can take (3) and ‘turn it into’ (1), using (5) and (6)
together with r =
√
x2 + y2, θ = arctan(y/x) to calculate the
necessary derivatives:
du = ur dr + uθ dθ = ur (rx dx+ ry dy) + uθ (θx dx+ θy dy)
= (ur rx + uθ θx) dx+ (ur ry + uθ θy) dy
=
(
ur cos θ − uθ
sin θ
r
)
dx+
(
ur sin θ + uθ
cos θ
r
)
dy.
(Alternatively, in this calculation we could’ve simply used the
Chain Rule, but it is always nice to recap where the Chain Rule
comes from.)
Thus, for du we have
du =
(
ur cos θ − uθ
sinθ
r
)
dx+
(
ur sin θ + uθ
cos θ
r
)
dy
1
2
and similarly for dv
dv =
(
vr cos θ − vθ
sinθ
r
)
dx+
(
vr sin θ + vθ
cos θ
r
)
dy.
The last two equations are simply (1) and (2) so that the ex-
pressions in brackets are ux, uy, vx and vy. Substituting these
expressions into the Cauchy-Riemann equations, we obtain
ur cos θ − uθ
sinθ
r
= vr sin θ + vθ
cos θ
r
,
ur sin θ + uθ
cos θ
r
= −
(
vr cos θ − vθ
sinθ
r
)
or, after resolving these equations with respect to ur and uθ, we
arrive at
ur =
vθ
r
, vr = −
uθ
r
.
2. Find an analytic function f(z) = u(x, y) + iv(x, y), z = x + iy, if
its real part is u(x, y) = 2x− x3 + 3xy2 and f(0) = 0.
Solution.
From the Cauchy-Riemann equations we get two equations for
the function v(x, y):
∂v
∂x
= −∂u
∂y
= −6xy, ∂v
∂y
=
∂u
∂x
= 2− 3x2 + 3y2.
After integrating the first of these equations, we arrive at
v = −3x2y + φ(y),
and, after using the second one, obtain
φ′(y) = 2 + 3y2, φ(y) = 2y + y3 + φ0,
where φ0 is a constant of integration. Using that f(0) = 0, i.e.
v(0, 0) = 0, we have φ0 = 0. Thus,
v = −3x2y + 2y + y3,
and hence
f = u+ iv = 2z − z3.
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3. Find all values of
(i) (−2)
√
2,
(ii) 1−i
Solution.
(i)
(−2)
√
2 =
[
elog(−2)
]√2
= e
√
2(ln |−2|+iπ+2πki) = e
√
2(ln 2+iπ+2πki), k = 0,±1,±2, . . .
(ii)
1−i =
(
elog 1
)−i
= 1−i = e(−i)(ln 1+2πki) = e−2πk, k = 0,±1,±2, . . .
4. (i) Find the derivative of f(z) = ez
3+z and find the points with
f ′(z) = 0.
(ii) Solve the equation e2z = 4i.
Solution.
(i) We have f ′(z) = (3z2 + 1)ez
3+z, therefore f ′(z) = 0 when
z = ±i/
√
3.
(ii) The modulus and argument of 4i are 4 and π/2, respectively.
Therefore, z = 12Log 4 + iπ/4 + iπn, where n is an integer.
5. Let z = x+ iy. Show that cos z = cos x cosh y− i sinx sinh y. Find
| cos z|2.
Solution. Put z = x+ iy. We have
2 cos z = eiz + e−iz
= e−y(cosx+ i sinx) + ey(cosx− i sinx)
= (ey + e−y) cos x− i(ey − e−y) sin x.
Recalling the definition of cosh z and sinh z, we find cos z = cos x cosh y−
i sinx sinh y.
Then
| cos z|2 = cos2 x cosh2 y + sin2 x sinh2 y
= cos2 x(cosh2 y − sinh2 y) + (sin2 x+ cos2 x) sinh2 y.
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Since (cos2 x + sin2 x) = 1 and (cosh2 y − sinh2 y) = 1, we find
| cos z|2 = cos2 x+ sinh2 y.
6. Find the value of (
√
3 + i)i/2.
Solution. Note that the argument and modulus of
√
3+ i are 2
and π/6, respectively. Therefore, from the definition (
√
3+ i)i/2 =
e
i
2 (Log 2+iπ/6). Simplifying, we get
(
√
3 + i)i/2 = e−π/12
(
cos
(1
2
Log 2
)
+ i sin
(1
2
Log 2
))
.
7. Find the points in the complex plane where w = Log
(
z − i
z − 4i
)
is
analytic.
Solution. The function z−iz−4i is analytic in C except at z = 4i.
Moreover Log z is analytic and defined in C except for ℜ(z) ≤
0,ℑ(z) = 0. Now,
z − i
z − 4i
=
(z − i)(z̄ + 4i)
|z − 4i|2
=
|z|2 + i(4z − z̄) + 4
|z − 4i|2
.
This expression has the imaginary part zero when ℜ(4z − z̄) = 0,
i.e. ℜ(z) = 0, so z = iy. For such z, note that
z − i
z − 4i
=
y − 1
y − 4
,
and this expression is ≤ 0 when 1 ≤ y < 4. Hence, w is analytic
in C except when ℜ(z) = 0, 1 ≤ ℑ(z) ≤ 4.
8. Consider the mapping w = (1 − i)z + 2. Find the images of the
following points and region.
(i) z = 0, z = −1, z = −2;
(ii) |z + 1| ≤ 1.
Solution. The mapping w = (1 − i)z + 2 is firstly a rotation
through −π/4, secondly an expansion by
√
2 and finally a trans-
lation through 2.
(i) w(0) = 2, w(−1) = 1 + i, w(−2) = 2i;
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(ii) |w − 1− i| ≤
√
2.
9. Find the image of the region {z = x + iy| 0 ≤ x ≤ 1, |y| ≤ π}
under the mapping w = ez.
Solution. To find the image, you parametrize the boundary
and look where the boundary is mapped. For example, you take
{z = x + iy |x = 0, y = t, −π 6 t 6 π} and consider its image
w = ez. By varying t, you see where this part of the boundary
is mapped. Then, you repeat this step with all other boundaries.
The answer to this problem is: the image is an annulus described
by 1 ≤ |w| ≤ e.
10. Find the region(s) where the function f(x+ iy) = |x2−y2|+2i|xy|
is analytic.
Solution. All what the problem requires is opening the moduli
and check the Cauchy-Riemann equations to find that the function
is analytic for 0 < arg(z) < π/4, π < arg(z) < 5π/4 (where
f(z) = z2) and π/2 < arg(z) < 3π/4, 3π/2 < arg(z) < 7π/4
(where f(z) = −z2).
11. Is the function f(x+ iy) =
√
|xy| analytic at x+ iy = 0?
Solution. No. The real and imaginary parts of this function do
satisfy the Cauchy-Riemann equations at x = y = 0 (the imaginary
part is zero everywhere whilst the real part is zero along the axes and
hence has both derivatives equal to zero) but u(x, y) =
√
|xy| is not
differentiable at the origin (approaching the origin along the line y = x
we have u = x as x2 + y2 → 0 whilst for a differentiable u with zero
partial derivatives we would have
u(x, y)−u(0, 0)︸ ︷︷ ︸
=0
= ux︸︷︷︸
=0
(x−0)+ uy︸︷︷︸
=0
(y−0)+o(
√
x2 + y2) as x2+y2 → 0,
i.e. u vanishing faster than linearly as the origin is approached).
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Last LATEXed on: 24:10:2019. Last edited on or after: Thursday, Oct 24, 2019 at 20:18 YDS, Oct 19