Lecture 1 Complex Numbers Definitions. Let i 2 = −1. ∴ i = √ −1. Complex numbers are often denoted by z . Just as R is the set of real numbers, C is the set of complex numbers. If z is a complex number, z is of the form z = x + iy ∈ C, for some x, y ∈ R. e.g. 3 + 4i is a complex number. z = x + iy ↑ real part imaginary part. If z = x + iy, x, y ∈ R, the real part of z = (z ) = Re(z )= x the imaginary part of z = (z ) = Im(z )= y. eg.z =3+4i (z )=3 (z )=4. If z = x + iy, then z (“z bar”) is given by z = x − iy and is called the complex conjugate of z . eg. If z =3+4i, then z =3 − 4i. Example. Solve x 2 − 2x +3=0. x = −(−2)± √ (−2) 2 −4(1)(3) 2(1) = 2± √ −8 2 = 2±2 √ −2 2 =1 ± √ 2 i.
Transcript
Lecture 1
Complex NumbersDefinitions.
Let i2 = −1.
∴ i =√−1.
Complex numbers are often denoted by z.
Just as R is the set of real numbers, C is the set of complex numbers. If z is a complexnumber, z is of the form
∴ when we multiply two complex conjugates, we get a real number.
Example 3. 2+3i1+4i = 2+3i
1+4i × 1−4i1−4i = (2+3i)(1−4i)
(1+4i)(1−4i) = 2−8i+3i−12i2
1−(4i)2 = 14−5i17
(realising the denominator)
Lecture 3Theorem. If two complex numbers are equal then their real parts are equal and theirimaginary parts are equal, i.e., if a+ ib = c+ id where a, b, c, d ∈ R, then a = c and b = d.
Example 1. Find x, y if (3 + 4i)2 − 2(x− iy) = x+ iy.
An integral domain is a set of elements with two binary operations defined for them, whichobey the laws obeyed by the integers.
A set S is an integral domain if its elements a, b, c, . . . obey the following laws.
1. Closure Law for Addition, i.e., a + b ∈ S
2. Closure Law for Multiplication, i.e., a × b ∈ S
3. Commutative Law for Addition, i.e., a + b = b + a4. Commutative Law for Multiplication, i.e., a × b = b × a5. Associative Law for Addition, i.e., a + (b + c) = (a + b) + c6. Associative Law for Multiplication, i.e., a × (b × c) = (a × b) × c7. Distributive Law of Multiplication over Addition, i.e., a × (b + c) = a × b + a × c8. There exists an additive identity (or zero element) 0, such that for every a,
a + 0 = 0 + a = a (Note 0 ∈ S)9. There exists a multiplicative identity (or unity element) 1, such that for every a,
a × 1 = 1 × a = a (Note 1 ∈ S)10. There exists an additive inverse (or opposite), −a, for each member a of the set such
that a + (−a) = (−a) + a = 0.11. Cancellation Law. If ab = ac and a �= 0, then b = c.
Example 1. Z, the set of the integers, is an integral domain.
The elements of a field F obey the above axioms 1-10 for integral domains, (where a, b, care elements of F) and instead of the cancellation law, there is a law about the existenceof a multiplicative inverse (or reciprocal):
11′. If a−1 and 1 are elements of F, and a × a−1 = a−1 × a = 1, where a �= 0, then a−1 isthe multiplicative inverse of a.
Example 2. C, the set of complex numbers is a field.
Example 3. The additive inverse of z = 2 + 3i is −z = −2 − 3i
Example 4. The multiplicative inverse of z = 2 + 3i is z−1 = 12+3i = 1
2+3i
(2−3i2−3i
)= 2−3i
13 .
Lecture 7
(∗){
cos(A+B) = cosA cosB − sinA sinBsin(A+B) = sinA cosB + sinB cosA
Mod-arg theorems
i. If z1 = r1(cos θ1 + i sin θ1) & z2 = r2(cos θ2 + i sin θ2)then if z1 = z2 then r1 = r2 & θ1 = θ2.
ii. |z1z2| = |z1||z2| and arg(z1z2) = arg z1 + arg z2 ± 2π.i.e., for example:
arg(z1z2) = 100◦ + 140◦ − 360◦
= −120◦
arg(
z1z2
)= arg z1 − arg z2 ± 2π.
Proof . If z1 = r1(cos θ1 + i sin θ1)
and z2 = r2(cos θ2 + i sin θ2)
then z1z2 = r1(cos θ1 + i sin θ1) · r2(cos θ2 + i sin θ2)
= r1r2(cos θ1 cos θ2 − sin θ1 sin θ2 + i sin θ2 cos θ1 + i sin θ1 cos θ2)
= r1r2(cos(θ1 + θ2) + i sin(θ1 + θ2)) − (see (∗) above)
N.B. |z − z1| = |z − z2| will always be a straight line. It will always be the perpendicularbisector of the interval joining z1 to z2.
Lecture 11(∗) Note. sin(A+B) = sinA cos B +sinB cos A & cos(A+B) = cos A cos B− sinA sinB.
De Moivres Theorem. (cos θ + i sin θ)n = cos nθ + i sinnθ.
Proof. (By mathematical induction for n = 0, 1, 2, . . . .)
Step 1. Test n = 0.
L.H.S. = (cos θ + i sin θ)0
= 1R.H.S. = cos 0 + i sin 0
= 1= L.H.S.
∴ it is true for n = 0.
Step 2. Assume true for n = k i.e., (cos θ + i sin θ)k = cos kθ + i sin kθ.
Test for n = k + 1.
i.e., L.H.S. = (cos θ + i sin θ)k+1 & R.H.S. = cos(k + 1)θ + i sin(k + 1)θ
= (cos θ + i sin θ)k(cos θ + i sin θ)1
= (cos kθ + i sin kθ)(cos θ + i sin θ)(since we have assumed it true for n = k)= cos kθ cos θ + i sin θ cos kθ + i sin kθ cos θ − sin kθ sin θ
= cos kθ cos θ − sin kθ sin θ + i(sin θ cos kθ + sin kθ cos θ)
= cos(kθ + θ) + i sin(kθ + θ) (see (∗) above)
= cos(k + 1)θ + i sin(k + 1)θ= R.H.S.
Step 3. If the result is true for n = 0, then true for n = 0 + 1, i.e., n = 1. If the result istrue for n = 1, then true for n = 1 + 1, i.e., n = 2 ans so on for all nonnegative integersn �
Example 1. Simplify:
(a) (cos θ − i sin θ)−4 (b) (sin θ − i cos θ)7 (c) (cos 2θ+i sin 2θ)3
(cos θ−i sin θ)4 .
(a) (cos θ − i sin θ)−4 = cos(−4θ) − i sin(−4θ)= cos 4θ + i sin 4θ �
(b) (sin θ − i cos θ)7 = (−i cos θ + sin θ)7
= −i7(cos θ − i sin θ)7
= i(cos 7θ − i sin 7θ)= sin 7θ + i cos 7θ �
(c) (cos 2θ+i sin 2θ)3
(cos θ−i sin θ)4 = (cos θ+i sin θ)6
(cos θ−i sin θ)4
= (cos θ+i sin θ)6
(cos(−θ)+i sin(−θ))4
= (cos θ+i sin θ)6
(cos θ+i sin θ)−4
= (cos θ + i sin θ)10
= cos 10θ + i sin 10θ �Example 2. Express in the form x + iy:
(a)(cos π
2 + i sin π2
)6(b)
(1 +
√3
)10.
(a) (cos π2 + i sin π
2 )6 = cos 6π2 + i sin 6π
2
= cos 3π + i sin 3π
= −1 + 0i
= −1 �(b) (1 +
√3)10 = (2(cos π
3 + i sin π3 )10
= 210(cos 10π3 + i sin 10π
3 )
= 210(− 1
2 − i√
32
)
= −512 − 512i√
3 �
Lecture 12De Moivre’s Theorem and the Argand Diagram
Example. If z =√
3 + i represent the following on the Argand Diagram:
z, iz, 1z ,−z, 2z, z, z2 + z, z3 − z
z = 2(cos π6 + i sin π
6 )
z−1 = (2(cos π6 + i sin π
6 ))−1
= 12 (cos−π
6 + i sin−π6 )
= 12 (cos π
6 − i sin π6 )
2z = 4(cos π6 + i sin π
6 )
z2 = (2(cos π6 + i sin π
6 ))2
= 4(cos π3 + i sin π
3 )z3 = (2(cos π
6 + i sin π6 ))3
= 8(cos π2 + i sin π
2 )
Solution on next page.
Lecture 13Trigonometric Identities and DeMoivre’s Theorem
Example. Obtain cos 6θ in terms of cos θ. Hence show that x = cos(2k + 1) π12 where
k = 0, 1, 2, 3, 4, 5 is a solution to the equation 32x6−48x4 +18x2−1 = 0 and hence deducethat cos π
