Homework #6 Solutions
Math 128, Fall 2013
Instructor: Dr. Doreen De Leon
1
HW #6(a)
1. p. 109: 14: Show that
(a) cos(iz) = cos(iz) for all z
cos(iz) =
=
=
=
=
ei(iz) + e−i(iz)
2
e−z + ez
2
e−(x+iy) + ex+iy
2
e−x e−iy + ex eiy
2
e−x
ex
(cos y − i sin y) + (cos y + i sin y).
2
2
So,
cos(iz) =
=
=
=
=
And, cos(iz) =
=
e−x
ex
(cos y + i sin y) + (cos y − i sin y)
2
2
−x
iy
x
−iy
e e
e e
+
2
2
−x+iy
x−iy
e
+e
2
e−(x−iy) + ex−iy
2
e−z + ez
.
2
ei(iz) + e−i(iz)
2
e−z + ez
= cos(iz). X
2
1
(b) sin(iz) = sin(iz) if and only if z = nπi (n = 0, ±1, ±2, . . . )
ei(iz) − e−i(iz)
2i
e−z − ez
=
2i
e−(x+iy) − ex+iy
=
2i
e−x e−iy − ex eiy
=
2i
e−x
ex
=
(cos y − i sin y) − (cos y + i sin y)
2i
2i
e−x
ex
=−
(sin y + i cos y) + (− sin y + i cos y).
2
2
sin(iz) =
So,
ex
e−x
(sin y − i cos y) − (sin y + i cos y)
2
2
ex
e−x
(cos y − i sin y) −
(cos y + i sin y)
2i
2
ex e−iy
ex eiy
−
2i
2i
ex−iy − e−x+iy
2i
ex−iy − e−(x−iy)
2i
ez − e−z
.
2i
sin(iz) = −
=
=
=
=
=
And,
ei(iz) − e−i(iz)
2i
e−z − ez
=
.
2i
sin(iz) =
We want
e−z − ez
ez − e−z
=
2i
2i
z
−z
=⇒ 2e − 2e = 0
=⇒ ez − e−z = 0
2
Multiply both sides by ez .
e2z − 1 = 0
e2z = 1
e2z = ei(0)
e2(x−iy) = ei(0)
e2x e−2y = ei(0)
2x = ln 1 and − 2y = 0 + 2nπ, n ∈ Z
2x = 0 and y = −nπ = nπ, n ∈ Z
z = nπi, n ∈ Z. X
2. Find all solutions of cos z = 2.
Solution:
eiz + e−iz
=2
2
eiz + e−iz = 4
Multiply by eiz .
ei(2z) + 1 = 4eiz
( iz )2
−4=0
e
iz
So, eiz = 2 ±
√
√
(−4)2 − 4(1)(1)
2
√
4 ± 12
=
2√
4±2 3
=
2√
= 2 ± 3.
e =
4±
3. Now, we have that eiz = ei(x+iy) = eix−y = e−y eix .
√
3 in exponential form. We observe that
√
√
(2 + 3)(2 − 3) = 1
√
1
√ .
2− 3=
2+ 3
√
Therefore, we only need to write 2 + 3 in exponential form. So,
√
√
√
|2 + 3| = 2 + 3 and Arg (2 + 3) = 0.
We need to write 2 ±
We thus have
√
√
e−y eix = (2 + 3)ei(0) =⇒ e−y = 2 + 3, x = 0 + 2nπ, n ∈ Z.
√
So, y = − ln(2 + 3), and therefore,
√
z = x + iy = 2nπ − i ln(2 + 3), n ∈ Z.
3
The second solution is given by doing the above with 2 −
therefore,
ln(2 −
√
√
3) = − ln(2 + 3),
we obtain
z = 2nπ + i ln(2 +
The solution is then
z = ±i ln(2 +
2
√
√
3, but since 2 − 3 =
√
3).
√
3) + 2nπ, n ∈ Z .
p. 111-112: 7(a), (b); 16
7. Show that
(a) sinh(z + πi) = − sinh z
ez+πi − e−(z+πi)
2i
ez eπi − e−z e−πi
=
2i
−ez + e−z
=
2i
ez − e−z
=−
2i
= − sinh z.
sinh(z + πi) =
(b) cosh(z + πi) = − cosh z
ez+πi + e−z+πi
2
ez eπi + e−z e−πi
=
2
−ez − e−z
=
2
ez + e−z
=−
2
= − cosh z.
cosh(z + πi) =
16. Find all roots of the equation cosh z = −2.
Solution:
ez + e−z
= −2
2
=⇒ ez + e−z = −4.
4
1
√ , and
2+ 3
Multiply both sides by ez .
e2z + 1 = −4ez
(ez )2 + 4ez + 1 = 0
z
So, ez = −2 ±
√
(4)2 − 4
√2
= −2 ± 3.
e =
−4 ±
√
3. Now, we have that ez = ex+iy = ex eiy .
We find the first solution by solving ez = −2 +
exponential form. So,
|−2+
We thus have
ex eiy =
So, x = − ln(2 +
√
√
3| = 2 − 3 =
√
√
3. To do this, we need to write −2 + 3 in
√
1
√ and Arg (−2 + 3) = π.
2+ 3
1
1
√ ei(π) =⇒ ex =
√ , y = π + 2nπ, n ∈ Z.
2+ 3
2+ 3
√
3), and therefore,
√
3) + i(π + 2nπ), n ∈ Z.
√
The second solution is given by doing the above with −2 − 3. Since
√
√
−2 − 3 = (2 + 3)eiπ ,
z = x + iy = − ln(2 +
we obtain
z = ln(2 +
√
3) + i(π + 2nπ), n ∈ Z.
The solution is then
z = ± ln(2 +
3
√
3) + (2n + 1)πi, n ∈ Z .
p. 121: 2, 4
2. Evaluate the following integrals.
)2
∫ 2(
1
(a)
−i
dt
t
1
(
1
−i
t
)2
(
=
) ( )
1
2
−1 −
i
2
t
t
5
So,
∫
1
∫
(b)
π
6
2(
1
−i
t
)2
)
∫ 2( )
1
2
−
1
dt
−
i
dt
2
t
t
1
1
2
1
= − − t − i( 2 ln |t||21 )
t
1
1
= − − 2 − (−1 − 1) − i(2 ln 2 − 2 ln 1)
2
1
= − − 2 ln 2 i
2
1
= − − i ln 4 .
2
∫
2(
dt =
ei(2t) dt
0
∫
π
6
e
i(2t)
0
∫
(c)
∞
π
1 i(2t) 6
dt = e
2i
0
)
1 ( iπ
=
e 3 − ei(0)
2i
)
1 (
π
π
=
cos + i sin − 1
2i ( 3
3
√ )
1
1
3
=
− +i
2i
2
2
√
1
3
=− +
4i
4
√
3 1
=
+ i.
4
4
e−zt dt, Re z > 0
0
∫
∞
e
−zt
∫
0
lim e−bz
b→∞
b
e−zt dt
1 −zt b
= lim − e b→∞
z
0
(
)
1 −bz 1
= lim − e
+
b→∞
z
z
1
1
= − lim e−bz + .
z b→∞
z
= lim e−b(x+iy)
dt = lim
b→∞ 0
b→∞
= lim e−bx e−i(by)
b→∞
= lim e−bx (cos(by) − i sin(by))
b→∞
= 0 since x > 0.
6
So,
∫
∞
e−zt dt =
0
1
.
z
4. According to definition (2), Sec. 38, of definite integrals of complex-valued functions of a real
variable
∫ π
∫ π
∫ π
e(1+i)x dx =
ex cos x dx + i
ex sin x dx.
0
0
0
Evaluate the two integrals on the right here by evaluating the single integral on the left and then
using the real and imaginary parts of the value found.
Solution:
∫ π
π
1
e(1+i)x dx =
e(1+i)x 1
+
i
0
0
)
1 ( (1+i)π
=
e
− e(1+i)0
1+i
)
1 ( π iπ
=
e e −1
1+i
1
=
(−eπ − 1)
1+i
1−i
=
(−eπ − 1)
2
−eπ − 1 eπ + 1
=
+
i.
2
2
So,
]
eπ + 1
e
dx = −
e cos x dx = Re
2
[∫0 π
]
∫0 π
π
e +1
ex sin x dx = Im
e(1+i)x dx =
.
2
0
0
∫
[∫
π
π
(1+i)x
x
4
p. 125-126: 2, 5
2. Let C denote the right-hand half of the circle |z| = 2, in the counterclockwise direction, and note
that two parametric representations of C are
( π
π)
z = z(θ) = 2eiθ − ≤ θ ≤
, and
2
2
√
z = Z(y) = 4 − y 2 + iy (−2 ≤ y ≤ 2).
Verify that Z(y) = z[ϕ(y)], where
(
ϕ(y) = arctan
√
)
y
4 − y2
7
( π
π)
− < arctan t <
.
2
2
Also, show ϕ′ (y) > 0.
Solution:
z[ϕ(y)] = 2 exp[iϕ(y)]
[
(
√
= 2 exp i arctan
{
(
(
)]
y
4 − y2
y
))
√
4 − y2
= 2 cos arctan
(
(
+ i sin arctan
√
y
))}
4 − y2
.
Now, using the right triangle on the next page, we can verify that z[ϕ(y)] = Z(y).
2
y
θ
√
4 − y2
{√
z[ϕ(y)] = 2
=
√
y
4 − y2
+i
2
2
}
4 − y 2 + iy
= Z(y). X
Then,
√
1
(
ϕ′ (y) =
1+
=√
)

y2
4−y 2
4 − y2 + √ y
2
4−y 2
4−
y2



1
4 − y2
> 0 since |y| < 2.
5. Suppose that a function f (z) is analytic at a point z0 = z(t0 ) lying on a smooth arc z = z(t) (a ≤
t ≤ b). Show that if w(t) = f [z(t)], then
w′ (t) = f ′ [z(t)]z ′ (t)
when t = t0 .
8
Solution: Let f (z) = u(x, y) + iv(x, y) and let z(t) = x(t) + iy(t). Then,
w(t) = u(x(t), y(t)) + iv(x(t), y(t))
(
)
dw
∂u dx ∂u dy
∂v dx ∂v dy
=
·
+
·
+i
·
+
·
dt
∂x dt
∂y dt
∂x dt
∂y dt
)
(
)
(
∂u
∂v
dx
∂u
∂v
dy
=
+i
·
+
+i
·
∂x
∂x
dt
∂y
∂y
dt
′
′
= (ux + ivx )x + (uy + ivy )y .
By the Cauchy-Riemann equations, uy = −vx and ux = vy .
=⇒ w′ (t) = (ux + ivx )x′ + (−vx + iux )y ′
= (ux + ivx )x′ + i(ux + ivx )y ′
= (ux + ivx )(x′ + iy ′ )
= f ′ [z(t)]z ′ (t). X
9