Linear Algebra
Hints and solutions to selected exercises
Homework 1.
1. Determine the following:
๐) ๐
๐ [
2 + 5๐
],
๐−1
๐ผ๐ [
2 + 5๐
]
๐−1
2 + 5๐ −1 − ๐ −2 − 2๐ − 5๐ − 5๐ 2 −2 + 5 − 7๐
⋅
=
=
;
−1 + ๐ −1 − ๐
1 − ๐2
2
๐
๐ [
2 + 5๐
3
2 + 5๐
7
] = ; ๐ผ๐ [
]=−
−1 + ๐
2
−1 + ๐
2
g) For real numbers the absolute value can be split up : e.g. |๐ฅ ⋅ (๐ฅ + 2)| = |๐ฅ| ⋅ |๐ฅ + 2| . The same
happens for complex numbers.
|
๐ 7 (1 + ๐)8
|๐|7 ⋅ |(1 + ๐)| 8
=
|
12
12
(√2 − ๐√6 )
|√2 − ๐√6|
We calculate the specific absolute values:
|๐| = 1
|1 + ๐| = √1 + 1 = √2
|√2 − ๐√6| = √2 + 6 = √8
now we see
|๐|7 ⋅ |(1 + ๐)| 8
12
8
=
1 ⋅ (√2)
|√2 − ๐√6|
โ) ๐ผ๐ [
=
12
=
|√8|
24
86
๐7
๐6 ๐
(๐ 2 )3 ๐
1
−๐
]
=
๐ผ๐
[
]
=
๐ผ๐
[
] = 4 ๐ผ๐ [
]=
4
4
4
4
2
2
[1 − 2๐ − 1]2
(2 − 2๐)
2 (1 − ๐)
2 [(1 − ๐) ]
2
1
−๐
1
−๐
11
1
๐ผ๐ [
] = 4 ๐ผ๐ [
]= 4 = 6
4
2
[−2๐]
2
2
4(−1)
2 4 2
2. Find the absolute value |๐ง|, the Real and Imaginary parts of ๐ง: ๐
๐(๐ง), ๐ผ๐(๐ง), for
๐) ๐ง =
(1 + ๐)(1 + ๐)2 (1 + ๐)3 … (1 + ๐)20
๐ 0 + ๐ 2 + ๐ 4 + ๐ 6 + โฏ + ๐ 20
๐๐ฎ๐ฆ๐๐ซ๐๐ญ๐จ๐ซ โถ (1 + ๐)(1 + ๐)2 (1 + ๐)3 … (1 + ๐)20 = (1 + ๐)1+2+3+โฏ+20 = (1 + ๐)10⋅21
= (1 + ๐)2⋅5⋅21
we know that (1 + ๐)2 = 1 + 2๐ − 1 = 2๐
105
thus the numerator = 2
and that
21
((2๐)5 )
21
= (25 ๐ )
= 25⋅21 ๐,
๐.
๐๐๐ง๐จ๐ฆ๐ข๐ง๐๐ญ๐จ๐ซ: there are different methods here , one of them is the following:
๐ 0 + ๐ 2 + ๐ 4 + ๐ 6 +. . +๐ 20 = 1 + (−1) + (−1)2 + (−1)3 +. . +(−1)10 .
We notice that the pairs โถ 0. i 1. ๐. ๐. 10 + (−1)1 ;
2. i 3. ๐. ๐. (−1)2 + (−1)3 ;
4. i 5. ๐. ๐. (−1)4 + (−1)5 ๐๐ก๐..
cancel out each other, leaving only (−1)10 = 1
Thus the denominator = 1, so ๐ง = 2105 ๐, ๐
๐ (๐ง) = 0, ๐ผ๐(๐ง) = 2105 ; |๐ง| = 2105 .
3. Solve the following equations for z ๏ C , it is possible that there are no solutions or there are more
than one.
๐
) ๐๐ง 2 − ๐ง + 2๐ = 0;
๐ง1 =
Δ = 1 − 4 ⋅ ๐ ⋅ 2๐ = 1 + 8 = 9;
1+3 2
1 − 3 −2
= = −2๐ ; ๐ง2 =
=
= ๐ − two solutions.
2๐
๐
2๐
2๐
g) ๐ง 2 = 3 + 4๐
(๐ฅ + ๐๐ฆ)2 = 3 + 4๐
๐ฅ 2 − 2๐๐ฅ๐ฆ − ๐ฆ 2 = 3 + 4๐
๐ฅ2 − ๐ฆ2 = 3
๐ฅ2 − ๐ฆ2 = 3
{
{
2
2๐ฅ๐ฆ = 4
๐ฆ=๐ฅ
4
4
{
2
๐ฅ − 3๐ฅ − 4 = 0
๐ฅ2 − ๐ฅ2 = 3
2
๐ฆ=๐ฅ
๐ฅ4 − 3๐ฅ2 − 4 = 0
2
๐ฆ=๐ฅ
{
(๐ฅ1 )2 = 4: ๐ฆ1,1 = 1; ๐ฆ1,2 = −1
{
(๐ฅ2 )2 ≠ −1 ๐๐๐๐๐ข๐ ๐ ๐ฅ ∈ ๐
๐๐๐ . โถ ๐ง1 = 2 + ๐ ; ๐ง2 = −2 − ๐
๐) 2๐ง + (1 + ๐)๐ง = 1 − 3๐
2(๐ฅ + ๐๐ฆ) + (1 + ๐)(๐ฅ − ๐๐ฆ) = |collect real and imaginary parts| = (3๐ฅ + ๐ฆ) + ๐(๐ฅ + ๐ฆ)
(3๐ฅ + ๐ฆ) + ๐(๐ฅ + ๐ฆ) = 1 − 3๐
real and imaginary parts on both sides of equation are equal:.
3๐ฅ + ๐ฆ = 1
now you have to solve this system
{
๐ฅ + ๐ฆ = −3
{
๐ฅ=2
.
๐ฆ = −5
๐) (๐ง + ๐ง) + 2(๐ง − ๐ง) = 3 + 8๐; let ๐ง = ๐ฅ + ๐๐ฆ
(๐ฅ + ๐๐ฆ + ๐ฅ − ๐๐ฆ) + 2(๐ฅ + ๐๐ฆ − ๐ฅ + ๐๐ฆ) = 3 + 8๐
2๐ฅ + 4๐๐ฆ = 3 + 8๐
2๐ฅ = 3
{
4๐ฆ = 8
3
2
๐ง = + 2๐.
๐) ๐ง − ๐ = 2๐ง + 1 ; let ๐ง = ๐ฅ + ๐๐ฆ
๐ฅ + ๐๐ฆ − ๐ = 2๐ฅ + 2๐๐ฆ + 1
๐ฅ + ๐(๐ฆ − 1) = (2๐ฅ + 1) + 2๐๐ฆ
๐ฅ − ๐(๐ฆ − 1) = (2๐ฅ + 1) + ๐๐ฆ
๐ฅ = −1
๐ฅ = 2๐ฅ + 1
1
{
{
−(๐ฆ − 1) = 2๐ฆ
๐ฆ=
3
1
๐ง = −1 + ๐
3
4. Sketch the following sets in the complex plane
−3 + 2๐
]}
2−๐
−3 + 2๐
−3 + 2๐ 2 + ๐
−6 − 3๐ + 4๐ − 2
8
๐
๐ [
⋅
] = ๐
๐ [
] = ๐
๐ [
]=−
2−๐
2−๐ 2+๐
4+1
5
๐ ) ๐ = {๐ง ∈ ๐ถ โถ ๐
๐[๐ง] ≤ ๐
๐ [
8
๐
๐[๐ง] = ๐ฅ, ๐ ๐ ๐กโ๐ ๐๐๐๐๐ก๐ ๐๐๐๐ ๐ ๐ ๐๐ก๐๐ ๐๐ฆ ๐กโ๐ ๐๐๐ข๐๐ก๐๐๐ ๐ฅ ≤ − 5 โถ
5. Calculate the argument arg (๐ง) and the main argument, ๐ด๐๐(๐ง), of ๐ง.
๐) ๐๐๐(1 − ๐) , ๐ด๐๐(1 − ๐)
๐
๐
๐ด๐๐ (1 − ๐) = − ; arg(1 − ๐) = − + 2๐ π, k ∈ I.
4
4
๐) ๐ด๐๐(√2 − ๐√6)
๐ = √2 + 6 = √8 = 2√2
1
√2
cos ๐ผ =
=
2√2 2
√6
√3
sin ๐ผ = −
=−
2
{
2√2
๐
๐ด๐๐(√2 − ๐√6) = − ;
3
→
๐ผ= −
๐
3
arg( √2 − ๐√6) = −๐/3 + 2๐๐
6. Plot the following points and find the polar form (trigonometric form) and the exponential form
of
2
2
๐
๐) ๐ง = −2 + 2๐; ๐ = √4 + 4 = 2 √2 ; cos ๐ผ = −
, sin ๐ผ =
→ ๐ผ=3
4
2 √2
2 √2
3๐
1
1
3๐
3๐
๐
๐ง = −2 + 2๐ = 2√2 (−
+
๐) = 2√2 (cos
+ ๐ sin ) = 2√2 ๐ 4
4
4
√2 √2
๐) ๐ง = −4 − ๐ 4√3;
๐ = √16 + 48 = 8 ; cos ๐ผ = −
4
4√3
2๐
, sin ๐ผ =
→ ๐ผ=−
8
8
3
−2๐
1 √3
−2๐
−2๐
๐ง = −4 − ๐ 4√3 = 8 (− −
๐) = 8 (cos
+ ๐ sin
)=8 ๐ 3 ๐
2
2
3
3
y
−4
x
−4√3
i) ๐ง = ๐ 33 + (1 + ๐)4 = ๐ − 4; |๐ง| = √17
Plot ๐ง on the complex plane
Calculate ๐ผ using inverse trigonometruc functions eg..
4
4
4
cos ๐ฝ = 17, giving ๐ฝ = arccos ( 17) and ๐ผ = ๐ − arccos ( 17) .
√
√
√
7. Sketch the following sets in the complex plane, mark the main points (wedges and circles)
๐) ๐ = {๐ง ∈ ๐ถ: |๐ง + 3 − 2๐| ≤ 2 }
|๐ง + 3 − 2๐| ≤ 2 first we can plot the points ๐งฬ
, next take a mirror projection about axis OX of these
points:
or transform the inequality :
|๐ง + 3 − 2๐| ≤ 2
|๐ง + 3 + 2๐| ≤ 2
|๐ง + 3 + 2๐| ≤ 2
|๐ง + 3 + 2๐| ≤ 2 ๐๐๐ ๐๐๐ ๐๐ ๐๐๐๐๐๐ , ๐๐๐๐ก๐๐ ๐(−3, −2), ๐๐๐๐๐ข๐ ๐ ≤ 2.
๐) ๐ = {๐ง ∈ ๐ถ: −
3๐
๐
≤ ๐๐๐(๐ง) ≤ }
4
2
g) ๐ = {๐ง ∈ ๐ถ: −
3๐
๐
≤ ๐๐๐( ๐ง ) ≤ 2 }. The set from f ) is the set of points ๐งฬ
, so now, to obtain S,
4
we should mirror this set with respect to the OX axis
h) ๐ = {๐ง ∈ ๐ถ: −
3๐
4
๐
≤ ๐๐๐(๐ง − 2 + ๐) ≤ 2 }. First plot the points ๐ง1 = ๐ง − 2 + ๐, next to obtain
S, points ๐ง are shifted by 2 − ๐:
๐) ๐ = {๐ง ∈ ๐ถ: −
๐
๐
≤ ๐ด๐๐((๐ + 1) ⋅ ๐ง)) ≤ }
2
4
The argument of a product of complex numbers is the sum of the arguments i.e.
if ๐ง1 = ๐1 ๐ ๐ ๐ผ1 ๐๐๐ ๐ง2 = ๐2 ๐ ๐ ๐ผ2 , then ๐ง1 ๐ง2 = ๐1 ๐ ๐ ๐ผ1 ๐2 ๐ ๐ ๐ผ2 = ๐1 ๐2 ๐ ๐ (๐ผ1 +๐ผ2) .
๐
So arg(๐ง1 ๐ง2 ) = ๐ผ1 + ๐ผ2 . Let ๐ด๐๐(๐ง) = ๐ผ, we know that Arg(1 + ๐) = 4 , and
๐
๐ด๐๐((๐ + 1) ⋅ ๐ง)) = 4 + ๐ผ.
๐
๐
− 2 ≤ ๐ด๐๐((๐ + 1) ⋅ ๐ง)) ≤ 4 ,
๐ ๐
๐
≤ + ๐ผ + 2๐๐ ≤ , ๐๐๐
2 4
4
๐
Condition for ๐ด๐๐((1 + ๐)๐ง) : − ๐ < + ๐ผ + 2๐๐ ≤ ๐, only for ๐ = 0
4
3๐
−
≤ ๐ผ ≤ 0. This is an infinite wedge
4
−
๐ ∗ ) ๐ = {๐ง ∈ ๐ถ:
๐
๐
≤ ๐ด๐๐(๐ง 3 ) ≤ }
2
2
Let ๐ด๐๐(๐ง) = ๐ผ, then
−
๐
๐
≤ 3๐ผ + 2๐๐ ≤
2
2
๐
๐
− 2๐๐ ≤ 3๐ผ ≤ − 2๐๐
2
2
๐ 2๐๐
๐ 2๐๐
− −
≤๐ผ≤ −
true only for ๐ = −1,0,1
6
3
6
3
−
3๐
5๐
≤๐ผ≤
6
6
๐
๐
๐ = 0: − ≤ ๐ผ ≤
6
6
5๐
3๐
๐ = 1: −
≤๐ผ≤−
6
6
๐ = −1:
๐ง
๐) ๐ = {๐ง ∈ ๐ถ: 0 ≤ ๐ด๐๐ ( ) ≤ ๐ด๐๐(3 + 3๐) }
๐
๐
๐
๐ด๐๐(๐ง) = ๐ผ, ๐ด๐๐(๐ ) =
and ๐ด๐๐(3 + 3๐) = ๐ด๐๐(1 + ๐) = , so
2
4
๐ง
0 ≤ ๐ด๐๐ ( ) ≤ ๐ด๐๐(3 + 3๐)
๐
๐ง
๐
Condition for ๐ด๐๐ ( ) : − ๐ < ๐ผ − + 2๐๐ ≤ ๐,
๐๐๐๐ฆ ๐๐๐ ๐ = 0
๐
2
๐ ๐
π
0≤๐ผ− ≤
add
to both sides.
2 4
2
๐
3๐
≤๐ผ≤
2
4
Plot a wedge.
๐)
๐ = {๐ง ∈ ๐ถ: ๐ด๐๐(1 − 3๐) ≤ ๐ด๐๐ ๐ง ≤ ๐ด๐๐(−2 + 5๐)}
Simply plot the numbers ๐ง1 = 1 − 3๐ oraz ๐ง2 = −2 + 5๐ i.e. points ๐1 (1, −3), ๐2 (−2,5) on the
plane, the region is an infinite wedge between the line passing through points (0,0), ๐1 (1, −3) and the
line passing through (0,0), ๐2 (−2,5).
8. Sketch the following sets in the complex plane, mark the main points
๐) ๐ = {๐ง ∈ ๐ถ: ๐ผ๐ [(1 + 2๐)๐ง − 3๐] < 0},
๐๐๐ . : let ๐ง = ๐ฅ + ๐๐ฆ, then
๐ผ๐ [(1 + 2๐)๐ง − 3๐] = ๐ผ๐ [(1 + 2๐)(๐ฅ + ๐๐ฆ) − 3๐] = ๐ผ๐ [(๐ฅ − 2๐ฆ) + ๐(2๐ฅ + ๐ฆ − 3)]:
2๐ฅ + ๐ฆ − 3 < 0 ⇔ ๐ฆ < −2๐ฅ + 3.
๐
) ๐ = {๐ง ∈ ๐ถ: ๐ง + ๐ = ๐ง − 1},
๐ง+๐ = ๐ง−1
๐ง−๐ =๐ง−1
๐ง − ๐ง = ๐ − 1, let ๐ง = ๐ฅ + ๐๐ฆ, then since the real part are equal and the imaginary parts are equal:
๐ฅ + ๐๐ฆ − (๐ฅ + ๐๐ฆ) = ๐ − 1
๐ฅ − ๐๐ฆ − (๐ฅ + ๐๐ฆ) = ๐ − 1
−2๐๐ฆ = ๐ − 1;
f)
๐ = ∅.
๐) ๐ = {๐ง ∈ ๐ถ: |๐๐ง + 1 − ๐| < 2},
1
|๐๐ง + 1 − ๐| = |๐| ⋅ |๐ง + − 1| = |๐| ⋅ |๐ง − ๐ − 1| = |๐ง − (๐ + 1)| < 2;
๐
inside of circle, centre ๐(1,1), radius ๐ = 2. Sketch it.
๐) ๐ = {๐ง ∈ ๐ถ: |๐ง − ๐ + 1| < 3}:
|๐ง − ๐ + 1| = |๐ง − ๐ + 1 | = |๐ง − (−1 + ๐) | < 3; inside of circle, centre c(-1,1) i radius r = 3.
Sketch it.
๐) ๐ = {๐ง ∈ ๐ถ: |๐ง − 2๐| + |๐ง + 2๐| = 4 }
The set consists of complex numbers which lie in distance equal to 4 from points -2i and 2i. This is an
interval on the imaginary axis with from -2i to 2i.
๐) ๐ = {๐ง ∈ ๐ถ: |๐งฬ
+ ๐| < 2}
Method 1: |๐งฬ
+ ๐| = |๐งฬ
− ๐| = |๐งฬ
ฬ
ฬ
ฬ
ฬ
ฬ
− ๐ | = |๐ง − ๐| < 2
ฬ
ฬ
ฬ
ฬ
ฬ
ฬ
ฬ
ฬ
or Method 2: |๐ฅ
+ ๐๐ฆ + ๐| = |๐ฅ − ๐๐ฆ + ๐| = |๐ฅ + ๐(−๐ฆ + 1) = √๐ฅ 2 + (๐ฆ − 1)2 < 2.
Inside of circle, centre c(0,1), radius r = 2. Sketch it
๐) ๐ = { ๐ง ∈ ๐ถ: Im[(2 + ๐)(3 + 5๐)] ≥ |๐ง − 3 + ๐| ≥ |√5 + 2๐ |
∧
๐
Arg(3 − ๐) ≤ arg(๐ง) ≤ Arg [๐ ๐ 2 ] }
First inequality: we calculate the values on the left- and right-hand sides of
Im[(2 + ๐)(3 + 5๐)] ≥ |๐ง − 3 + ๐| ≥ |√5 + 2๐ |
LHS: Im[(2 + ๐)(3 + 5๐)] = ๐ผ๐ [6 + 10๐ + 3๐ + 5๐ 2 ] = 13
2
RHS: |√5 + 2๐ | = √(√5) + 22 = √5 + 4 = 3
The "in-between"
|๐ง − 3 + ๐| = |๐ง − (3 − ๐)| ๐กโ๐ ๐๐๐ ๐ก๐๐๐๐ ๐๐๐ก๐ค๐๐๐ ๐ ๐๐๐ 'z" ๐๐๐ ๐ง0 = 3 − ๐
Im[(2 + ๐)(3 + 5๐)] ≥ |๐ง − 3 + ๐| ≥ |√5 + 2๐ | means
13 ≥ |๐ง − (3 − ๐)| ≥ 3 this is a ring with centre at P(3,-1) and radii between 3, 13.
The second inequality:
๐
Arg(3 − ๐) ≤ arg(๐ง) ≤ Arg [๐ ๐ 2 ]
Arg(3 − ๐) angle between the positive OX axis and the point P(3,-1), we don not calculate the
argument – ONLY plot it.
๐
๐
Arg [๐ ๐ 2 ] = 2 − simple,
๐
so Arg(3 − ๐) ≤ arg(๐ง) ≤ Arg [๐ ๐ 2 ] is a infinite wedge between line P(0,0) to P(3,-1), and line
P(0,0) to P(0,1).
The intersection of both sets is ๐, top, right part of ring.
9. Find Arg(z), |z| for the following complex numbers
๐ ๐ 15
๐) ๐ง = (๐ 5 )
๐๐๐(๐ง) = 15
๐
= 3๐ โถ ๐ด๐๐(๐ง) = ๐; |๐ง| = 1
5
๐๐
๐๐
3
3
10
๐๐
๐ 3๐ ๐ ๐
๐) ๐ง = (1 + ๐)3 ⋅ ๐ 4 = (√2 ๐ 4 ) ⋅ ๐ 4 = 2 √2 ๐ 4 + 4 = 2 √2 ๐ ๐ ๐ โถ
๐ด๐๐(๐ง) = ๐; |๐ง| = 2 √2
(3 √2 ๐ 4๐๐ )
10
๐) ๐ง =
(−3 + 3๐)
=
4
๐๐
๐๐
(๐ 3 )
10
= (3 √2 )
4
30
4
37
๐ 4 ๐ ๐−3๐ ๐ = 310 25 ๐ 6 ๐ ๐ ;
(๐ 3 )
37
36 1
๐
๐
๐ = ( + ) ๐ = + 2 ⋅ 3๐; โถ ๐ด๐๐(๐ง) = ; |๐ง| = 310 25
6
6 6
6
6
10. Let ๐ง = −1 + ๐, write the following complex numbers in exponential form
3๐
First write ๐ง in exponential form: ๐ง = −1 + ๐ = √2 ๐ 4 ๐ ;
3๐
3๐
7๐
๐
๐)
− ๐ง = −1 ⋅ √2 ๐ 4 ๐ = ๐ ๐๐ √2 ๐ 4 ๐ = √2 ๐ 4 ๐ = √2 ๐ − 4 ๐
๐)
๐๐ง = ๐ 2 ๐ √2 ๐ 4 ๐ = √2 ๐ 4 ๐ = √2 ๐ − 4 ๐
๐)
1
=
๐ง
๐
3๐
1
=
3๐
๐
√2 ๐ 4
5๐
1
√2
๐
3๐
3๐
๐− 4 ๐
๐
11. Let ๐ง = 2 (cos 7 + ๐ sin 7 ), write in exponential form
๐
๐
๐
First write ๐ง in exponential form ๐ง = 2 (cos + ๐ sin ) = 2 ๐ 7 ๐
7
7
๐๐
๐
8๐ ๐
๐) − ๐ง = −1 ⋅ 2 ๐ 7 = 2 ๐ ๐๐ ๐ 7 ๐ = 2 ๐ 7 = | ๐๐๐๐ ๐๐๐๐ข๐๐๐๐ก | = 2๐
๐๐
๐
๐
9๐ ๐
๐) ๐๐ง = ๐ ⋅ 2 ๐ 7 = 2 ๐ 2 ๐ ๐ 7 ๐ = 2 ๐ 14
1
1
1 −๐ ๐
๐) =
๐ 7
๐ =
๐ง 2 ๐7๐
2
๐
๐) ๐ง = 2 ๐ −7 ๐
๐๐
๐
๐
10๐๐
๐) (1 + ๐√3) ๐ง = (1 + ๐√3) ⋅ 2 ๐ 7 = 2 ๐ 3 ๐ 2 ๐ 7 ๐ = 4 ๐ 21
−6๐ ๐
7
๐๐
10๐
−4๐
๐) ๐ง10 = 210 ๐ 7 ⋅10 = 210 ๐ 7 ๐ = |๐๐๐๐ ๐๐๐๐ข๐๐๐๐ก | = 210 ๐ 7 ๐
12. First express the complex number z in exponential, and polar form, next express it in
algebraical/canonical form ๐ง = ๐ฅ + ๐๐ฆ.
22
(1 + ๐)22
๐ 22๐
๐6๐
๐
๐
๐
(√2 (๐ ๐ ๐/4 )
−(−
)
5
4
3 = 25 ๐ − 2 ๐ = 25 (cos (− ) + ๐ sin (− ))
๐) ๐ง =
=
=
2
๐
6
−๐ ๐/3 )6
(2
2
2
๐
(1 − ๐√3)
= 0 − 32๐
13*. Calculate the Cartesian coordinates of the point Q obtained by rotating point ๐(2,3) by 60°
around (0,0) (hint: use the multiplication of complex numbers).
๐๐
๐
๐
The point obtained by rotation is ๐′ = (2 + 3๐) ⋅ ๐ 3 = (2 + 3๐) (cos 3 + ๐ sin 3 ) =
1
2
3
2
(2 + 3๐) ( + ๐ √ ),
3√3
3
so ๐′ (1 − 2 , √3 + 2).
14. Calculate and plot in the complex plane, the real and imaginary parts of the indicated complex
numbers, remember there might be more than one value. Where possible find the algebraic values of
the coordinates
๐) √2 √3 − 2๐ ; represent the number in exponential form
2 √3 − 2๐; ๐ = 4 , cos ๐ผ =
๐
√3
,
2
sin ๐ผ = −
1
π
⇔α= −
2
6
2 √3 − 2๐ = 4 ๐ − 6 ๐
√2 √3 − 2๐ ∈ {๐ง0 , ๐ง1 };
๐
๐) √5 + 12๐
(๐ฅ + ๐๐ฆ)2 = 5 + 12๐
๐ฅ 2 − ๐ฆ 2 + 2๐ฅ๐ฆ๐ = 5 + 12๐
๐ฅ2 − ๐ฆ2 = 5
{
2๐ฅ๐ฆ = 12
skฤ
d
11
๐ง0 = 2 ๐ −12 ๐ ,
๐ง0 = 2 ๐ 12 ๐ ๐
two equations are obtained:
6
๐ฆ
๐ฅ = ; ๐ฆ 4 + 5๐ฆ 2 − 36 = 0 so ๐ง0 = 3 + 2๐, ๐ง1 = −3 − 2๐ .
√5 + 12๐ ∈ {3 + 2๐, −3 − 2๐}
โ) √8 + 6๐
Take ๐ง = ๐ฅ + ๐๐ฆ, then ๐ง 2 = 8 + 6๐ and solve for ๐ฅ, ๐ฆ
15. Solve for z:
๐) ๐ง (1 + ๐)2 = 1,
1
1
๐
๐ง=
=
=
−
(1 + ๐)2
2๐
2
๐
1
๐) 3 −
=0
๐ง
27๐
๐
1
=
⇔ ๐ง 3 = −27
๐ง 3 27๐
๐
5๐
⇔ ๐ง = √−27 , ๐ง ∈ { 3 ๐ ๐ 3 , 3 ๐ ๐ ๐ , 3 ๐ ๐ 3 }
3
๐)
๐
๐ง4
= √2 ๐ ๐ 4
๐+1
๐
๐ง4
๐
๐
√2
√2
= √2 ๐ ๐ 4 = √2 (cos + ๐ sin ) = √2 ( + ๐ ) = 1 + ๐
๐+1
4
4
2
2
๐ง4
=1+๐
๐+1
thus:
๐ง 4 = (1 + ๐)(1 + ๐) = 1 + 2๐ + ๐ 2 = 2๐
4
๐ง = √2๐
We need the n-th root formula:
๐
√๐ง = {๐ง0 , ๐ง1 , ๐ง2 , … , ๐ง๐ };
๐ผ 2๐
๐ผ
2๐
๐ผ
2๐
๐
๐
๐( +๐ )
๐ง๐ = √๐ (cos ( + ๐ ) + ๐ sin ( + ๐ )) = √๐ ⋅ ๐ ๐ ๐
๐
๐
๐
๐
๐๐
๐
First we have to express 2๐ in exponential form: 2๐ = 2๐ 2 , i.e. ๐ผ = 2 , ๐ = 2.
4
4
√๐ง = {๐ง0 , ๐ง1 , ๐ง2 , ๐ง3 }; ๐ง๐ = √2 ⋅ ๐
So
4
๐
2๐
๐( 2 +0 )
4
4
4
๐
2๐
๐( 2 +1 )
4
4
4
๐( 2 +2
4
๐( 2 +3
๐ง0 = √2 ⋅ ๐
๐ง1 = √2 ⋅ ๐
๐ง2 = √2 ⋅ ๐
๐ง3 = √2 ⋅ ๐
๐
4
๐
4
2๐
)
4
2๐
)
4
๐/2 2๐
๐(
+๐ )
4
4
๐๐
4
= √2 ⋅ ๐ 8 (๐๐ ๐ด๐๐)
4
๐ ๐
5๐
4
๐( + )
๐( )
8 2 = √2 ⋅ ๐ 8 (๐๐ ๐ด๐๐)
4
๐(
9๐
)
8
4
๐(
13๐
)
8
= √2 ⋅ ๐
= √2 ⋅ ๐
= √2 ⋅ ๐
−7๐
= (๐๐ ๐ด๐๐) = √2 ⋅ ๐ ๐ 8 .
4
−3๐
= (๐๐ ๐ด๐๐) = √2 ⋅ ๐ ๐ 8 .
4
We sketch 4 vertices of a square: ๐ง0 , ๐ง1 , ๐ง2 , ๐ง3 .
2
16. Let ๐ง1 = 3๐ + ๐ 2 , ๐ง2 = 1−๐
e) give the geometric interpretation of the above operations (sum, product, cubic root, power) and plot
the results.
ans to e) :
a) when adding two complex numbers we add corresponding two vectors with beginning at 0 and end
points at these numbers;
b) all the n-th roots of a complex number z, lie on a circle of centre 0 and of radius ๐√|๐ง| . Each one
1
of them is obtained by rotating a chosen root (e.g. one with the argument ๐ด๐๐ ๐ง) by the angle equal
๐
to multiplicity of 2 ๐. The roots are the vertices of a regular n-gon .
๐
c) when multipying complex numbers we multiply the absolute values and add the arguments
d) when taking the n-the power of a complex number we the n-the power of the absolute value and the
n-th multiplicty of the argument.
17*. Use the de Moivre’a Formula to determine the dependence of sin 2๐ผ and cos 2๐ผ on the sin ๐ผ
and cos ๐ผ (i.e. formulas for sin 2๐ผ and cos 2๐ผ which contain only ‘sin ๐ผ’ and ‘cos ๐ผ’).
Let ๐ง = cos ๐ผ + ๐ sin ๐ผ , then form the de Moivre′ a Formula:
๐ง 2 = cos 2๐ผ + ๐ sin 2๐ผ
On the other hand, if we square both sides of the equation ๐ง = cos ๐ผ + ๐ sin ๐ผ ,
then
๐ง 2 = (cos ๐ผ + ๐ sin ๐ผ)2 = cos2 ๐ผ − sin2 ๐ผ + 2๐ sin ๐ผ cos ๐ผ, thus
cos 2๐ผ + ๐ sin 2๐ผ = cos 2 ๐ผ − sin2 ๐ผ + 2๐ sin ๐ผ cos ๐ผ , real and imaginary parts should be equal:
cos 2๐ผ = cos 2 ๐ผ − sin2 ๐ผ ; sin 2๐ผ = 2 sin ๐ผ cos ๐ผ .
18*. Use the exponential form to solve
๐)
|๐ง|2 ๐ง
๐ง
3
= −1, ๐ง ≠ 0
|๐ง|2 |๐ง|๐ ๐ ๐ผ
|๐ง|๐ ๐ ๐ผ
3
i niech ๐ง = |๐ง|๐ ๐ ๐ผ . then the equation becomes:
= −1
|๐ง|2 |๐ง|๐ ๐ ๐ผ
= −1 ⇔ ๐ ๐4 ๐ผ = −1
|๐ง|3 ๐ −๐3 ๐ผ
⇔ ๐ ๐4 ๐ผ = ๐ ๐(๐+2๐π) ⇔ 4 ๐ผ = ๐ + 2๐π
and finally
we obtain z number with an arbitrary |๐ง|a nd four different arguments:
๐ผ=
๐ ๐π
+
4
2
๐ = 0,1,2,3.
1
√3
19*. Write ๐ง = 2 + ๐ 2 in exponential form
a) calculate all the possible integer powers ๐ง ๐ , ๐ ∈ ๐ผ, (๐ผ = Integers)
1
2
First we express ๐ง = + ๐
√3
2
1
2
in exponential form: ๐ง = + ๐
√3
2
๐
๐
= ๐ ๐( 3 +2๐๐) , so ๐ง ๐ = ๐ ๐ ๐ 3 ,
(๐ ⋅ 2๐๐ is a multiplicity of the period 2๐๐). The periodicity of sine and cosine causes that there are
only 6 different values of the powers of n i.e. for ๐ = 6๐, 6๐ + 1, 6๐ + 2, … , 6๐ + 5, ๐ ∈ ๐ผ
so
1
๐๐๐ ๐ = 6๐
1 ๐√3
+
๐๐๐ ๐ = 6๐ + 1
2
2
1 ๐√3
− +
๐๐๐ ๐ = 6๐ + 2
๐
2
2
๐ง๐ = ๐๐ ๐ 3 =
−1
๐๐๐ ๐ = 6๐ + 3
1 ๐√3
− −
๐๐๐ ๐ = 6๐ + 4
2
2
1 ๐√3
๐๐๐ ๐ = 6๐ + 5
{ 2− 2
๐) ๐ง ๐ ,
where ๐ is the imaginary unit ๐ 2 = −1.
1
√3
1
๐
√3
Express ๐ง = 2 + ๐ 2 in exponential form; ๐ง = 2 + ๐ 2 = ๐ ๐( 3 +2๐๐) ,
๐
๐
1
√3
๐ ∈ ๐ผ,
๐
So ๐ง ๐ = ๐ ๐⋅๐ ( 3 +2๐๐) = ๐ − ( 3 +2๐๐) . This means that (2 + ๐ 2 ) are infinitely many real numbers. .
20 . Calculate the power ๐ ๐ , where ๐ is the imaginary unit ๐ 2 = −1.
๐ ๐ = cos 1 + ๐ sin 1 .
21. Find all the complex roots of the equations
a) According to the Rational Root Test, it is easily seen that -1 is a root of the polynomial. We divide
the polynomial by (x + 1) to obtain that ๐ฅ 3 − ๐ฅ 2 + 3๐ฅ + 5 = (๐ฅ + 1)(๐ฅ 2 − 2๐ฅ + 5), next we find the
roots of the quadratic polynomial ๐ฅ 2 − 2๐ฅ + 5.
๐) 2๐ง 3 + 4๐ง 2 + 3๐ง + 6 = (๐ง + 2)(2๐ง 3 + 3) ,
3
3
roots ๐ง1 = ๐ √ , ๐ง2 = −๐√ , ๐ง3 = −2.
2
2
7
3
1
๐
) ๐ง 3 − ๐ง 2 − ๐ง − = 0 | ⋅ 6 ⇔ 6๐ง 3 − 7๐ง 2 − 9๐ง − 2 = 0
6
2
3
6๐ง 3 − 7๐ง 2 − 9๐ง − 2 = (๐ง − 2)(6๐ง 2 + 5๐ง + 1)
22. Let
๐) ๐ง1 = −๐√2, ๐ง2 = ๐ be two of the roots of ๐ง 6 − 2๐ง 5 + 5๐ง 4 − 6๐ง 3 + 8๐ง 2 − 4๐ง + 4 = 0 find all the
other roots.
๐ง 6 − 2๐ง 5 + 5๐ง 4 − 6๐ง 3 + 8๐ง 2 − 4๐ง + 4 = (๐ง + ๐ √2)(๐ง − ๐ √2)(๐ง − 1)(๐ง + ๐)(๐ง 2 − 2๐ง + 2)
The equation ๐ง 2 − 2๐ง + 2 has the following two roots ๐ง5 = 1 + ๐, ๐ง6 = 1 − ๐ .
23. Write a polynomial with real coefficients of the fourth degree which has the following roots:
๐ง1 = 1 − ๐, ๐ง2 = 3๐.
e. g. (๐ง − (1 − ๐))(๐ง − (1 + ๐))(๐ง − 3๐)(๐ง + 3๐) = (๐ง 2 − 2๐ง + 2)(๐ง 2 + 9)
= ๐ง 4 − 2๐ง 3 + 11๐ง 2 − 18๐ง + 18.
0
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