Linear Algebra
Hints and solutions to selected exercises
Homework 1.
1. Determine the following:
𝑏) 𝑅𝑒 [
2 + 5𝑖
],
𝑖−1
𝐼𝑚 [
2 + 5𝑖
]
𝑖−1
2 + 5𝑖 −1 − 𝑖 −2 − 2𝑖 − 5𝑖 − 5𝑖 2 −2 + 5 − 7𝑖
⋅
=
=
;
−1 + 𝑖 −1 − 𝑖
1 − 𝑖2
2
𝑅𝑒 [
2 + 5𝑖
3
2 + 5𝑖
7
] = ; 𝐼𝑚 [
]=−
−1 + 𝑖
2
−1 + 𝑖
2
g) For real numbers the absolute value can be split up : e.g. |𝑥 ⋅ (𝑥 + 2)| = |𝑥| ⋅ |𝑥 + 2| . The same
happens for complex numbers.
|
𝑖 7 (1 + 𝑖)8
|𝑖|7 ⋅ |(1 + 𝑖)| 8
=
|
12
12
(√2 − 𝑖√6 )
|√2 − 𝑖√6|
We calculate the specific absolute values:
|𝑖| = 1
|1 + 𝑖| = √1 + 1 = √2
|√2 − 𝑖√6| = √2 + 6 = √8
now we see
|𝑖|7 ⋅ |(1 + 𝑖)| 8
12
8
=
1 ⋅ (√2)
|√2 − 𝑖√6|
ℎ) 𝐼𝑚 [
=
12
=
|√8|
24
86
𝑖7
𝑖6 𝑖
(𝑖 2 )3 𝑖
1
−𝑖
]
=
𝐼𝑚
[
]
=
𝐼𝑚
[
] = 4 𝐼𝑚 [
]=
4
4
4
4
2
2
[1 − 2𝑖 − 1]2
(2 − 2𝑖)
2 (1 − 𝑖)
2 [(1 − 𝑖) ]
2
1
−𝑖
1
−𝑖
11
1
𝐼𝑚 [
] = 4 𝐼𝑚 [
]= 4 = 6
4
2
[−2𝑖]
2
2
4(−1)
2 4 2
2. Find the absolute value |𝑧|, the Real and Imaginary parts of 𝑧: 𝑅𝑒(𝑧), 𝐼𝑚(𝑧), for
𝑒) 𝑧 =
(1 + 𝑖)(1 + 𝑖)2 (1 + 𝑖)3 … (1 + 𝑖)20
𝑖 0 + 𝑖 2 + 𝑖 4 + 𝑖 6 + ⋯ + 𝑖 20
𝐍𝐮𝐦𝐞𝐫𝐚𝐭𝐨𝐫 ∶ (1 + 𝑖)(1 + 𝑖)2 (1 + 𝑖)3 … (1 + 𝑖)20 = (1 + 𝑖)1+2+3+⋯+20 = (1 + 𝑖)10⋅21
= (1 + 𝑖)2⋅5⋅21
we know that (1 + 𝑖)2 = 1 + 2𝑖 − 1 = 2𝑖
105
thus the numerator = 2
and that
21
((2𝑖)5 )
21
= (25 𝑖 )
= 25⋅21 𝑖,
𝑖.
𝐃𝐞𝐧𝐨𝐦𝐢𝐧𝐚𝐭𝐨𝐫: there are different methods here , one of them is the following:
𝑖 0 + 𝑖 2 + 𝑖 4 + 𝑖 6 +. . +𝑖 20 = 1 + (−1) + (−1)2 + (−1)3 +. . +(−1)10 .
We notice that the pairs ∶ 0. i 1. 𝑖. 𝑒. 10 + (−1)1 ;
2. i 3. 𝑖. 𝑒. (−1)2 + (−1)3 ;
4. i 5. 𝑖. 𝑒. (−1)4 + (−1)5 𝑒𝑡𝑐..
cancel out each other, leaving only (−1)10 = 1
Thus the denominator = 1, so 𝑧 = 2105 𝑖, 𝑅𝑒 (𝑧) = 0, 𝐼𝑚(𝑧) = 2105 ; |𝑧| = 2105 .
3. Solve the following equations for z  C , it is possible that there are no solutions or there are more
than one.
𝒅) 𝑖𝑧 2 − 𝑧 + 2𝑖 = 0;
𝑧1 =
Δ = 1 − 4 ⋅ 𝑖 ⋅ 2𝑖 = 1 + 8 = 9;
1+3 2
1 − 3 −2
= = −2𝑖 ; 𝑧2 =
=
= 𝑖 − two solutions.
2𝑖
𝑖
2𝑖
2𝑖
g) 𝑧 2 = 3 + 4𝑖
(𝑥 + 𝑖𝑦)2 = 3 + 4𝑖
𝑥 2 − 2𝑖𝑥𝑦 − 𝑦 2 = 3 + 4𝑖
𝑥2 − 𝑦2 = 3
𝑥2 − 𝑦2 = 3
{
{
2
2𝑥𝑦 = 4
𝑦=𝑥
4
4
{
2
𝑥 − 3𝑥 − 4 = 0
𝑥2 − 𝑥2 = 3
2
𝑦=𝑥
𝑥4 − 3𝑥2 − 4 = 0
2
𝑦=𝑥
{
(𝑥1 )2 = 4: 𝑦1,1 = 1; 𝑦1,2 = −1
{
(𝑥2 )2 ≠ −1 𝑏𝑒𝑐𝑎𝑢𝑠𝑒 𝑥 ∈ 𝑅
𝑎𝑛𝑠. ∶ 𝑧1 = 2 + 𝑖 ; 𝑧2 = −2 − 𝑖
𝒇) 2𝑧 + (1 + 𝑖)𝑧 = 1 − 3𝑖
2(𝑥 + 𝑖𝑦) + (1 + 𝑖)(𝑥 − 𝑖𝑦) = |collect real and imaginary parts| = (3𝑥 + 𝑦) + 𝑖(𝑥 + 𝑦)
(3𝑥 + 𝑦) + 𝑖(𝑥 + 𝑦) = 1 − 3𝑖
real and imaginary parts on both sides of equation are equal:.
3𝑥 + 𝑦 = 1
now you have to solve this system
{
𝑥 + 𝑦 = −3
{
𝑥=2
.
𝑦 = −5
𝒉) (𝑧 + 𝑧) + 2(𝑧 − 𝑧) = 3 + 8𝑖; let 𝑧 = 𝑥 + 𝑖𝑦
(𝑥 + 𝑖𝑦 + 𝑥 − 𝑖𝑦) + 2(𝑥 + 𝑖𝑦 − 𝑥 + 𝑖𝑦) = 3 + 8𝑖
2𝑥 + 4𝑖𝑦 = 3 + 8𝑖
2𝑥 = 3
{
4𝑦 = 8
3
2
𝑧 = + 2𝑖.
𝒋) 𝑧 − 𝑖 = 2𝑧 + 1 ; let 𝑧 = 𝑥 + 𝑖𝑦
𝑥 + 𝑖𝑦 − 𝑖 = 2𝑥 + 2𝑖𝑦 + 1
𝑥 + 𝑖(𝑦 − 1) = (2𝑥 + 1) + 2𝑖𝑦
𝑥 − 𝑖(𝑦 − 1) = (2𝑥 + 1) + 𝑖𝑦
𝑥 = −1
𝑥 = 2𝑥 + 1
1
{
{
−(𝑦 − 1) = 2𝑦
𝑦=
3
1
𝑧 = −1 + 𝑖
3
4. Sketch the following sets in the complex plane
−3 + 2𝑖
]}
2−𝑖
−3 + 2𝑖
−3 + 2𝑖 2 + 𝑖
−6 − 3𝑖 + 4𝑖 − 2
8
𝑅𝑒 [
⋅
] = 𝑅𝑒 [
] = 𝑅𝑒 [
]=−
2−𝑖
2−𝑖 2+𝑖
4+1
5
𝑎 ) 𝑆 = {𝑧 ∈ 𝐶 ∶ 𝑅𝑒[𝑧] ≤ 𝑅𝑒 [
8
𝑅𝑒[𝑧] = 𝑥, 𝑠𝑜 𝑡ℎ𝑒 𝑝𝑜𝑖𝑛𝑡𝑠 𝑓𝑟𝑜𝑚 𝑆 𝑠𝑎𝑡𝑖𝑠𝑓𝑦 𝑡ℎ𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑥 ≤ − 5 ∶
5. Calculate the argument arg (𝑧) and the main argument, 𝐴𝑟𝑔(𝑧), of 𝑧.
𝑎) 𝑎𝑟𝑔(1 − 𝑖) , 𝐴𝑟𝑔(1 − 𝑖)
𝜋
𝜋
𝐴𝑟𝑔 (1 − 𝑖) = − ; arg(1 − 𝑖) = − + 2𝑘 π, k ∈ I.
4
4
𝑐) 𝐴𝑟𝑔(√2 − 𝑖√6)
𝑟 = √2 + 6 = √8 = 2√2
1
√2
cos 𝛼 =
=
2√2 2
√6
√3
sin 𝛼 = −
=−
2
{
2√2
𝜋
𝐴𝑟𝑔(√2 − 𝑖√6) = − ;
3
→
𝛼= −
𝜋
3
arg( √2 − 𝑖√6) = −𝜋/3 + 2𝑘𝜋
6. Plot the following points and find the polar form (trigonometric form) and the exponential form
of
2
2
𝜋
𝒄) 𝑧 = −2 + 2𝑖; 𝑟 = √4 + 4 = 2 √2 ; cos 𝛼 = −
, sin 𝛼 =
→ 𝛼=3
4
2 √2
2 √2
3𝜋
1
1
3𝜋
3𝜋
𝑖
𝑧 = −2 + 2𝑖 = 2√2 (−
+
𝑖) = 2√2 (cos
+ 𝑖 sin ) = 2√2 𝑒 4
4
4
√2 √2
𝒉) 𝑧 = −4 − 𝑖 4√3;
𝑟 = √16 + 48 = 8 ; cos 𝛼 = −
4
4√3
2𝜋
, sin 𝛼 =
→ 𝛼=−
8
8
3
−2𝜋
1 √3
−2𝜋
−2𝜋
𝑧 = −4 − 𝑖 4√3 = 8 (− −
𝑖) = 8 (cos
+ 𝑖 sin
)=8 𝑒 3 𝑖
2
2
3
3
y
−4
x
−4√3
i) 𝑧 = 𝑖 33 + (1 + 𝑖)4 = 𝑖 − 4; |𝑧| = √17
Plot 𝑧 on the complex plane
Calculate 𝛼 using inverse trigonometruc functions eg..
4
4
4
cos 𝛽 = 17, giving 𝛽 = arccos ( 17) and 𝛼 = 𝜋 − arccos ( 17) .
√
√
√
7. Sketch the following sets in the complex plane, mark the main points (wedges and circles)
𝒄) 𝑆 = {𝑧 ∈ 𝐶: |𝑧 + 3 − 2𝑖| ≤ 2 }
|𝑧 + 3 − 2𝑖| ≤ 2 first we can plot the points 𝑧̅ , next take a mirror projection about axis OX of these
points:
or transform the inequality :
|𝑧 + 3 − 2𝑖| ≤ 2
|𝑧 + 3 + 2𝑖| ≤ 2
|𝑧 + 3 + 2𝑖| ≤ 2
|𝑧 + 3 + 2𝑖| ≤ 2 𝑖𝑛𝑠𝑖𝑑𝑒 𝑜𝑓 𝑐𝑖𝑟𝑐𝑙𝑒 , 𝑐𝑒𝑛𝑡𝑟𝑒 𝑐(−3, −2), 𝑟𝑎𝑑𝑖𝑢𝑠 𝑟 ≤ 2.
𝑓) 𝑆 = {𝑧 ∈ 𝐶: −
3𝜋
𝜋
≤ 𝑎𝑟𝑔(𝑧) ≤ }
4
2
g) 𝑆 = {𝑧 ∈ 𝐶: −
3𝜋
𝜋
≤ 𝑎𝑟𝑔( 𝑧 ) ≤ 2 }. The set from f ) is the set of points 𝑧̅, so now, to obtain S,
4
we should mirror this set with respect to the OX axis
h) 𝑆 = {𝑧 ∈ 𝐶: −
3𝜋
4
𝜋
≤ 𝑎𝑟𝑔(𝑧 − 2 + 𝑖) ≤ 2 }. First plot the points 𝑧1 = 𝑧 − 2 + 𝑖, next to obtain
S, points 𝑧 are shifted by 2 − 𝑖:
𝑘) 𝑆 = {𝑧 ∈ 𝐶: −
𝜋
𝜋
≤ 𝐴𝑟𝑔((𝑖 + 1) ⋅ 𝑧)) ≤ }
2
4
The argument of a product of complex numbers is the sum of the arguments i.e.
if 𝑧1 = 𝑟1 𝑒 𝑖 𝛼1 𝑎𝑛𝑑 𝑧2 = 𝑟2 𝑒 𝑖 𝛼2 , then 𝑧1 𝑧2 = 𝑟1 𝑒 𝑖 𝛼1 𝑟2 𝑒 𝑖 𝛼2 = 𝑟1 𝑟2 𝑒 𝑖 (𝛼1 +𝛼2) .
𝜋
So arg(𝑧1 𝑧2 ) = 𝛼1 + 𝛼2 . Let 𝐴𝑟𝑔(𝑧) = 𝛼, we know that Arg(1 + 𝑖) = 4 , and
𝜋
𝐴𝑟𝑔((𝑖 + 1) ⋅ 𝑧)) = 4 + 𝛼.
𝜋
𝜋
− 2 ≤ 𝐴𝑟𝑔((𝑖 + 1) ⋅ 𝑧)) ≤ 4 ,
𝜋 𝜋
𝜋
≤ + 𝛼 + 2𝑘𝜋 ≤ , 𝑎𝑛𝑑
2 4
4
𝜋
Condition for 𝐴𝑟𝑔((1 + 𝑖)𝑧) : − 𝜋 < + 𝛼 + 2𝑘𝜋 ≤ 𝜋, only for 𝑘 = 0
4
3𝜋
−
≤ 𝛼 ≤ 0. This is an infinite wedge
4
−
𝑙 ∗ ) 𝑆 = {𝑧 ∈ 𝐶:
𝜋
𝜋
≤ 𝐴𝑟𝑔(𝑧 3 ) ≤ }
2
2
Let 𝐴𝑟𝑔(𝑧) = 𝛼, then
−
𝜋
𝜋
≤ 3𝛼 + 2𝑘𝜋 ≤
2
2
𝜋
𝜋
− 2𝑘𝜋 ≤ 3𝛼 ≤ − 2𝑘𝜋
2
2
𝜋 2𝑘𝜋
𝜋 2𝑘𝜋
− −
≤𝛼≤ −
true only for 𝑘 = −1,0,1
6
3
6
3
−
3𝜋
5𝜋
≤𝛼≤
6
6
𝜋
𝜋
𝑘 = 0: − ≤ 𝛼 ≤
6
6
5𝜋
3𝜋
𝑘 = 1: −
≤𝛼≤−
6
6
𝑘 = −1:
𝑧
𝑚) 𝑆 = {𝑧 ∈ 𝐶: 0 ≤ 𝐴𝑟𝑔 ( ) ≤ 𝐴𝑟𝑔(3 + 3𝑖) }
𝑖
𝜋
𝜋
𝐴𝑟𝑔(𝑧) = 𝛼, 𝐴𝑟𝑔(𝑖 ) =
and 𝐴𝑟𝑔(3 + 3𝑖) = 𝐴𝑟𝑔(1 + 𝑖) = , so
2
4
𝑧
0 ≤ 𝐴𝑟𝑔 ( ) ≤ 𝐴𝑟𝑔(3 + 3𝑖)
𝑖
𝑧
𝜋
Condition for 𝐴𝑟𝑔 ( ) : − 𝜋 < 𝛼 − + 2𝑘𝜋 ≤ 𝜋,
𝑜𝑛𝑙𝑦 𝑓𝑜𝑟 𝑘 = 0
𝑖
2
𝜋 𝜋
π
0≤𝛼− ≤
add
to both sides.
2 4
2
𝜋
3𝜋
≤𝛼≤
2
4
Plot a wedge.
𝑛)
𝑆 = {𝑧 ∈ 𝐶: 𝐴𝑟𝑔(1 − 3𝑖) ≤ 𝐴𝑟𝑔 𝑧 ≤ 𝐴𝑟𝑔(−2 + 5𝑖)}
Simply plot the numbers 𝑧1 = 1 − 3𝑖 oraz 𝑧2 = −2 + 5𝑖 i.e. points 𝑃1 (1, −3), 𝑃2 (−2,5) on the
plane, the region is an infinite wedge between the line passing through points (0,0), 𝑃1 (1, −3) and the
line passing through (0,0), 𝑃2 (−2,5).
8. Sketch the following sets in the complex plane, mark the main points
𝒃) 𝑆 = {𝑧 ∈ 𝐶: 𝐼𝑚 [(1 + 2𝑖)𝑧 − 3𝑖] < 0},
𝑎𝑛𝑠. : let 𝑧 = 𝑥 + 𝑖𝑦, then
𝐼𝑚 [(1 + 2𝑖)𝑧 − 3𝑖] = 𝐼𝑚 [(1 + 2𝑖)(𝑥 + 𝑖𝑦) − 3𝑖] = 𝐼𝑚 [(𝑥 − 2𝑦) + 𝑖(2𝑥 + 𝑦 − 3)]:
2𝑥 + 𝑦 − 3 < 0 ⇔ 𝑦 < −2𝑥 + 3.
𝒅) 𝑆 = {𝑧 ∈ 𝐶: 𝑧 + 𝑖 = 𝑧 − 1},
𝑧+𝑖 = 𝑧−1
𝑧−𝑖 =𝑧−1
𝑧 − 𝑧 = 𝑖 − 1, let 𝑧 = 𝑥 + 𝑖𝑦, then since the real part are equal and the imaginary parts are equal:
𝑥 + 𝑖𝑦 − (𝑥 + 𝑖𝑦) = 𝑖 − 1
𝑥 − 𝑖𝑦 − (𝑥 + 𝑖𝑦) = 𝑖 − 1
−2𝑖𝑦 = 𝑖 − 1;
f)
𝑆 = ∅.
𝒈) 𝑆 = {𝑧 ∈ 𝐶: |𝑖𝑧 + 1 − 𝑖| < 2},
1
|𝑖𝑧 + 1 − 𝑖| = |𝑖| ⋅ |𝑧 + − 1| = |𝑖| ⋅ |𝑧 − 𝑖 − 1| = |𝑧 − (𝑖 + 1)| < 2;
𝑖
inside of circle, centre 𝑐(1,1), radius 𝑟 = 2. Sketch it.
𝒉) 𝑆 = {𝑧 ∈ 𝐶: |𝑧 − 𝑖 + 1| < 3}:
|𝑧 − 𝑖 + 1| = |𝑧 − 𝑖 + 1 | = |𝑧 − (−1 + 𝑖) | < 3; inside of circle, centre c(-1,1) i radius r = 3.
Sketch it.
𝒊) 𝑆 = {𝑧 ∈ 𝐶: |𝑧 − 2𝑖| + |𝑧 + 2𝑖| = 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.
𝒋) 𝑆 = {𝑧 ∈ 𝐶: |𝑧̅ + 𝑖| < 2}
Method 1: |𝑧̅ + 𝑖| = |𝑧̅ − 𝑖| = |𝑧̅̅̅̅̅̅
− 𝑖 | = |𝑧 − 𝑖| < 2
̅̅̅̅̅̅̅̅
or Method 2: |𝑥
+ 𝑖𝑦 + 𝑖| = |𝑥 − 𝑖𝑦 + 𝑖| = |𝑥 + 𝑖(−𝑦 + 1) = √𝑥 2 + (𝑦 − 1)2 < 2.
Inside of circle, centre c(0,1), radius r = 2. Sketch it
𝒌) 𝑆 = { 𝑧 ∈ 𝐶: Im[(2 + 𝑖)(3 + 5𝑖)] ≥ |𝑧 − 3 + 𝑖| ≥ |√5 + 2𝑖 |
∧
𝜋
Arg(3 − 𝑖) ≤ arg(𝑧) ≤ Arg [𝑒 𝑖 2 ] }
First inequality: we calculate the values on the left- and right-hand sides of
Im[(2 + 𝑖)(3 + 5𝑖)] ≥ |𝑧 − 3 + 𝑖| ≥ |√5 + 2𝑖 |
LHS: Im[(2 + 𝑖)(3 + 5𝑖)] = 𝐼𝑚 [6 + 10𝑖 + 3𝑖 + 5𝑖 2 ] = 13
2
RHS: |√5 + 2𝑖 | = √(√5) + 22 = √5 + 4 = 3
The "in-between"
|𝑧 − 3 + 𝑖| = |𝑧 − (3 − 𝑖)| 𝑡ℎ𝑒 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑠𝑜𝑚𝑒 'z" 𝑎𝑛𝑑 𝑧0 = 3 − 𝑖
Im[(2 + 𝑖)(3 + 5𝑖)] ≥ |𝑧 − 3 + 𝑖| ≥ |√5 + 2𝑖 | means
13 ≥ |𝑧 − (3 − 𝑖)| ≥ 3 this is a ring with centre at P(3,-1) and radii between 3, 13.
The second inequality:
𝜋
Arg(3 − 𝑖) ≤ arg(𝑧) ≤ Arg [𝑒 𝑖 2 ]
Arg(3 − 𝑖) angle between the positive OX axis and the point P(3,-1), we don not calculate the
argument – ONLY plot it.
𝜋
𝜋
Arg [𝑒 𝑖 2 ] = 2 − simple,
𝜋
so Arg(3 − 𝑖) ≤ arg(𝑧) ≤ Arg [𝑒 𝑖 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 𝑆, top, right part of ring.
9. Find Arg(z), |z| for the following complex numbers
𝑖 𝜋 15
𝑎) 𝑧 = (𝑒 5 )
𝑎𝑟𝑔(𝑧) = 15
𝜋
= 3𝜋 ∶ 𝐴𝑟𝑔(𝑧) = 𝜋; |𝑧| = 1
5
𝑖𝜋
𝑖𝜋
3
3
10
𝑖𝜋
𝑖 3𝜋 𝑖 𝜋
𝑏) 𝑧 = (1 + 𝑖)3 ⋅ 𝑒 4 = (√2 𝑒 4 ) ⋅ 𝑒 4 = 2 √2 𝑒 4 + 4 = 2 √2 𝑒 𝑖 𝜋 ∶
𝐴𝑟𝑔(𝑧) = 𝜋; |𝑧| = 2 √2
(3 √2 𝑒 4𝜋𝑖 )
10
𝑐) 𝑧 =
(−3 + 3𝑖)
=
4
𝑖𝜋
𝑖𝜋
(𝑒 3 )
10
= (3 √2 )
4
30
4
37
𝑒 4 𝜋 𝑖−3𝜋 𝑖 = 310 25 𝑒 6 𝜋 𝑖 ;
(𝑒 3 )
37
36 1
𝜋
𝜋
𝜋 = ( + ) 𝜋 = + 2 ⋅ 3𝜋; ∶ 𝐴𝑟𝑔(𝑧) = ; |𝑧| = 310 25
6
6 6
6
6
10. Let 𝑧 = −1 + 𝑖, write the following complex numbers in exponential form
3𝜋
First write 𝑧 in exponential form: 𝑧 = −1 + 𝑖 = √2 𝑒 4 𝑖 ;
3𝜋
3𝜋
7𝜋
𝜋
𝑎)
− 𝑧 = −1 ⋅ √2 𝑒 4 𝑖 = 𝑒 𝜋𝑖 √2 𝑒 4 𝑖 = √2 𝑒 4 𝑖 = √2 𝑒 − 4 𝑖
𝑏)
𝑖𝑧 = 𝑒 2 𝑖 √2 𝑒 4 𝑖 = √2 𝑒 4 𝑖 = √2 𝑒 − 4 𝑖
𝑐)
1
=
𝑧
𝜋
3𝜋
1
=
3𝜋
𝑖
√2 𝑒 4
5𝜋
1
√2
𝜋
3𝜋
3𝜋
𝑒− 4 𝑖
𝜋
11. Let 𝑧 = 2 (cos 7 + 𝑖 sin 7 ), write in exponential form
𝜋
𝜋
𝜋
First write 𝑧 in exponential form 𝑧 = 2 (cos + 𝑖 sin ) = 2 𝑒 7 𝑖
7
7
𝜋𝑖
𝜋
8𝜋 𝑖
𝑎) − 𝑧 = −1 ⋅ 2 𝑒 7 = 2 𝑒 𝜋𝑖 𝑒 7 𝑖 = 2 𝑒 7 = | 𝑚𝑎𝑖𝑛 𝑎𝑟𝑔𝑢𝑚𝑒𝑛𝑡 | = 2𝑒
𝜋𝑖
𝜋
𝜋
9𝜋 𝑖
𝑏) 𝑖𝑧 = 𝑖 ⋅ 2 𝑒 7 = 2 𝑒 2 𝑖 𝑒 7 𝑖 = 2 𝑒 14
1
1
1 −𝜋 𝑖
𝑐) =
𝑒 7
𝜋 =
𝑧 2 𝑒7𝑖
2
𝜋
𝑑) 𝑧 = 2 𝑒 −7 𝑖
𝜋𝑖
𝜋
𝜋
10𝜋𝑖
𝑒) (1 + 𝑖√3) 𝑧 = (1 + 𝑖√3) ⋅ 2 𝑒 7 = 2 𝑒 3 𝑖 2 𝑒 7 𝑖 = 4 𝑒 21
−6𝜋 𝑖
7
𝜋𝑖
10𝜋
−4𝜋
𝑓) 𝑧10 = 210 𝑒 7 ⋅10 = 210 𝑒 7 𝑖 = |𝑚𝑎𝑖𝑛 𝑎𝑟𝑔𝑢𝑚𝑒𝑛𝑡 | = 210 𝑒 7 𝑖
12. First express the complex number z in exponential, and polar form, next express it in
algebraical/canonical form 𝑧 = 𝑥 + 𝑖𝑦.
22
(1 + 𝑖)22
𝑖 22𝜋
𝑖6𝜋
𝜋
𝜋
𝜋
(√2 (𝑒 𝑖 𝜋/4 )
−(−
)
5
4
3 = 25 𝑒 − 2 𝑖 = 25 (cos (− ) + 𝑖 sin (− ))
𝒄) 𝑧 =
=
=
2
𝑒
6
−𝑖 𝜋/3 )6
(2
2
2
𝑒
(1 − 𝑖√3)
= 0 − 32𝑖
13*. Calculate the Cartesian coordinates of the point Q obtained by rotating point 𝑃(2,3) by 60°
around (0,0) (hint: use the multiplication of complex numbers).
𝑖𝜋
𝜋
𝜋
The point obtained by rotation is 𝑃′ = (2 + 3𝑖) ⋅ 𝑒 3 = (2 + 3𝑖) (cos 3 + 𝑖 sin 3 ) =
1
2
3
2
(2 + 3𝑖) ( + 𝑖 √ ),
3√3
3
so 𝑃′ (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
𝑓) √2 √3 − 2𝑖 ; represent the number in exponential form
2 √3 − 2𝑖; 𝑟 = 4 , cos 𝛼 =
𝜋
√3
,
2
sin 𝛼 = −
1
π
⇔α= −
2
6
2 √3 − 2𝑖 = 4 𝑒 − 6 𝑖
√2 √3 − 2𝑖 ∈ {𝑧0 , 𝑧1 };
𝜋
𝑔) √5 + 12𝑖
(𝑥 + 𝑖𝑦)2 = 5 + 12𝑖
𝑥 2 − 𝑦 2 + 2𝑥𝑦𝑖 = 5 + 12𝑖
𝑥2 − 𝑦2 = 5
{
2𝑥𝑦 = 12
skąd
11
𝑧0 = 2 𝑒 −12 𝑖 ,
𝑧0 = 2 𝑒 12 𝜋 𝑖
two equations are obtained:
6
𝑦
𝑥 = ; 𝑦 4 + 5𝑦 2 − 36 = 0 so 𝑧0 = 3 + 2𝑖, 𝑧1 = −3 − 2𝑖 .
√5 + 12𝑖 ∈ {3 + 2𝑖, −3 − 2𝑖}
ℎ) √8 + 6𝑖
Take 𝑧 = 𝑥 + 𝑖𝑦, then 𝑧 2 = 8 + 6𝑖 and solve for 𝑥, 𝑦
15. Solve for z:
𝒄) 𝑧 (1 + 𝑖)2 = 1,
1
1
𝑖
𝑧=
=
=
−
(1 + 𝑖)2
2𝑖
2
𝑖
1
𝒆) 3 −
=0
𝑧
27𝑖
𝑖
1
=
⇔ 𝑧 3 = −27
𝑧 3 27𝑖
𝜋
5𝜋
⇔ 𝑧 = √−27 , 𝑧 ∈ { 3 𝑒 𝑖 3 , 3 𝑒 𝑖 𝜋 , 3 𝑒 𝑖 3 }
3
𝒇)
𝜋
𝑧4
= √2 𝑒 𝑖 4
𝑖+1
𝜋
𝑧4
𝜋
𝜋
√2
√2
= √2 𝑒 𝑖 4 = √2 (cos + 𝑖 sin ) = √2 ( + 𝑖 ) = 1 + 𝑖
𝑖+1
4
4
2
2
𝑧4
=1+𝑖
𝑖+1
thus:
𝑧 4 = (1 + 𝑖)(1 + 𝑖) = 1 + 2𝑖 + 𝑖 2 = 2𝑖
4
𝑧 = √2𝑖
We need the n-th root formula:
𝑛
√𝑧 = {𝑧0 , 𝑧1 , 𝑧2 , … , 𝑧𝑘 };
𝛼 2𝜋
𝛼
2𝜋
𝛼
2𝜋
𝑛
𝑛
𝑖( +𝑘 )
𝑧𝑘 = √𝑟 (cos ( + 𝑘 ) + 𝑖 sin ( + 𝑘 )) = √𝑟 ⋅ 𝑒 𝑛 𝑛
𝑛
𝑛
𝑛
𝑛
𝑖𝜋
𝜋
First we have to express 2𝑖 in exponential form: 2𝑖 = 2𝑒 2 , i.e. 𝛼 = 2 , 𝑟 = 2.
4
4
√𝑧 = {𝑧0 , 𝑧1 , 𝑧2 , 𝑧3 }; 𝑧𝑘 = √2 ⋅ 𝑒
So
4
𝜋
2𝜋
𝑖( 2 +0 )
4
4
4
𝜋
2𝜋
𝑖( 2 +1 )
4
4
4
𝑖( 2 +2
4
𝑖( 2 +3
𝑧0 = √2 ⋅ 𝑒
𝑧1 = √2 ⋅ 𝑒
𝑧2 = √2 ⋅ 𝑒
𝑧3 = √2 ⋅ 𝑒
𝜋
4
𝜋
4
2𝜋
)
4
2𝜋
)
4
𝜋/2 2𝜋
𝑖(
+𝑘 )
4
4
𝑖𝜋
4
= √2 ⋅ 𝑒 8 (𝑖𝑛 𝐴𝑟𝑔)
4
𝜋 𝜋
5𝜋
4
𝑖( + )
𝑖( )
8 2 = √2 ⋅ 𝑒 8 (𝑖𝑛 𝐴𝑟𝑔)
4
𝑖(
9𝜋
)
8
4
𝑖(
13𝜋
)
8
= √2 ⋅ 𝑒
= √2 ⋅ 𝑒
= √2 ⋅ 𝑒
−7𝜋
= (𝑖𝑛 𝐴𝑟𝑔) = √2 ⋅ 𝑒 𝑖 8 .
4
−3𝜋
= (𝑖𝑛 𝐴𝑟𝑔) = √2 ⋅ 𝑒 𝑖 8 .
4
We sketch 4 vertices of a square: 𝑧0 , 𝑧1 , 𝑧2 , 𝑧3 .
2
16. Let 𝑧1 = 3𝑖 + 𝑖 2 , 𝑧2 = 1−𝑖
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 𝑛√|𝑧| . Each one
1
of them is obtained by rotating a chosen root (e.g. one with the argument 𝐴𝑟𝑔 𝑧) by the angle equal
𝜋
to multiplicity of 2 𝑛. The roots are the vertices of a regular n-gon .
𝑛
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 2𝛼 and cos 2𝛼 on the sin 𝛼
and cos 𝛼 (i.e. formulas for sin 2𝛼 and cos 2𝛼 which contain only ‘sin 𝛼’ and ‘cos 𝛼’).
Let 𝑧 = cos 𝛼 + 𝑖 sin 𝛼 , then form the de Moivre′ a Formula:
𝑧 2 = cos 2𝛼 + 𝑖 sin 2𝛼
On the other hand, if we square both sides of the equation 𝑧 = cos 𝛼 + 𝑖 sin 𝛼 ,
then
𝑧 2 = (cos 𝛼 + 𝑖 sin 𝛼)2 = cos2 𝛼 − sin2 𝛼 + 2𝑖 sin 𝛼 cos 𝛼, thus
cos 2𝛼 + 𝑖 sin 2𝛼 = cos 2 𝛼 − sin2 𝛼 + 2𝑖 sin 𝛼 cos 𝛼 , real and imaginary parts should be equal:
cos 2𝛼 = cos 2 𝛼 − sin2 𝛼 ; sin 2𝛼 = 2 sin 𝛼 cos 𝛼 .
18*. Use the exponential form to solve
𝒃)
|𝑧|2 𝑧
𝑧
3
= −1, 𝑧 ≠ 0
|𝑧|2 |𝑧|𝑒 𝑖 𝛼
|𝑧|𝑒 𝑖 𝛼
3
i niech 𝑧 = |𝑧|𝑒 𝑖 𝛼 . then the equation becomes:
= −1
|𝑧|2 |𝑧|𝑒 𝑖 𝛼
= −1 ⇔ 𝑒 𝑖4 𝛼 = −1
|𝑧|3 𝑒 −𝑖3 𝛼
⇔ 𝑒 𝑖4 𝛼 = 𝑒 𝑖(𝜋+2𝑘π) ⇔ 4 𝛼 = 𝜋 + 2𝑘π
and finally
we obtain z number with an arbitrary |𝑧|a nd four different arguments:
𝛼=
𝜋 𝑘π
+
4
2
𝑘 = 0,1,2,3.
1
√3
19*. Write 𝑧 = 2 + 𝑖 2 in exponential form
a) calculate all the possible integer powers 𝑧 𝑛 , 𝑛 ∈ 𝐼, (𝐼 = Integers)
1
2
First we express 𝑧 = + 𝑖
√3
2
1
2
in exponential form: 𝑧 = + 𝑖
√3
2
𝜋
𝜋
= 𝑒 𝑖( 3 +2𝑘𝜋) , so 𝑧 𝑛 = 𝑒 𝑖 𝑛 3 ,
(𝑛 ⋅ 2𝑘𝜋 is a multiplicity of the period 2𝑘𝜋). The periodicity of sine and cosine causes that there are
only 6 different values of the powers of n i.e. for 𝑛 = 6𝑘, 6𝑘 + 1, 6𝑘 + 2, … , 6𝑘 + 5, 𝑘 ∈ 𝐼
so
1
𝑓𝑜𝑟 𝑛 = 6𝑘
1 𝑖√3
+
𝑓𝑜𝑟 𝑛 = 6𝑘 + 1
2
2
1 𝑖√3
− +
𝑓𝑜𝑟 𝑛 = 6𝑘 + 2
𝜋
2
2
𝑧𝑛 = 𝑒𝑖 𝑛 3 =
−1
𝑓𝑜𝑟 𝑛 = 6𝑘 + 3
1 𝑖√3
− −
𝑓𝑜𝑟 𝑛 = 6𝑘 + 4
2
2
1 𝑖√3
𝑓𝑜𝑟 𝑛 = 6𝑘 + 5
{ 2− 2
𝒃) 𝑧 𝑖 ,
where 𝑖 is the imaginary unit 𝑖 2 = −1.
1
√3
1
𝜋
√3
Express 𝑧 = 2 + 𝑖 2 in exponential form; 𝑧 = 2 + 𝑖 2 = 𝑒 𝑖( 3 +2𝑘𝜋) ,
𝜋
𝜋
1
√3
𝑘 ∈ 𝐼,
𝑖
So 𝑧 𝑖 = 𝑒 𝑖⋅𝑖 ( 3 +2𝑘𝜋) = 𝑒 − ( 3 +2𝑘𝜋) . This means that (2 + 𝑖 2 ) are infinitely many real numbers. .
20 . Calculate the power 𝑒 𝑖 , where 𝑖 is the imaginary unit 𝑖 2 = −1.
𝑒 𝑖 = cos 1 + 𝑖 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 𝑥 3 − 𝑥 2 + 3𝑥 + 5 = (𝑥 + 1)(𝑥 2 − 2𝑥 + 5), next we find the
roots of the quadratic polynomial 𝑥 2 − 2𝑥 + 5.
𝒃) 2𝑧 3 + 4𝑧 2 + 3𝑧 + 6 = (𝑧 + 2)(2𝑧 3 + 3) ,
3
3
roots 𝑧1 = 𝑖 √ , 𝑧2 = −𝑖√ , 𝑧3 = −2.
2
2
7
3
1
𝒅) 𝑧 3 − 𝑧 2 − 𝑧 − = 0 | ⋅ 6 ⇔ 6𝑧 3 − 7𝑧 2 − 9𝑧 − 2 = 0
6
2
3
6𝑧 3 − 7𝑧 2 − 9𝑧 − 2 = (𝑧 − 2)(6𝑧 2 + 5𝑧 + 1)
22. Let
𝑏) 𝑧1 = −𝑖√2, 𝑧2 = 𝑖 be two of the roots of 𝑧 6 − 2𝑧 5 + 5𝑧 4 − 6𝑧 3 + 8𝑧 2 − 4𝑧 + 4 = 0 find all the
other roots.
𝑧 6 − 2𝑧 5 + 5𝑧 4 − 6𝑧 3 + 8𝑧 2 − 4𝑧 + 4 = (𝑧 + 𝑖 √2)(𝑧 − 𝑖 √2)(𝑧 − 1)(𝑧 + 𝑖)(𝑧 2 − 2𝑧 + 2)
The equation 𝑧 2 − 2𝑧 + 2 has the following two roots 𝑧5 = 1 + 𝑖, 𝑧6 = 1 − 𝑖 .
23. Write a polynomial with real coefficients of the fourth degree which has the following roots:
𝑧1 = 1 − 𝑖, 𝑧2 = 3𝑖.
e. g. (𝑧 − (1 − 𝑖))(𝑧 − (1 + 𝑖))(𝑧 − 3𝑖)(𝑧 + 3𝑖) = (𝑧 2 − 2𝑧 + 2)(𝑧 2 + 9)
= 𝑧 4 − 2𝑧 3 + 11𝑧 2 − 18𝑧 + 18.