### Complex Numbers

E.g.1

As with everything in maths, complex numbers also have its own birth, may be even in multiple forms! This is one form of it.

Suppose we want to solve the quadratic equation, x2+ x + 6 = 0. If we stick to the formula method, the process is as follows:

x = [-1 + √(1 - 24)]/2 or x = [-1 - √(1 - 24)]/2

x = [-1 + √( - 23)]/2 or x = [-1 - √(- 23)]/2

We cannot go beyond this point, as we have come to a stage where we have a square root of a negative number; we have been told it does not exist and we wind up our calculation with familiar conclusions - there are no real roots or solutions. That's it.

However, that was not the end of the world for mathematicians; they have gone much further than that and the concept of complex numbers was born.

In the above example we came across √(-23). Using the rules on indices, we can write it as follows:

√ -23 = √ -1 √ 23
Now we call √-1 as j, and suddenly √-23 becomes something that can be handled.
√-23 = j√ 23
Now, the above quadratic equation can be solved!
x = (-1 + j√23)/2 or (-1 - j√23)/2
Solutions are expressed in terms of j. These are called complex numbers.
Complex numbers are expressed in terms of j or √-1.

• j = √-1
• j2 = √-1.√-1 = -1
• j3 = √-1.√-1.√-1 = -j
• j4 = √-1.√-1.√-1.√-1 = -1 x -1 = 1

#### The Form of a Complex Number

Z = a + jb
a = the real part of the complex number
jb = imaginary part of the complex number
j = √-1

A complex number can be represented in a x-y grid. Then it is called an Argon Diagram: the real part is shown in the x-axis and the imaginary part in the y-axis.

This is the Cartesian representation of a complex number: the x-axis represents the real part; y-axis represents the imaginary part.

Complex numbers can be added, subtracted, multiplied and even divided like any other number, with a bit of caution though.

E.g.
Z1 = 8 + j6 ; Z2 = 2 - j4

Z1 + Z2 = (8 + 2) + j (6 - 4) = 10 + j2

#### Subtraction

Z1 - Z2 = (8 - 2) + j (6 - -4) = 10 + j10

#### Multiplication

Z1.Z2
= (8+j6).(2-j4)
= (8X2) - (8X4j) + (2X6j) - 24j2
= 16 - j32 + j12 - 24(-1) ; note j2 = -1
= 16 -j20 + 24
= 40 - j20

#### Division

Z1 / Z2 = (8 + J6) / (2-j4)

In this case, we need a special way of dealing with the division; we need a special form of denominator, called conjugate. The conjugate of (2 - j4) is (2 + j4) - just change the sign!.

Then multiply both the top and the bottom by the conjugate. That's it.

Z1 / Z2
= (8 + j6) * (2 + j4) / [(2 - j4)(2 + j4)]
= (16 + j32 + j12 + j224) / (4 + j8 - j8 - j216)
= (-8 + j44) / (20)
= -2/5 + j(11/5)

#### The modulus of a complex number - |Z|

If z = a + jb, then modulus is defined as,
√(a2 + b2)
|Z| =√(a2 + b2)

#### The polar representation of a complex number

Here, the angle t in radians is called the argument. Therefore, the complex number can be written in terms of its modulus and the argument. This is called the polar representation of the complex number

a = |Z| cos θ  :    b = |Z| sin t   : θ = tan-1(b/a) ;   -π< θ <= π
Z = |Z| cos θ + j|Z| sin θ

If the modulus of a complex number is r and the argument is t, it can be written as (r,θ) in the polar form.

E.g.1

Write the complex number, Z = 3 + j4 in the polar form.
|Z| = √(32 + 42) = 5
t = tan-1(4/3) = 0.92
Z = (5,0.92).

E.g.2

Write Z = (10,0.3) in Cartesian form.
a = 10 cos 0.3 = 9.5
b= 10 sin 0.3 = 2.9
Z = 9.5 + J2.9

E.g.3

If Z = -√3 + i, write it in the form of r(cos θ + i sin θ).
tan k = 1/√3
k = π/6
θ = π - k = 5π/6
|Z| = √(3 + 1) = 2
Z = 2(cos 5π/6 + i sin 5π/6)

E.g.4

If Z = -1 - i, write it in the form of r(cos θ + i sin θ).
tan k = -1/-1 = 1
k = π/4
θ = π + k or -(π - k);
Since -π< θ <=π
θ = -(π - π/4) = -3π/4 |Z| = √(1 + 1) = √2
Z = √2(cos -3π/4 + i sin -3π/4)

#### Maclaurin Series

f(x) = f(0) + f'(0)x/1! + f"(0)x2/2! + f"'(0)x3/3! + ....

sin(x) = sin(0) + cos(0)x/1! + -sin(0)x2/2! - cos(0)x3/3! + ...
= x -x3/3! + x5/5!

cos(x) = cos(0) - sin(0)x/1! - cos(0)x2/2! + sin(0)x3/3! + ...
= 1 -x2/3! + x4/4!

ex = 1 + e0x/1! + e0x2/2! + e0x3x/3!+...
= 1 + x + x2/2! + x3/3! + ...

#### Complex Numbers in Exponential Forms - Euler's Relation

From Maclaurin Series,
e = 1 + iθ/1! + (iθ)2/2! + (iθ)3/3! + ...
= 1 + iθ - θ2 -iθ3/3! + θ4/4! + iθ5/5! - θ6/6!
= (1 - θ2 + θ4/4! - θ6/6!... ) + i(iθ -iθ3/3! + iθ5/5!...)
= cos θ + i sin θ

e = cos θ + i sin θ

Z = r[cos θ + i sin θ ]
Z= re

E.g.1

Express Z = √3 + i in the form of Z = re

r = √(3 + 1) = 2
tan θ = 1/√3 => θ = π/6
Z = eiπ/6

E.g.2

Express Z = 3 cos π/3 + i sin π/3 in the form of Z = e.

r = 3; θ=π/3
Z = 3eiπ/3

E.g.3

Express Z = 5 cos π/4 - i sin π/4 in the form of Z = e.

Z = 5 cos (-π/4) + i sin (-π/4), because cos(-x) = cos(x) and sin(-x) = -sin(x)
r = 3; and θ=-π/4
Z = 3e-iπ/4

E.g.4

Express Z = 4ei5π/4 in the form of Z = r[cos θ + i sinθ].

Since -π< θ <=π, θ = -π/4; r = 4
So, Z = 2[cos -π/4 + i sin -π/4]

#### Product and Division of Complex Numbers

Let Z1 = r1[cosθ1; + i sinθ1;] = e1;    ;   Z2 = r2[cosθ2; + i sinθ2;] = e2;

Z1 X Z2 = r1e2; X r2e2;
= r1r2ei[θ1 + θ2]
= r1r2[cos(θ1 + θ1) + i sin(θ1 + θ2)]

Z1 / Z2 = r1e2; / r2e2;
= (r1/r2) ei[θ1 - θ2]
= (r1/r2) [cos(θ1 - θ1) + i sin(θ1 - θ2)]

Based on the above proofs, the following set of rules applies in dealing with the product and division of complex numbers:

|Z1 Z2 | = |Z1||Z1|

arg(Z1Z2) = arg(Z1) + arg(Z2)

|Z1 / Z2 | = |Z1|/|Z1|

arg(Z1 / Z2) = arg(Z1) - arg(Z2)

E.g.1

Express 2(cosπ/6 + i sinπ/6) X 3(cosπ/6 + i sinπ/6)in the form of x + iy.
= 6(cos(π/6+π/6) + i sin(π/6+π/6)
= 6(cosπ/3 + i sinπ/3)
= 3 + 3√3i

E.g.2

Express 12(cos5π/6 + i sin5π/6) / 3(cos2π/6 + i sin2π/6)in the form of x + iy.
= 4(cos(5π/6-2π/6) + i sin(5π/6-2π/6)
= 4(cos3π/6 + i sin3π/6)
= 4(cosπ/2 + i sinπ/2)
= 4(0 + i)
= 4i

E.g.3

Express 5(cos5π/6 + i sin5π/6) X 2(cos2π/6 - i sin2π/6)in the form of x + iy.
= 5(cos5π/6 + i sin5π/6) X 2(cos(-2π/6) + i sin(-2π/6)) , because, cos-θ = cosθ and sin-θ = -sinθ
= 10(cos3π/6 + i sin3π/6)
= 10(cosπ/2 + i sinπ/2)
= 10(0 + 1i)
= 10i

#### de Moivre's Theorem

[r(cosθ + isinθ)]n = rn[cos nθ + i sin nθ]

de Moivre Theorem in exponential form

Z = r[cosθ + isinθ]
Zn = rn[cosθ + isinθ]n
Zn = rn[cos nθ + isin nθ]
Zn = rnei(nθ)

#### The path leading to de Moivre's Theorem

By multiplying two complex numbers together, it is easy to show the path that leads to de Moivre Theorem, which is as follows:

Z = r[cosθ + i sinθ]
Z1 = r1[cos1θ + i sin1θ]   1
Z2 = Z X Z = r2[cos2θ + i sin2θ]   2
Z3 = Z2 X Z = r2[cos2θ + i sin2θ] X [cos1θ + i sin1θ]
Z3 = r3[cos3θ + i sin3θ]   3
Z4 = Z3 X Z = r3[cos3θ + i sin3θ] X [cos1θ + i sin1θ]
Z4 = r4[cos4θ + i sin4θ]   4
Z5 = Z4 X Z = r4[cos4θ + i sin4θ] X [cos1θ + i sin1θ]
Z5 = r5[cos5θ + i sin5θ]   5
...
...
...
...

Zn = rn[cos nθ + i sin nθ]   n
This is de Moivre Theorem.

de Moivre's Theorem is valid for any rational number - negative and positive integers as well as fractions. Normally, it is proved by the mathematical induction method.

It is true for any positive integer, n

[r(cos θ + i sinθ)]n = rn[cos nθ + i sin nθ]
n=1
LHS: [r(cos θ + i sinθ)]1 = r[cosθ + i sinθ]
RHS: r1[(cos1θ + i sin1θ)] = r[cosθ + i sinθ]
Since LHS: = RHS:, de Moivre Theorem is true for n=1.
Assume that de Moivre's Theorme is true for n=k. So,
Zk = rk[cos kθ + i sin kθ]
Now. let's try it for n = k+1
[r(cos θ + i sinθ)]k+1 = [r(cosθ + i sinθ)]k[r(cos nθ + i sin nθ)]
= [rk(cos kθ + i sin kθ)][r(cosθ + i sinθ)]
= [rk+1(cos (k+1)θ + i sin (k+1)θ)]

Therefore, if de Moivre' Theorem is true for n=k, it is true for n=k+1; Since it is true for n=1 too, de Moivre's Theorem is true for any integer, n>=1.

n=0

LHS: [r(cos θ + i sinθ)]0 = 1
RHS: r0[(cos0θ + i sin 0θ)] = 1[1 + 0] = 1
Since LHS: = RHS:, de Moivre's Theorem is true when n=0.

n is negative =>n<0

Let n=-m, where m is a positive number.
Since the theorem is true for positive integers,
LHS: [r(cos θ + i sinθ)]n = [r(cos θ + i sinθ)]-m
= 1 / [r(cos θ + i sinθ)]m
= 1 / [r(cos mθ + i msinθ)]   (since it is valid for positive integers)
X [cos mθ - i sin mθ] / [cos mθ - i sin mθ] =>
= [cos mθ - i sin mθ] / [(cos mθ - i sin mθ)rm(cos mθ + i msinθ)] =>
= [cos mθ - i sin mθ] / rm[cos2mθ - i2 sin 2mθ]
= [cos mθ - i sin mθ] / rm[cos2mθ + sin 2mθ]
= [cos mθ - i sin mθ] / rm X 1
=r-m[cos mθ - i sin mθ]
=r-m[cos (-mθ) + i sin (-mθ)],    since cos-θ= cosθ and sin-θ=-sinθ
Since -m = n
rn[cos nθ) + i sin nθ]
RHS:

For fractions

Let n=p/q, where p and q are integers.
[r(cos θ + i sinθ)]p = rp[cos pθ + i sin pθ]
= rp[cos pq(1/q)θ + i sin pq(1/q)θ]
= rp[cos p(1/q)θ + i sin p(1/q)θ]q
Now take the qth root of both sides
=rp/q[cos p(1/q)θ + i sin p(1/q)θ]
=rp/q[cos (p/q)θ + i sin (p/q)θ]
= rn[cos nθ + i sin nθ]
Therefore, deMoivre's Theorem is valid for fractions as well.

#### Derivation of Trigonometric Identities by de Moivre's Theorem

In order to derive the trigonometric identities, we are going to use the binomial expansion along with de Moivre Theorem.

(a + x)n = Σ nCr an-r xr

E.g.1

Find an identity for cos5θ.

[cosθ + i sinθ]5 = Σ 5Cr cosθ5-r (i sinθ)r
= cos5θ + 5 cos4θ(i sinθ) + 10 cos3θ(i sinθ)2 + 10 cos2θ(i sinθ)3 + 5 cosθ(i sinθ)4 + (i sinθ)5
= cos5θ + 5 cos4θ(i sinθ) + 10 cos3θ(-sin2θ) + 10 cos2θ(-i sin3θ) + 5 cosθ(sin4θ) + (i sin5θ)
From de Moivre't Theorem,
[cosθ + sinθ]5 = [cos5θ + i sin5θ]
So, [cos5θ + i sin5θ] = cos5θ + 5 cos4θ(i sinθ) + 10 cos3θ(-sin2θ) + 10 cos2θ(-i sin3θ) + 5 cosθ(sin4θ) + (i sin5θ)
Now equate the real parts on both sides:
cos5θ = cos5θ + 10 cos3θ(-sin2θ) + 5 cosθsin4θ
= cos5θ + 10 cos3θ(-sin2θ) + 5 cosθ(sin2θ)2
= cos5θ -10 cos3θsin2θ + 5 cosθ(1 - cos2θ)2
= cos5θ -10 cos3θ(1 - cos2θ) + 5 cosθ(1 -2cos2θ + cos4θ)
= cos5θ -10 cos3θ + 10 cos5θ + 5 cosθ -10cos3θ +5cos5θ
= 16cos5θ -20 cos3θ + 5 cosθ

cos 5θ = 16cos5θ -20 cos3θ + 5 cosθ

E.g.2

Find an identity for sin4θ.

[cosθ + i sinθ]4 = Σ 4Cr cosθ4-r (i sinθ)r
= cos4θ + 4 cos3θ(i sinθ) + 6 cos2θ(i sinθ)2 + 4 cosθ(i sinθ)3 + (i sinθ)4
= cos4θ + 4 cos3θ(i sinθ) + 6 cos2θ(-sin2θ) + 4 cosθ(-i sin3θ) + (i sin4θ)
From de Moivre't Theorem,
[cosθ + sinθ]4 = [cos4θ + i sin4θ]
So, [cos4θ + i sin4θ] = = cos4θ + 4 cos3θ(i sinθ) + 6 cos2θ(-sin2θ) + 4 cosθ(-i sin3θ) + (i sin4θ)
Now equate the imaginary parts on both sides:
sin4θ = 4cos3θsinθ - 4 cosθ sin3θ + sin4θ

sin 4θ = 4cos3θsinθ - 4 cosθ sin3θ + sin4θ

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