Download A Solutions to exercises on complex numbers.

A
A.1
Solutions to exercises on complex numbers.
addition and multiplication
1. Do problems 1-4, 11, 12 from appendix G in the book (page A47).
Evaluate the expression and write your answer in the form a + bi.
(1.) (5 − 6i) + (3 + 2i)
Solution. 8 − 4i.
(2.) (4 − 21 i) − (9 + 52 i)
Solution. −5 − 3i.
(3.) (2 + 5i)(4 − i)
Solution. (2 + 5i)(4 − i) = 8 − 2i + 20i − 5i2 = 13 + 18i.
(4.) (1 − 2i)(8 − 3i)
Solution. (1 − 2i)(8 − 3i) = 8 − 3i − 16i + 6i2 = 2 − 19i.
(11.) i3
Solution. i3 = i2 (i) = (−1)i = −i.
(12.) i100
Solution. i100 = i2·50 = (i2 )50 = (−1)50 = 1.
2. Find 2 distinct complex numbers z1 and z2 with the property that zj2 = −1 for
both j = 1 and j = 2.
Solution. i and −i.
3. Find 4 distinct complex numbers z1 , z2 , z3 , and z4 with the property that zj4 = 1
for all j = 1, 2, 3, 4.
Solution. 1, i, −1, and −i.
4. Write the product (
√
3
2
+ 12 i)( 21 +
√
3
i)
2
in the form a + bi.
Solution.
√
!
3 1
+ i
2
2
A.2
√ ! √
√
3
3 3
3 2
1
1
+
i =
+ i+ i+
i = i.
2
2
4
4
4
4
conjugates and multiplicative inverses
1. Do problems 5-10 from appendix G in the book (page A47).
Evaluate the expression and write your answer in the form a + bi.
(5.) 12 + 7i.
Solution. 12 − 7i.
(6.) 2i( 21 − i)
Solution. Since 2i( 21 − i) = i − 2i2 = 2 + i we have 2i( 21 − i) = 2 − i.
1
(7.)
1+4i
3+2i
Solution.
(8.)
1
1+i
Solution.
(10.)
=
1+4i
3+2i
·
3−2i
3−2i
=
3−2i+12i−8i2
32 +22
=
11+10i
13
3+2i
1−4i
=
3+2i
1−4i
·
1+4i
1+4i
=
3+12i+2i+8i2
12 +42
=
−5+12i
17
3+2i
1−4i
Solution.
(9.)
1+4i
3+2i
3
4−3i
Solution.
1
1+i
3
4−3i
=
=
1
1+i
·
3
4−3i
1−i
1−i
·
=
4+3i
4+3i
1−i
12 +12
=
=
12+9i
42 +32
1−i
2
=
=
1
2
12+9i
25
=
11
13
+
10
i.
13
5
= − 17
+
12
i.
17
− 12 i.
12
25
=
+
9
i.
25
2. Find the multiplicative inverse of each of the complex numbers:
√
√
√
2
2
−1 − i, 3 − 2i, i, 2 3 − 2i, −
+
i
2
2
Solution.
• (−1 − i)−1 =
• (3 − 2i)−1 =
1
(−1
(−1−i)(−1+i)
1
(3
(3−2i)(3+2i)
+ i) =
+ 2i) =
1
(−1
12 +12
1
(3
32 +22
+ i) = − 21 + 12 i.
+ 2i) =
1
(−i) = 11 (−i) = −i.
• i−1 = i(−i)
√
√
1 √
• (2 3 − 2i)−1 = (2√3−2i)(2
3 + 2i) =
(2
3+2i)
3
13
+
2
i.
13
√ 1
(2
(2 3)2 +22
√
√
3
2 3
2
+
i
=
+ 18 i.
16
16
8
√
√
√
√
(− 22 + 22 i)−1 = √2 √2 1 √2 √2 (− 22 − 22 i)
(− 2 + 2 i)(− 2 − 2 i)
√
√
2
2
= − 2 − 2 i.
√
3 + 2i)
=
•
=
(
√
1
√
2 2
) +( 22 )2
2
(−
√
2
− 22 i)
2
√
3. Find the multiplicative inverse of bi where b is an arbitrary nonzero real number.
Solution.
(bi)−1 =
1
1
i
1
(bi) =
(−bi) = 2 (−bi) = − .
(bi)(−bi)
b
b
(bi)(bi)
4. Show that z + w = z̄ + w̄ for any complex numbers z and w.
[Hint: Set z = a + bi and w = c + di and compute each side of the equation
separately.]
Solution. Set z = a + bi and w = c + di. Then we have
z + w = (a + bi) + (c + di) = (a + c) − (b + d)i = (a + c) − (b + d)i,
z̄ + w̄ = a + bi + c + di = a − bi + c − di = (a + c) − (b + d)i.
Thus z + w = z̄ + w̄.
2
5. Show that z · w = z̄ · w̄ for any complex numbers z and w.
[Hint: Set z = a + bi and w = c + di and compute each side of the equation
separately.]
Solution. Set z = a + bi and w = c + di. Then we have
z · w = (a + bi)(c + di) = (ac − bd) + (ad + bc)i = (ac − bd) − (ad + bc)i,
z̄·w̄ = (a + bi)(c + di) = (a−bi)(c−di) = ac−adi−bci+bdi2 = (ac−bd)−(ad+bc)i.
Thus z · w = z̄ · w̄.
A.3
the complex plane
1. (Proving the parallelogram rule). Let z = a + bi and w = c + di and assume
that no two of the points 0, z and w lie on a single vertical line.
(a) Find a formula for the slope of the line through 0 and z.
b−0
Answer. a−0
(b) Find a formula for the slope of the line through w and z + w.
Answer.
(b+d)−d
(a+c)−c
(c) Show that the formulas you found in parts (a) and (b) are equal.
Solution. Both are equal to ab .
2. Consider the complex numbers z, w, and v pictured below:
3
Draw a separate picture for each of the following:
(a) Plot the complex numbers z + z̄, w + w̄ and v + v̄ in the complex plane.
Solution.
(b) Plot the complex number 2w + z̄ + v in the complex plane.
Solution.
4
(c) Plot the complex number v − z − w in the complex plane.
Solution.
3. (a) Shade the region of the complex plane consisting of all complex numbers
whose real part is greater than or equal to 1.
Solution.
(b) Shade the region of the complex plane consisting of all complex numbers
whose imaginary part is less than or equal to 4.
Solution.
5
4. Let z = 2 − 2i and w = −1 − 3i.
(a) Carefully plot z and w in the complex plane.
(b) Use the parallelogram rule and your answer from part (a) to plot the
complex number z̄ − w.
(c) Check your answer for part (b) by writing 2 − 2i − (−1 − 3i) in the form
a + bi.
Solution.
z̄ − w = 2 − 2i − (−1 − 3i) = 2 + 2i + 1 + 3i = 3 + 5i.
6
A.4
moduli, limits, and series
1. Assume z 6= 0 is a complex number. Show that z −1 =
z̄
.
|z|2
√
Solution. Let z = a + bi. Then z · z̄ = a2 + b2 = ( a2 + b2 )2 = |z|2 . Thus
z̄
z −1 = z·z̄
= |z|z̄ 2 .
2. Show that |z · w| = |z| · |w| for any complex numbers z and w.
[Hint: Set z = a + bi and w = c + di and compute both sides of the equation
separately.]
Solution. Set z = a + bi and w = c + di. Then
p
|z · w| = |(a + bi)(c + di)| = |(ac − bd) + (ad + bc)i| = (ac − bd)2 + (ad + bc)2
√
√
= a2 c2 − 2abcd + b2 d2 + a2 d2 + 2abcd + b2 c2 = a2 c2 + b2 d2 + a2 d2 + b2 c2 .
Whereas
p
√
√
|z| · |w| = |a + bi| · |c + di| = a2 + b2 c2 + d2 = (a2 + b2 )(c2 + d2 )
√
= a2 c2 + b2 d2 + a2 d2 + b2 c2 .
3. For each of the given series
(I) Determine whether or not the series is convergent, and if the series is
convergent find its sum (and write your answer in the form a + bi).
(II) Find the first six partial sums (s0 , s1 , s2 , s3 , s4 , s5 ).
(III) Plot the first six partial sums in the complex plane as in example 1 to
verify your answer for part (I).
7
(a)
∞
P
n=0
2i n
3
Solution. The series is geometric with |r| = | 2i3 | =
Moreover, we have
n
∞ X
2i
n=0
Since
3
=
1
1−
2i
3
=
n
∞ X
2i
n=0
3
2
3
< 1 so it converges.
3
3
3 + 2i
9 + 6i
9
6
=
·
= 2
=
+ i.
2
3 − 2i
3 − 2i 3 + 2i
3 +2
13 13
4
8
16
32
2
+
i−···
=1+ i− − i+
3
9 27
81 243
we see the first few partial sums are
s0 = 1
s1 = 1 + 32 i
s2 =
5
9
+ 23 i
s3 =
5
9
+
s4 =
61
81
+
10
i
27
s5 =
61
81
+
122
i
243
10
i
27
Here are is a picture of the first several partial sums plotted in the complex
plane:
8
(b)
∞
P
n=0
− 2i3
n
Solution. The series is geometric with |r| = |− 2i3 | =
Moreover, we have
n
∞ X
1
2i
=
−
3
1+
n=0
Since
2i
3
=
2
3
< 1 so it converges.
3
3
3 − 2i
9 − 6i
9
6
=
·
= 2
=
− i.
2
3 + 2i
3 + 2i 3 − 2i
3 +2
13 13
n
∞ X
4
8
16
32
2
2i
−
i −···
=1− i− + i+
−
3
3
9
27
81
243
n=0
we see the first few partial sums are
s0 = 1
s1 = 1 − 23 i
s2 =
5
9
− 23 i
s3 =
5
9
−
s4 =
61
81
−
10
i
27
s5 =
61
81
−
122
i
243
10
i
27
Here are is a picture of the first several partial sums plotted in the complex
plane:
9
(c)
∞
P
(2i)n
n=0
Solution. The series is geometric with |r| = |2i| = 2 > 1 so it diverges.
Since
∞
X
n=0
(2i)n = 1 + 2i − 4 − 8i + 16 + 32i − 64 − 128i + · · ·
we see the first few partial sums are
s0 = 1
s1 = 1 + 2i
s2 = −3 + 2i
s3 = −3 − 6i
s4 = 13 − 6i
s5 = 13 + 26i
Here are is a picture of the first several partial sums plotted in the complex
plane:
4. Consider the series
∞
P
( 21 + 21 i)n .
n=0
(I) Determine whether or not the series is convergent, and if the series is
convergent find its sum (and write your answer in the form a + bi).
(II) Find the first 8 partial sums (s0 , s1 , s2 , s3 , s4 , s5 , s6 , s7 ).
10
(III) Plot the first 8 partial sums in the complex plane as in example 1 to verify
your answer for part (I).
Solution. This a geometric series with
s 2 2 r
1 1 1
1
1
<1
+
=
|r| = + i =
2 2
2
2
2
so it converges. Moreover,
∞ X
1
n=0
n
1
1
=
+ i =
1
2 2
1 − 2 + 12 i
1
2
2
2 1+i
2 + 2i
1
=
·
=
= 1+i.
1 =
1−i
1−i 1+i
2
− 2i
To check that our answer makes sense, let’s look at some of the partial sums:
∞ X
1
n=0
n
2 3 4
1
1 1
1 1
1 1
1 1
+ i = 1+
+ i +
+ i +
+ i +
+ i +···
2 2
2 2
2 2
2 2
2 2
1
2
Since
∞ X
1
n=0
2
+ 12 i = 21 i we have
n
1 1
1
1 1
1
1
1 1
+ − − i
+ i = 1+
+ i +
i + − + i + −
2 2
2 2
2
4 4
4
8 8
1
1
1
− i +···
+ − i +
8
16 16
so the partial sums look like:
s0 = 1
s1 =
3
2
+ 21 i
s2 =
3
2
+i
s3 =
5
4
+ 45 i
s4 = 1 + 45 i
s5 =
7
8
+ 89 i
s6 =
7
8
+i
s7 =
15
16
11
..
.
+
15
i
16
Here’s a picture of the first several partial sums plotted in the complex plane,
illustrating our answer.
5. For each of the following power series find the radius of convergence and draw
a picture of the disk where the series converges as in example 2.
(a)
∞
P
n=0
nxn
.
n+1
n
nx
we have
Solution. Using the ratio test with an = n+1
an+1 (n + 1)xn+1 n + 1 x(n2 + 2n + 1) → |x| as n → ∞
=
·
an = n + 2
nxn n2 + 2
So the series converges when |x| < 1 and diverges when |x| > 1. Thus the
radius of convergence is 1 as pictured below.
(b)
∞
P
n=1
(x−2+i)n
(n+1)3n
[Hint: To see where the power series is centered you need to write x − 2 + i
in the form x − (a + bi).]
12
n
we have
Solution. Using the ratio test with an = (x−2+i)
(n+1)3n
an+1 (x − 2 + i)n+1 (n + 1)3n an = (n + 2)3n+1 · (x − 2 + i)n (x − 2 + i)(n + 1) → |x − 2 + i| as n → ∞
= (n + 2)3
3
So the series converges when |x − 2 + i| = |x − (2 − i)| < 3 and diverges
when |x − 2 + i| = |x − (2 − i)| > 3. Thus the radius of convergence is 3
as pictured below.
(c)
∞
P
n=1
n!
(x
nn
+ 2i)n
Solution. Using the ratio test with an = nn!n (x + 2i)n we have
an+1 (n + 1)!(x + 2i)n+1
nn
=
·
an (n + 1)n+1
n!(x + 2i)n n
nn
n
=
(x + 2i) = |x + 2i|
(n + 1)n
n+1
”
“
n+1
1
n
(
)
ln
n
2
)
(
n
(n+1)
n
−n
n+1
Since lim ln n+1
= lim
= lim
= −1,
= lim n+1
1
− 12
n→∞
n→∞
n→∞
n→∞
n
n
n
n
n
n
= eln( n+1 ) → e−1 as n → ∞. Thus
we see n+1
n
n
|x + 2i|
|x + 2i|
as n → ∞
→
n+1
e
So the series converges when |x + 2i| = |x − (−2i)| < e and diverges when
|x + 2i| = |x − (−2i)| > e. Thus the radius of convergence is e as pictured
below.
13
A.5
exponentials and Euler’s formula
1. Write each of the following in the form a + bi.
(a) e
−2π
i
3
Solution. e
−2π
i
3
π
(b) 12e 6 i
= cos( −2π
) + i sin( −2π
) = − 12 −
3
3
π
Solution. 12e 6 i = 12[cos( π6 ) + i sin( π6 )] = 12[
2π
4π
√
3
2
√
3
i.
2
√
+ 12 i] = 6 3 + 6i.
6π
(c) e 3 i + e 3 i + e 3 i
Solution.
2π
) + i sin( 2π
) = − 12 +
• e 3 i = cos( 2π
3
3
4π
• e 3 i = cos( 4π
) + i sin( 4π
) = − 12 −
3
3
√
3
i,
2
√
3
i,
2
6π
• e 3 i = e2πi = 1.
Thus
e
2π
i
3
2. Show that cos(θ) =
+e
4π
i
3
eiθ +e−iθ
2
+e
6π
i
3
√
√
1
1
3
3
i− −
i + 1 = 0.
=− +
2
2
2
2
and sin(θ) =
eiθ −e−iθ
2i
for all real numbers θ.
[Hint: Use Euler’s formula and the negative angle identities for sine and cosine.]
Solution.
eiθ + e−iθ cos(θ) + i sin(θ) + cos(−θ) + sin(−θ)
=
2
2
cos θ + i sin θ + cos θ − i sin θ
=
2
2 cos θ
=
= cos θ.
2
eiθ − e−iθ cos(θ) + i sin(θ) − [cos(−θ) + sin(−θ)]
=
2i
2i
cos θ + i sin θ − [cos θ − i sin θ]
=
2i
cos θ + i sin θ − cos θ + i sin θ]
=
2i
2i sin θ
= sin θ.
=
2i
14
3. Let S be the set of all complex numbers with real part 3.
(a) Draw a picture of S in the complex plane and label at least one point in
S using the form a + bi.
(b) Draw a picture of the image of S under the map z 7→ ez in the complex
plane and label at least one point in the image of S using the form a + bi.
Solution.
4. Let T be the set of all complex numbers a + bi with a = b > 0.
(a) Draw a picture of T in the complex plane and label at least one point in
T using the form a + bi.
(b) Draw a picture of the preimage of T under the map z 7→ ez in the complex
plane and label at least one point in the preimage of T using the form
a + bi.
Solution.
15
5. Solve the equation ex = e for x.
[Find ALL complex solutions.]
Solution. Since ea+bi = ea ebi we know that ea+bi = e if and only if ea = e and
ebi = 1. Thus the solutions to the equation ex = e are all complex numbers
1 + 2πki where k is any integer.
6. Show that |ea+bi | = ea .
Solution.
|ea+bi | = |ea ebi | = |ea |·|ebi | = ea | cos(b)+i sin(b)| = ea
p
√
cos2 b + sin2 b = ea 1 = ea .
7. Let z be an arbitrary complex number. Explain why ez 6= 0.
[Hint: Use the result from the previous exercise.]
Solution. Let z = a + bi for a and b real. Then from exercise 6 we know
|ez | = ea > 0. However, |0| = 0. So it’s impossible for ez to be equal to 0.
A.6
polar form and multiplication.
1. Write the following numbers in standard (rectangular) form (a + bi).
3π
(a) 3e 4 i
√
3π
Solution. 3e 4 i = 3 cos( 3π
) + 3i sin( 3π
) = −3 22 +
4
4
√
3 2
i.
2
22π
(b) 12e− 3 i
= 2π
− 8π we have
Solution. Since − 22π
3
3
22π
2π
2π
22π
− 22π
i
12e 3 = 12 cos −
+ 12i sin −
= 12 cos
+ i sin
3
3
3
3
√
= −6 + 6 3i.
(c) 19e
14π
i
2
Solution. 19e
14π
i
2
= 19e7πi = 19eπi = 19 cos(π) + 19i sin(π) = −19.
2. Write the following numbers in polar form (reiθ ).
(a) 4 − 4i
p
√
√
Solution. r = |4 − 4i| = 42 + (−4)2 = 32 = 4 2. Also
√
√
2
2
4
4
cos θ = √ =
and sin θ = − √ = −
2
2
4 2
4 2
. (This is not the only possible correct choice of θ. θ
so we choose θ = − 3π
4
√
3π
3π
could be any − 4 +2πk as long as k is an integer.) Thus 4−4i = 4 2e− 4 i .
16
(b) −32i
0
= 0 and sin θ = −32
= −1
Solution. r = | − 32i| = 32. Also, cos θ = 32
32
π
so we choose θ = − 2 (again, this is not the only θ which works). Thus
π
−32i = 32e− 2 i .
√
(c) 7 3 − 7i
q
√
√
√
√
Solution. r = |7 3 − 7i| = 72 + (7 3)2 = 49 + 147 = 196 = 14.
Also
√
√
3
7
1
−7 3
cos θ =
=
and sin θ =
=−
.
14
2
14
2
√
π
We choose θ = − π3 (other correct θ exist). Thus 7 3 − 7i = 14e− 3 i .
√
(d) −2 − 2 3i
q
√
√
√
Solution. r = | − 2 − 2 3i| = (−2)2 + (−2 3)2 = 4 + 12 = 4. Also,
√
√
3
−2 3
and sin θ =
=−
.
4
2
√
2π
(other correct θ exist). So −2 − 2 3i = 2e− 3 i .
−2
1
cos θ =
=−
4
2
So we choose θ = − 2π
3
√
3. Write (− 3 + i)8 in the form a + bi.
[Hint: use example 3 as a guide.]
√
Solution. First√we write − 3 + i in polar form. You should get something
5π
equivalent to − 3 + i = 2e 6 i . Thus
√
5π
40π
2π
5π
2π
(− 3 + i)8 = (2e 6 i )8 = 28 e8· 6 i = 256e 6 i = 256e6πi+ 3 i = 256e 3 i .
Now we convert back to rectangular coordinates (a + bi). You should get
√
2π
256e 3 i = −128 + 128 3.
5π
4. Let z = e 6 i . Plot the complex numbers z, z 2 , z 3 , z 4 , z 5 , and z 6 in the complex
plane.
Solution.
17
2π
5. Let z = e− 3 i . Plot the complex numbers z, z 2 , z 3 , z 4 , z 5 , and z 6 in the complex
plane.
Solution.
6. Plot the complex number zw where z and w are the complex numbers given
below.
Solution.
7. Plot the complex number iw + z w̄ where z and w are the complex numbers
given below.
Solution.
8. Suppose z = reiθ is an arbitrary complex number.
(a) Write −z in polar form.
Solution. −z = rei(θ+π)
18
(b) Write z̄ in polar form.
Solution. z̄ = re−iθ
(c) Write z −1 in polar form (assuming z 6= 0).
Solution. z −1 = 1r e−iθ
9. The nth roots of unity are all the complex solutions to the equation xn = 1.
(a) Write down all 12th roots of unity in polar form (reiθ ).
Solution.
2π
e0i ,
e 12 i ,
4π
e 12 i ,
6π
e 12 i ,
8π
e 12 i ,
2π
e 6 i,
e 12 i ,
10π
12π
e 12 i ,
14π
16π
e 12 i ,
e 12 i ,
18π
e 12 i ,
20π
22π
e 12 i ,
e 12 i ,
or simplified a bit
1,
π
e 6 i,
π
e 3 i,
π
e 2 i,
e 3 i,
5π
eπi ,
7π
e 6 i,
4π
e 3 i,
3π
e 2 i,
5π
e 3 i,
e
11π
i
6
(b) Write down all 12th roots of unity in standard (rectangular) form (a + bi).
Solution.
You should
get ±1, ±i and all complex numbers of the form
√
√
± 23 ± 12 i or ± 21 ± 23 i.
(c) Plot the 12th roots of unity in the complex plane.
(d) Indicate which of the 12th roots of unity are also 6th roots of unity in your
answer for part (c).
(e) Indicate which of the 12th roots of unity are also 3rd roots of unity in your
answer for part (c).
(f) Indicate which of the 12th roots of unity are also 4th roots of unity in your
answer for part (c).
Solution.
19
.
10. (a) Find all complex roots of the polynomial x5 − 1.
Solution. The roots are the five 5th roots of unity:
1,
4π
2π
e 5 i,
6π
e 5 i,
8π
e 5 i,
e 5 i.
(b) Find all complex roots of the polynomial x4 + x3 + x2 + x + 1.
[Hint: expand (x − 1)(x4 + x3 + x2 + x + 1) and use your answer from part
(a).]
Solution. Since (x − 1)(x4 + x3 + x2 + x + 1) = x5 − 1 we know the roots
of x4 + x3 + x2 + x + 1 are all the fifth roots of unity except 1, namely
2π
e 5 i,
4π
e 5 i,
6π
e 5 i,
8π
e 5 i.
11. Solve for x (give all complex solutions):
(a) x3 = −1.
π
Solution. x is one of the following: e 3 i ,
unity which are not 3rd roots of unity).
(b) x4 = 2.
Solution. x is one of the following:
√
4
2,
−1,
√
4
2i,
(c) x5 = 32.
Solution. x is one of the following: 2,
20
2π
2e 5 i ,
5π
e 3 i (the 6th roots of
√
− 4 2,
4π
2e 5 i ,
√
− 4 2i
6π
2e 5 i ,
8π
2e 5 i