Download 1 The problem of square roots of negative numbers

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The problem of square roots of negative numbers
Exercise 1.1
Let a and b be two numbers such that a + b = p and a · b = q Write the equation in terms of p
and q that has roots a and b
Hint: if you don’t know how to start, try with a + b = 8 and a · b = 15 and other obvious
situations, then generalise.
√
√
Use this formula to find two numbers whose sum is 5 and product is − 3
Exercise 1.2
Using the formula above, find two numbers such that their sum is 2 and their product is 5.
Exercise 1.3
√
Using the usual algebraic rules, and not trying to find out what −3 could be, check that
q
2
q
√
√
1 + −3 + 1 − −3
is a real number (∈ R)
Exercise 1.4
If a + b = 4 and a · b = 16, find a and b.
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Complex numbers
Definition 1.1 (i)
i2 = −1
or
√
−1 = i
√
Hence − −1 = −i
Definition 1.2 (square root symbol)
√
The symbol will be reserved for real positive numbers, i.e:
√
If a is a real positive number, a is the real positive number which squared is a. And
√
− a is the real negative number which squared is a
p
√
If a is a negative real number, then a must be rewritten i |a|.
√
√
If a calculation yields, say, −5 this must be rewritten i 5
√
√ √
The rule a · b = a · b is only valid for positive values of a and b.
Exercise 1.5
p
√
√
Check that −4 · −4 6= (−4) · (−4)
Definition 1.3 (Imaginary numbers)
i is called an imaginary number.
Multiples of i are also imaginary numbers.
Definition 1.4 (Complex numbers)
The sum of a real number and an imaginary number form a complex number.
if a, b ∈ R
then
z = a + ib is a complex number
The set of complex numbers is written C
Let z = a + ib
Im(z) = b is the imaginary part of z
Re(z) = a is the real part of z
If b = 0 then z ∈ R i.e: R ⊂ C
In general: z without index is any complex (undetermined) number.
zi – with index is a given (determined) number.
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Operations in C
Exercise 2.1
Prove that for z1 , z2 ∈ C z1 · z2 = 0 ⇒ z1 = 0 OR z2 = 0
i.e. C has no zero divisors.
Exercise 2.2
What is (−i)2
Exercise 2.3
√
√
Find the (real) value of −2 · −3
Exercise 2.4
Compute the value of i2 , i3 , i4 , i5 , i6 , i7 , i8 . . .
Exercise 2.5
Calculate:
√ !2
2
2
+i
2
2
√
• (3 + 5i) + (9 − 2i)
•
• (3 − 4i) − (1 + i)
√ !3
2
2
+i
2
2
√
• 4(2 + 7i)
•
• (4 + 3i)(2 + i)
√
• (−2 − i)(3 + 2i)
•
• (a + ib) + (a − ib)
•
• (a + ib)(a − ib)
3
√ !4
2
2
+i
2
2
3 + 2i
3 + 4i
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Representing Complex Numbers
A Cartesian coordinate system can be used to represent complex numbers on a two dimensional
plane. The horizontal axis is the real numbers axis, positive to the right. The vertical axis is the
imaginary numbers axis, positive pointing upward.
A number a + ib becomes a point (a, b)
iR
a + ib
b
a
R
Exercise 3.1
On the complex plane, situate:
√
2
2
+i
2
2
√
• 1
•
• i
√
• −i
•
• 1+i
√ !2
2
2
+i
2
2
√
• 4(2 − i)
•
• 2 + 3i
√ !3
2
2
+i
2
2
√
• −1 − 2i
•
• −5
√ !4
2
2
+i
2
2
Exercise 3.2
Plot each number on the complex plane and plot the result of the operations on the complex
plane.
• (3 + 5i) + (9 − 2i)
• (4 + 3i) · (2 + i)
• (3 − 4i) − (1 + i)
• (−2 − i) · (3 + 2i)
• 4 · (2 + 7i)
•
−2 − i
3 + 2i
Can you give a graphical rule for addition, for subtraction, for multiplication, for division?
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Conjugate
Definition 4.1 (Conjugate)
Let z0 = a + ib the conjugate is z0 = a − ib
i.e: Re(z) = Re(z)
Im(z) = −Im(z)
On the Complex plane, z is the reflection of z through the real axis.
−z
z
−z
z
Note that z = z
Exercise 4.1
Calculate (z + z) and (z − z) for the following value of z
• 3 + 2i
• 5
• −1 − i
• −6 + 3i
• i
• 102 − 2.5 · 104 i
Then generalise for z + z
Exercise 4.2
Calculate
• (5 + 2i) · (5 − 2i)
• (−3 − 10i) · (−3 + 10i)
Then generalise for z · z
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Exercise 4.3
What can be said of z if z = z?
Same question if z = −z
Same question if z = 0
Exercise 4.4
Prove that
z1 + z2 = z1 + z2
and
z 1 · z2 = z1 · z2
Exercise 4.5
Prove that
∀n
zn = zn
Hint: induction proof.
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Modulus
The modulus (absolute value) of a real number is its distance to zero. This definition remains
valid in C. Pythagoras’ theorem yields
| z0 |=| a + ib |=
p
a2 + b2
a + ib
b
√
a2 + b2
a
This definition remains true whether a, b are positive or negative.
Exercise 5.1
The reciprocal of a number z is the number (noted z −1 or z1 ) such that z · z −1 = 1.
Prove that
1
z
=
z
| z |2
Exercise 5.2
If z = 0, is z1 defined in C?
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Division is defined by multiplying the reciprocal.
z1
1
= z1 ·
z2
z2
Exercise 5.3
Show that
1
1
=
z
z
and
z1
z2
=
z1
z2
Exercise 5.4
On the complex plane, for each of the following, plot the number, its conjugate, its opposite and
1
i
its reciprocal. z1 = + 4i z2 = −1 − i z3 = −i z4 = 2 +
2
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Exercise 5.5
Calculate 1i
Exercise 5.6
Show that:
• | z |=| −z |=| z |
• | z1 · z2 |=| z1 | · | z2 |
• | z n |=| z |n
Induction proof required!
• | z1 + z2 |≤| z1 | + | z2 | This is called the Triangle Inequality
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1
• If z 6= 0 then =
z
|z|
z1 | z1 |
• =
z2
| z2 |
Exercise 5.7
If z is the vertex of a square in the C-plane centred on 0, what are the coordinates of the other
three vertices?
Exercise 5.8
What is the (complex) equation of the circle of radius 3 centred on the origin?
What is the (complex) equation of the circle of radius 5 centred on 5 + 2i?
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Polar Coordinates
a + ib
b
ρ
θ
a
The length of the line is the modulus, here given by ρ (1 )
The angle is the argument, noted by θ = arg(z) (2 )
!
In the complex plane the angles are never measured in degrees. Always in radian measure.
Exercise 6.1
From one representation to the other:
√
Starting from z = a + ib it is easy to see that ρ =| z |= a2 + b2
How do you calculate θ?
Exercise 6.2
Converse: Given z = (ρ; θ) what is Re(z) and Im(z), i.e: how is it written in the form a + ib?
Exercise 6.3
Express the following complex numbers in polar form:
√
• 3 + i4
•
√
2
2
+i
2
2
• i
• 6 − 3i
• 2
• 3 − 4i
Exercise 6.4
Express the following complex numbers in Cartesian coordinate form a + ib
1
2
∀x
• (2; π)
• (0; x)
π
• (0.5; − )
2
π
• (−2; − )
4
• (1; 1)
• (π; π)
The Greek letter ρ is pronounced row as in to row a boat.
The Greek letter θ is pronounced theeta with the th sound as in think
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Trigonometric Form
Given z and its two forms a + ib and (ρ; θ) it follows that
z =| z | ·(cos(θ) + i sin(θ))
Exercise 7.1
Prove the following:
• arg(z1 · z2 ) = arg(z1 ) + arg(z2 )
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• arg
= − arg(z)
z
z1
• arg
= arg(z1 ) − arg(z2 )
z2
• arg (z n ) = n · arg(z)
Exercise 7.2
Using exercise 5.6 which gives | z |n =| z n | and exercise 7.1 which gives arg(z n ) = n · arg(z)
what is, in polar coordinates, (ρ; θ)n ?
Exercise 7.3
Draw the following points in the complex plane (scaled horizontally and vertically from -10 to
+10)
• (2; 0)
π
• (1; )
4
π
• (0.5; )
4
π
• (1.5; )
4
!
(2; 0)2
(2; 0)3
π
(1; )2
4
π
(0.5; )2
4
π 2
(1.5; )
4
π
(1; )3
4
π
(1; )4
4
π
(0.5; )3
4
π 3
(1.5; )
4
π
(1; )5
4
π
(1.5; )4
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For a given modulus ρ, an angle θ + k · 2π
k ∈ Z yields the same point as θ, So
arg(z) = arg(z) + k · 2π k ∈ Z
Exercise 7.4
Let | z |= 1 z = cos(θ) + i sin(θ)
Consider θ to be a variable.
1. Calculate
dz
. Write the result with respect to z
dθ
2. Can you find another function f which would yield z 0 = iz (remember that i is a constant),
hence in general, a function such that z 0 = kz.
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The Exponential Form
Consider the exponential eiθ and its multiplication by a real positive number ρ · eiθ
Exercise 8.1
Differentiate
f : θ 7→
eiθ
cos(θ) + i sin(θ)
df (θ)
with repect to θ i.e., calculate
dθ
Explain why this proves that this ratio is a constant and show that this constant is 1.
Why does it prove the following theorem?
Theorem 8.1 (Exponential Form)
For a general complex number, its modulus | z |= ρ and its argument arg(z) = θ
z = ρ · eiθ
Exercise 8.2
Check that:
iπ
• e−i = ei
• e2 =i
• e−ix = eix
• e2iπ = e0 = 1
• eiπ = −1
• e 2 iπ = −i
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Exercise 8.3
Calculate ln(i), i.e. for a certain a ⇔ ea = i
Hint: a is necessarily a complex number (why?) so it can be written u + iv.
(Assume that ln(z) is defined for complex numbers... there are certain delicate things to
consider, which we will not... but the log is defined under certain conditions.)
Exercise 8.4
Prove the following:
Theorem 8.2 (Real Power of a Complex Number)
z n = ρn · einθ
Exercise 8.5
Calculate ii
Exercise 8.6
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1
Prove or check that = e−iθ
z
ρ
Exercise 8.7
Draw on a C-plane the roots of:
10
• z2 = 1
• z6 = 1
• z3 = 1
• z 3 = −1
• z 6 = −1
Exercise 8.8
Solve the following equations:
• 2iz − 3 = z + i
• −
• 3z(z + i) = −iz
z
3z
+
=3+i
iz + 1 z − 1
z−1
= 4i
• −i(1 + 5i)z + (z − 1)(i − 2) = 3z − i
iz + 3
Exercise 8.9
Write the following numbers in exponential form and also in trigonometric form:
√
• a + ib
• 1+i 3
•
•
1
i
• √ +√
2
2
• e3iπ
• 4 + 2ei
• 5e−2iπ
• −3e2
1+i
2
Exercise 8.10
Write the following numbers in Cartesian form:
Exercise 8.11
Prove the Euler formulae in two ways: 1) expanding eix in trigonometric form; 2) by differentiating. 3
eix + e−ix
cos(x) =
2
ix
e − e−ix
sin(x) =
2i
Exercise 8.12
Solve the following equations and plot the solutions in the complex plane:
• z 2 + (i − 1)z + (2 − 2i) = 0
• z2 + z + i = 0
• z 2 − 3iz − 3 + i = 0
• z 2 − 5z + 6 = 0
Exercise 8.13
Solve and plot the solutions of:
• z3 + 1 = 0
• z 5 + 32 = 0
• z3 − 1 = 0
√
• z 4 − 8 = 8i 3
• z4 − 1 = 0
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Euler was a XVIIth century Swiss mathematician. He invented the “e” notation, as in ... Euler...
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Exercise 8.14
The complex solution of z 2 + z + 1 = 0 with positive imaginary part is called j
• Give the solutions of the equation in algebraic form, trigonometric form and exponential
form.
• Prove that j = j 2 =
1
j
• Calculate j 1999 j 2000 and j 2001
Exercise 8.15
Find all points of the plane such that:
π
+ k · 2π
6
π
• arg(z) = + k · 2π
2
1
2π
• arg
=
+ k · 2π
z
3
• arg(z) =
Exercise 8.16
• For what complex values of z does z 2 + 2z + 1 yield a real number?
• For what complex values of z does (z − 1)(2iz + 3) yield an imaginary number?
Exercise 8.17
Simplify the following:
π
π
• e−i 2 · ei 3
π 3
ei 3
•
5π
2ei 8
√
Exercise 8.18
1
3
Let z1 = −1 − i and z2 = + i
2
2
z1
z2
and
• Evaluate
z2
z1
• Write z1 and z2 in exponential form.
• Give the modulus and argument of
z1
z2
• From the former questions, deduce the exact values of cos
• What is the smallest positive integer n such that
12
z1
z2
11π
12
and sin
n
is a real number?
11π
12
Exercise 8.19
Let p be a polynomial with real coefficients.
1. Show that p (z) = p(z)
2. Show that of z0 is a root, then z0 is also a root.
3. Explain why if the order of p is odd, then p necessarily has at least one real root.
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Functions of Complex Variables
The main difficulty with functions from f : C → C is the impossibility to plot the points as
(z; f (z) because this would need 4 dimensions.
So the Domain plane (input) and Range plane (output) are sketched separately.
Let f (z) = z 2 and a = 2 + 2i b = 2i c = −2 + 2i d = −2 j = 2
f(a)
f(a)
f(z)=z 2
b
b
a
c
i
j
d
−2
a
c
i
−1
1
f(b)
2
3
i
j
d
4
−2
−1
1
f(j)
2
3
f(j)
f(b)
4
−2
−1
1
f(d)
f(c)
where f (a) = 8i
f (b) = −4 f (c) = −8i
f (d) = 4 f (j) = f (d) = 4
Question: What does the rectangle with a, c, d, j as vertices become?
Exercise 9.1
Let f (z) = z 2
• Draw the image of the real line.
• Draw the image of the imaginary line.
• Draw the image of the line t 7→ t + i for t ∈ R.
• Draw the image of the line t 7→ 1 + t · i for t ∈ R
• Draw the image of the square with vertexes at 0, 1, 1 + i and i.
The unit grid is the grid with vertical lines and horizontal lines for every unit.
Exercise 9.2
Draw the unit grid after transformation through z 7→ z 2
Exercise 9.3
Draw the unit grid after transformation through z 7→ z 3
Exercise 9.4
Draw the unit grid after transformation through z 7→ z −1
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2
3
4
f(d)
f(c)
Definition 9.1 (Fixed Point)
A fixed point z0 of a function f is a value (or point or number depending on how the
elements are considered) such that
f (z0 ) = z0
Exercise 9.5
Determine the image of the two following geometric domains by the functions
1
2
1
f : z 7→ 1 +
2
2
z−1
and
g : z 7→ (z − 1)2 + 1
(Four separate drawings!)
Do these functions have fixed points? (Whether yes or no, prove your answer.)
Exercise 9.6
Using exercise 8.11 study the sine and cosine functions. For these drawings, evaluate for certain
numbers, then try to join smoothly the image points.
• Draw the image of the real line by the function sin(z) and cos(z)
• Same thing with the imaginary line.
• Add in the horizontal lines t ± i and vertical lines ±1 + it for t ∈ R
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Exercise 9.7
The following is the normal grid with unit measures. Shade the squares as if it were a small
chessboard.
then draw the same chessboard on the following grid (after transformation z 7→ z 2
Notice that the “lines” still intersect at right angles.
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then draw the same figure on the following grid (after transformation z 7→
1
0.5
-1
-0.5
0.5
-0.5
-1
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1
1
z
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The Mandelbrot set
A very special case of functions is the iterated functions.
Definition 10.1 (Iterated functions)
An iterated function is a repeated function, i.e: where the value for zn depends on the
value of zn−1
Consider the following iterated function:
2
zn = zn−1
+ z0
with any complex number as z0
This yields; z0 z1 = z02 + z0
z2 = z12 + z0
...
0 is a fixed point.
= 1 + 1 = 2 22 + 1 = 5 52 + 1 = 26
and as the iterations continue, the image of
1 is sent of to infinity.
−1 = 1 − 1 = 0 02 − 1 = −1
so the image of -1 oscillates between 0 and -1.
i = −1 + i (−1 + i)2 + i = −i − i2 + i = −1 + i
again an oscillatory situation.
The question is: For what complex numbers does the image stay bounded however many
iterations are performed – or conversely, for what complex numbers do the images go off to
infinity?
It is easy to prove that if the modulus becomes greater than 2 then it will be ever increasing,
so testing if the modulus is less than 2 after, say, 100 or 10 000 iterations gives a good view of
the problem.
A computer is used to plot the calculation results. The point is black if its image stays
bounded under a given number of iterations. The point is grey if it goes off to infinity (here: if
the modulus has reached 2 after 1000 iterations). The Mandelbrot set is the set of all points
who’s image stays bounded: the black points.
The shade of grey shows how fast (after how many iterations) the image became greater
than 2.
The result is a quite spectacular drawing with surprising properties such as: the border of
this set is infinitely long! If one zooms on a part of the set near the border, the same shape is
found again.
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Here is a zoom, somewhere near the border...
It is a chaotic system, in the sense that two points very close can have very different (and
not easily predictable) destinations.
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