Embed Size (px)
Citation preview
A Solutions to exercises on complex numbers.
A.1 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 − 12i) − (9 + 5
2i)
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 z2j = −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 z4j = 1
for all j = 1, 2, 3, 4.
Solution. 1, i,−1, and −i.
4. Write the product (√
32
+ 12i)(1
2+
√3
2i) in the form a + bi.
Solution.(√
3
2+
1
2i
)(
1
2+
√3
2i
)
=
√3
4+
3
4i +
1
4i +
√3
4i2 = i.
A.2 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(12− i)
Solution. Since 2i(12− i) = i− 2i2 = 2 + i we have 2i(1
2− i) = 2 − i.
1
(7.) 1+4i3+2i
Solution. 1+4i3+2i
= 1+4i3+2i
· 3−2i3−2i
= 3−2i+12i−8i2
32+22 = 11+10i13
= 1113
+ 1013
i.
(8.) 3+2i1−4i
Solution. 3+2i1−4i
= 3+2i1−4i
· 1+4i1+4i
= 3+12i+2i+8i2
12+42 = −5+12i17
= − 517
+ 1217
i.
(9.) 11+i
Solution. 11+i
= 11+i
· 1−i1−i
= 1−i12+12 = 1−i
2= 1
2− 1
2i.
(10.) 34−3i
Solution. 34−3i
= 34−3i
· 4+3i4+3i
= 12+9i42+32 = 12+9i
25= 12
25+ 9
25i.
2. Find the multiplicative inverse of each of the complex numbers:
−1 − i, 3 − 2i, i, 2√
3 − 2i, −√
2
2+
√2
2i
Solution.
• (−1 − i)−1 = 1(−1−i)(−1+i)
(−1 + i) = 112+12 (−1 + i) = −1
2+ 1
2i.
• (3 − 2i)−1 = 1(3−2i)(3+2i)
(3 + 2i) = 132+22 (3 + 2i) = 3
13+ 2
13i.
• i−1 = 1i(−i)
(−i) = 11(−i) = −i.
• (2√
3 − 2i)−1 = 1(2√
3−2i)(2√
3+2i)(2√
3 + 2i) = 1(2√
3)2+22 (2√
3 + 2i)
= 2√
316
+ 216
i =√
38
+ 18i.
• (−√
22
+√
22
i)−1 = 1
(−√
22
+√
22
i)(−√
22−
√
22
i)(−
√2
2−
√2
2i) = 1
(√
22
)2+(√
22
)2(−
√2
2−
√2
2i)
= −√
22−
√2
2i.
3. Find the multiplicative inverse of bi where b is an arbitrary nonzero real number.
Solution.
(bi)−1 =1
(bi)(bi)(bi) =
1
(bi)(−bi)(−bi) =
1
b2(−bi) = − i
b.
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 equationseparately.]
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 equationseparately.]
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 assumethat 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.
Answer. b−0a−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 ba.
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 numberswhose real part is greater than or equal to 1.
Solution.
(b) Shade the region of the complex plane consisting of all complex numberswhose 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 thecomplex number z̄ − w.
(c) Check your answer for part (b) by writing 2 − 2i − (−1 − 3i) in the forma + 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. Thusz−1 = z̄
z·z̄ = z̄|z|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 equationseparately.]
Solution. Set z = a + bi and w = c + di. Then
|z ·w| = |(a + bi)(c + di)| = |(ac− bd) + (ad + bc)i| =√
(ac − bd)2 + (ad + bc)2
=√
a2c2 − 2abcd + b2d2 + a2d2 + 2abcd + b2c2 =√
a2c2 + b2d2 + a2d2 + b2c2.
Whereas
|z| · |w| = |a + bi| · |c + di| =√
a2 + b2√
c2 + d2 =√
(a2 + b2)(c2 + d2)
=√
a2c2 + b2d2 + a2d2 + b2c2.
3. For each of the given series
(I) Determine whether or not the series is convergent, and if the series isconvergent 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 toverify your answer for part (I).
7
(a)∞∑
n=0
(
2i3
)n
Solution. The series is geometric with |r| = |2i3| = 2
3< 1 so it converges.
Moreover, we have
∞∑
n=0
(
2i
3
)n
=1
1 − 2i3
=3
3 − 2i=
3
3 − 2i· 3 + 2i
3 + 2i=
9 + 6i
32 + 22=
9
13+
6
13i.
Since ∞∑
n=0
(
2i
3
)n
= 1 +2
3i − 4
9− 8
27i +
16
81+
32
243i − · · ·
we see the first few partial sums are
s0 = 1
s1 = 1 + 23i
s2 = 59
+ 23i
s3 = 59
+ 1027
i
s4 = 6181
+ 1027
i
s5 = 6181
+ 122243
i
Here are is a picture of the first several partial sums plotted in the complexplane:
8
(b)∞∑
n=0
(
−2i3
)n
Solution. The series is geometric with |r| = |− 2i3| = 2
3< 1 so it converges.
Moreover, we have
∞∑
n=0
(
−2i
3
)n
=1
1 + 2i3
=3
3 + 2i=
3
3 + 2i· 3 − 2i
3 − 2i=
9 − 6i
32 + 22=
9
13− 6
13i.
Since ∞∑
n=0
(
−2i
3
)n
= 1 − 2
3i − 4
9+
8
27i +
16
81− 32
243i − · · ·
we see the first few partial sums are
s0 = 1
s1 = 1 − 23i
s2 = 59− 2
3i
s3 = 59− 10
27i
s4 = 6181
− 1027
i
s5 = 6181
− 122243
i
Here are is a picture of the first several partial sums plotted in the complexplane:
9
(c)∞∑
n=0
(2i)n
Solution. The series is geometric with |r| = |2i| = 2 > 1 so it diverges.Since
∞∑
n=0
(2i)n = 1 + 2i − 4 − 8i + 16 + 32i − 64 − 128i + · · ·
we see the first few partial sums are
s0 = 1
s1 = 1 + 2is2 = −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 complexplane:
4. Consider the series∞∑
n=0
(12
+ 12i)n.
(I) Determine whether or not the series is convergent, and if the series isconvergent 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 verifyyour answer for part (I).
Solution. This a geometric series with
|r| =
∣
∣
∣
∣
1
2+
1
2i
∣
∣
∣
∣
=
√
(
1
2
)2
+
(
1
2
)2
=
√
1
2< 1
so it converges. Moreover,
∞∑
n=0
(
1
2+
1
2i
)n
=1
1 −(
12
+ 12i) =
112− 1
2i
=2
1 − i=
2
1 − i·1 + i
1 + i=
2 + 2i
2= 1+i.
To check that our answer makes sense, let’s look at some of the partial sums:
∞∑
n=0
(
1
2+
1
2i
)n
= 1+
(
1
2+
1
2i
)
+
(
1
2+
1
2i
)2
+
(
1
2+
1
2i
)3
+
(
1
2+
1
2i
)4
+ · · ·
Since(
12
+ 12i)2
= 12i we have
∞∑
n=0
(
1
2+
1
2i
)n
= 1+
(
1
2+
1
2i
)
+
(
1
2i
)
+
(
−1
4+
1
4i
)
+
(
−1
4
)
+
(
−1
8− 1
8i
)
+
(
−1
8i
)
+
(
1
16− 1
16i
)
+ · · ·
so the partial sums look like:
s0 = 1
s1 = 32
+ 12i
s2 = 32
+ i
s3 = 54
+ 54i
s4 = 1 + 54i
s5 = 78
+ 98i
s6 = 78
+ i
s7 = 1516
+ 1516
i...
11
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 drawa picture of the disk where the series converges as in example 2.
(a)∞∑
n=0
nxn
n+1.
Solution. Using the ratio test with an = nxn
n+1we have
∣
∣
∣
∣
an+1
an
∣
∣
∣
∣
=
∣
∣
∣
∣
(n + 1)xn+1
n + 2· n + 1
nxn
∣
∣
∣
∣
=
∣
∣
∣
∣
x(n2 + 2n + 1)
n2 + 2
∣
∣
∣
∣
→ |x| as n → ∞
So the series converges when |x| < 1 and diverges when |x| > 1. Thus theradius of convergence is 1 as pictured below.
(b)∞∑
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
Solution. Using the ratio test with an = (x−2+i)n
(n+1)3n we have∣
∣
∣
∣
an+1
an
∣
∣
∣
∣
=
∣
∣
∣
∣
(x − 2 + i)n+1
(n + 2)3n+1· (n + 1)3n
(x − 2 + i)n
∣
∣
∣
∣
=
∣
∣
∣
∣
(x − 2 + i)(n + 1)
(n + 2)3
∣
∣
∣
∣
→ |x − 2 + i|3
as n → ∞
So the series converges when |x − 2 + i| = |x − (2 − i)| < 3 and divergeswhen |x − 2 + i| = |x − (2 − i)| > 3. Thus the radius of convergence is 3as pictured below.
(c)∞∑
n=1
n!nn (x + 2i)n
Solution. Using the ratio test with an = n!nn (x + 2i)n we have
∣
∣
∣
∣
an+1
an
∣
∣
∣
∣
=
∣
∣
∣
∣
(n + 1)!(x + 2i)n+1
(n + 1)n+1· nn
n!(x + 2i)n
∣
∣
∣
∣
=
∣
∣
∣
∣
nn
(n + 1)n(x + 2i)
∣
∣
∣
∣
= |x + 2i|(
n
n + 1
)n
Since limn→∞
ln(
nn+1
)n= lim
n→∞ln( n
n+1)1n
= limn→∞
(n+1n
)“
1(n+1)2
”
− 1n2
= limn→∞
−nn+1
= −1,
we see(
nn+1
)n= eln( n
n+1)n
→ e−1 as n → ∞. Thus
|x + 2i|(
n
n + 1
)n
→ |x + 2i|e
as n → ∞
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 picturedbelow.
13
A.5 exponentials and Euler’s formula
1. Write each of the following in the form a + bi.
(a) e−2π
3i
Solution. e−2π
3i = cos(−2π
3) + i sin(−2π
3) = −1
2−
√3
2i.
(b) 12eπ
6i
Solution. 12eπ
6i = 12[cos(π
6) + i sin(π
6)] = 12[
√3
2+ 1
2i] = 6
√3 + 6i.
(c) e2π
3i + e
4π
3i + e
6π
3i
Solution.
• e2π
3i = cos(2π
3) + i sin(2π
3) = −1
2+
√3
2i,
• e4π
3i = cos(4π
3) + i sin(4π
3) = −1
2−
√3
2i,
• e6π
3i = e2πi = 1.
Thus
e2π
3i + e
4π
3i + e
6π
3i = −1
2+
√3
2i − 1
2−
√3
2i + 1 = 0.
2. Show that cos(θ) = eiθ+e−iθ
2and sin(θ) = eiθ−e−iθ
2ifor all real numbers θ.
[Hint: Use Euler’s formula and the negative angle identities for sine and cosine.]
Solution.
eiθ + e−iθ
2=
cos(θ) + i sin(θ) + cos(−θ) + sin(−θ)
2
=cos θ + i sin θ + cos θ − i sin θ
2
=2 cos θ
2= cos θ.
eiθ − e−iθ
2i=
cos(θ) + i sin(θ) − [cos(−θ) + sin(−θ)]
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
2i= sin θ.
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 inS using the form a + bi.
(b) Draw a picture of the image of S under the map z 7→ ez in the complexplane 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 inT using the form a + bi.
(b) Draw a picture of the preimage of T under the map z 7→ ez in the complexplane and label at least one point in the preimage of T using the forma + bi.
Solution.
15
5. Solve the equation ex = e for x.
[Find ALL complex solutions.]
Solution. Since ea+bi = eaebi we know that ea+bi = e if and only if ea = e andebi = 1. Thus the solutions to the equation ex = e are all complex numbers1 + 2πki where k is any integer.
6. Show that |ea+bi| = ea.
Solution.
|ea+bi| = |eaebi| = |ea|·|ebi| = ea| cos(b)+i sin(b)| = ea√
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).
(a) 3e3π
4i
Solution. 3e3π
4i = 3 cos(3π
4) + 3i sin(3π
4) = −3
√2
2+ 3
√2
2i.
(b) 12e−22π
3i
Solution. Since −22π3
= 2π3− 8π we have
12e−22π
3i = 12 cos
(
−22π
3
)
+12i sin
(
−22π
3
)
= 12 cos
(
2π
3
)
+ i sin
(
2π
3
)
= −6 + 6√
3i.
(c) 19e14π
2i
Solution. 19e14π
2i = 19e7πi = 19eπi = 19 cos(π) + 19i sin(π) = −19.
2. Write the following numbers in polar form (reiθ).
(a) 4 − 4i
Solution. r = |4 − 4i| =√
42 + (−4)2 =√
32 = 4√
2. Also
cos θ =4
4√
2=
√2
2and sin θ = − 4
4√
2= −
√2
2
so we choose θ = −3π4
. (This is not the only possible correct choice of θ. θ
could be any −3π4
+2πk as long as k is an integer.) Thus 4−4i = 4√
2e−3π
4i.
16
(b) −32i
Solution. r = | − 32i| = 32. Also, cos θ = 032
= 0 and sin θ = −3232
= −1so we choose θ = −π
2(again, this is not the only θ which works). Thus
−32i = 32e−π
2i.
(c) 7√
3 − 7i
Solution. r = |7√
3 − 7i| =√
72 + (7√
3)2 =√
49 + 147 =√
196 = 14.
Also
cos θ =7
14=
1
2and sin θ =
−7√
3
14= −
√3
2.
We choose θ = −π3
(other correct θ exist). Thus 7√
3 − 7i = 14e−π
3i.
(d) −2 − 2√
3i
Solution. r = | − 2 − 2√
3i| =√
(−2)2 + (−2√
3)2 =√
4 + 12 = 4. Also,
cos θ =−2
4= −1
2and sin θ =
−2√
3
4= −
√3
2.
So we choose θ = −2π3
(other correct θ exist). So −2 − 2√
3i = 2e−2π
3i.
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 somethingequivalent to −
√3 + i = 2e
5π
6i. Thus
(−√
3 + i)8 = (2e5π
6i)8 = 28e8· 5π
6i = 256e
40π
6i = 256e6πi+ 2π
3i = 256e
2π
3i.
Now we convert back to rectangular coordinates (a + bi). You should get
256e2π
3i = −128 + 128
√3.
4. Let z = e5π
6i. Plot the complex numbers z, z2, z3, z4, z5, and z6 in the complex
plane.
Solution.
17
5. Let z = e−2π
3i. Plot the complex numbers z, z2, z3, z4, z5, and z6 in the complex
plane.
Solution.
6. Plot the complex number zw where z and w are the complex numbers givenbelow.
Solution.
7. Plot the complex number iw + zw̄ where z and w are the complex numbersgiven 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 = 1re−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.
e0i, e2π
12i, e
4π
12i, e
6π
12i, e
8π
12i, e
10π
12i, e
12π
12i, e
14π
12i, e
16π
12i, e
18π
12i, e
20π
12i, e
22π
12i,
or simplified a bit
1, eπ
6i, e
π
3i, e
π
2i, e
2π
3i, e
5π
6i, eπi, e
7π
6i, e
4π
3i, e
3π
2i, e
5π
3i, e
11π
6i.
(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±
√3
2± 1
2i or ±1
2±
√3
2i.
(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 youranswer for part (c).
(e) Indicate which of the 12th roots of unity are also 3rd roots of unity in youranswer for part (c).
(f) Indicate which of the 12th roots of unity are also 4th roots of unity in youranswer 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, e2π
5i, e
4π
5i, e
6π
5i, e
8π
5i.
(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 rootsof x4 + x3 + x2 + x + 1 are all the fifth roots of unity except 1, namely
e2π
5i, e
4π
5i, e
6π
5i, e
8π
5i.
11. Solve for x (give all complex solutions):
(a) x3 = −1.
Solution. x is one of the following: eπ
3i, −1, e
5π
3i (the 6th roots of
unity which are not 3rd roots of unity).
(b) x4 = 2.
Solution. x is one of the following: 4√
2, 4√
2i, − 4√
2, − 4√
2i
(c) x5 = 32.
Solution. x is one of the following: 2, 2e2π
5i, 2e
4π
5i, 2e
6π
5i, 2e
8π
5i
20
