Math 267 : Exercises 1 Solutions
All questions are due Monday January 14th
1. Simplify each number to the form a + ib
(b)
1
3−i
4+i
2−3i
(c)
(12+5i)2
2−4i
(a)
Answer
(b)
1
1 3+i
3+i
3
3−i = 3−i 3+i = 32 +12 = 10
4+i
4+i 2+3i
5
14
2−3i = 2−3i 2+3i = 13 + i 13
(c)
(12+5i)2
2−4i
(a)
=
119+120i
2−4i
1
+ i 10
179
= − 121
10 + i 5
2. Find all complex numbers z with z 6 = 27 i.
Answer
π
Substitute z = r eiθ −→ r6 ei6θ = 27 i = 27 ei 2 .
√
Match r6 = 27, gives r = 3.
π
Match ei6θ = ei 2 , gives 6θ = π2 , 6θ = π2 + 2π, ..., 6θ = π2 + 10π.
√ π
√ i( π + π ) √ i( π + 2π ) √ i( π +π) √ i( π + 4π )
z = √ 3ei 12 ,
3e 12 3 ,
3e 12 3 ,
3e 12
3e 12 3 ,
,
π
5π
and 3ei( 12 + 3 ) .
P+∞
3. Use Euler’s theorem to rewrite k=−∞ sin(kx)
k2 +1 as a sum of complex exponentials.
Answer Note that sin θ =
1
iθ
2i (e
− e−iθ ). Therefore,
+∞
+∞
X
X
sin(kx)
eikx − e−ikx
=
k2 + 1
2i(k 2 + 1)
k=−∞
k=−∞
+∞
X
=
k=−∞
+∞
X
=
k=−∞
+∞
X
=
k=−∞
+∞
X
eikx
e−ikx
−
2i(k 2 + 1)
2i(k 2 + 1)
k=−∞
+∞
X
e
eilx
−
2i(k 2 + 1)
2i((−l)2 + 1)
ikx
l=−∞
+∞
X
eikx
eikx
−
2i(k 2 + 1)
2i(k 2 + 1)
k=−∞
=0
1
letting l = −k
4. Substitute u(x, t) = X(x)T (t) into,
(
∂t u = ∂x2 u − ∂x u
u(0, t) = u(5, t) = 0
Find an ODE for X(x), and another for T (t). Do not solve the ODEs.
Note: you should have boundary conditions for X(x).
Answer
T 0 (t)
X 00 (x) − X 0 (x)
=
T (t)
X(x)
X(x) T 0 (t) = X 00 (x) T (t) − X 0 (x) T (t) ⇒
Both sides must equal a constant −σ. Now, from the constraints:
u(0, t) = X(0)T (t) = 0 ⇒
X(0) = 0
Together:



X 00 − X 0 + σX = 0
X(0) = 0
X(5) = 0,


where σ is an unknown constant. For T (t), we have the ODE:
T 0 (t) + σT (t) = 0.
with no initial condition provided.
5. Find all numbers σ such that,

2
∂x
X(x)

 −σ = X(x)
X(0) = 0


∂x X(1) = 0
has a solutions X(x) 6≡ 0.
Answer
X 00 + σX = 0 has characteristic polynomial r2 + σ = 0.
√
Case 1 : σ < 0, eg: r = ± −σ
√
X(x) = c1 e
−σx
√
+ c2 e−
−σx
0 = X(0) = c1 + c2
√
√
√
√
0 = X 0 (1) = c1 −σe −σ − c2 −σe− −σ
)
⇒
c1 = −c2 ,
√
√
√
c1 −σ e −σ + e− −σ = 0
This means, c1 = c2 = 0, so the only solution is X(x) = 0.
Case 2 : σ = 0, eg: r = 0 twice
2
X(x) = c1 + c2 x
0 = X(0) = c1 , and 0 = X 0 (1) = c2 , so the only solution is X(x) = 0.
√
Case 3 : σ > 0, eg: r = ±i σ
√
√
X(x) = c1 cos( σx) + c2 sin( σx)
)
0 = X(0) = c1
√
⇒ X(x) = 0 or cos( σ) = 0
√
√
0
0 = X (1) = 0 + c2 σ cos( σ)
√
This means σ = π2 + kπ, where k is an integer.
2
2
2
σ = π 2 12 , π 2 32 , π 2 52 , . . .
6. Fully solve:









∂t2 u = 16∂x2 u
u(0, t) = u(π, t) = 0
u(x, 0) = 0
∂t u(x, 0) = cos(x) sin(x)
Use the general solution formula from lecture.
Answer
The general solution (for the given wave equation and the boundary conditions) is
u(x, t) =
+∞
X
sin(kx) (αk cos(4kt) + βk sin(4kt))
k=1
Now, use the initial conditions. From
0 = u(x, 0) =
∞
X
αk sin(kx)
k=1
we see αk = 0 for k = 1, 2, 3, · · · .. To use ∂t u(x, 0) = cos(x) sin(x) first
notice that sin(2x) = 2 sin x cos x. So, we get
∞
X
1
sin(2x) = ut (x, 0) =
βk 4k sin(kx)
2
k=1
Therefore, for k = 2 8β2 = 21 , so β2 =
1
16 ,
and for k 6= 2, bk = 0.
Finally, the solution is
u(x, t) =
1
sin(2x) sin(8t).
16
Alternative (but, equivalent) solution:
3
Recognize constants c0 = 4 and L = π. General solution is then:
u(x, t) =
+∞
X
eikx − e−ikx
αk ei4kt + βk e−i4kt
k=1
Using first piece of data:
0 = u(x, 0) =
+∞
X
eikx − e−ikx (αk + βk )
k=1
That means αk + βk = 0, for every k.
Using second piece of data:
cos(x) sin(x) =
cos(x) sin(x) =
e
ix
+e
2
−ix
ix
e
−ix
−e
2i
=
1 i2x
e − e−i2x =
4i
+∞
X
k=1
+∞
X
k=1
+∞
X
k=1
+∞
X
eikx − e−ikx (αk (i4k) + βk (−i4k))
eikx − e−ikx (i4k) (αk − βk )
eikx − e−ikx (i4k) (αk − βk )
eikx − e−ikx (i4k) (αk − βk )
k=1
Note, the term on the left shows up when k = 2.
That means:
1
4i
= (i 4 · 2) (α2 − β2 ), and 0 = (i4k) (αk − βk ) for all k 6= 2.


1


αk = −βk for all k
α2 = −




64


1
⇒ β =+ 1
α2 = β2 −
2


32


64




αk = +βk for all k 6= 2
αk = βk = 0 for all k 6= 2
u(x, t) = ei2x − e−i2x
4
1
1 −i4·2t
− ei4·2t +
e
64
64