Download For example, i= 1 i^2= -1 i^3= (i^2 * i) = (-1 * i) =

Complex Numbers by Lissy Barrow and Anthony Boyd
Definition of complex numbers: for real numbers a and b, the number a+bi is a complex number. If a is
equal to 0, and b is not equal to 0 then the complex number bi is a purely imaginary number (the real part
vanishes).
A complex number can be visually represented as a pair of numbers forming a vector on an Argand
diagram, representing the complex plane. Re is the real axis, Im is the imaginary axis, and i is such that
i^2=-1.
This graph of z=a+bi shows that from the origin that a represents the length on the real axis, b is the
height on the imaginary axis, and r is the modulus or length of the vector z.
For example,
i= 1
i^2= -1
i^3= (i^2 * i) = (-1 * i) = -i
And i^4= 1 (i^2 * i^2) = (-1 * -1) = 1
i^0= 1
i^4= 1
i^1= i
i^5= i
i^2= -1
i^6= -1
i^3= -i
i^7= -i
Characteristics of a position vector are that it has magnitude and direction, which is emphasized in
complex number’s Polar Form. Addition and multiplication take on geometric characteristics when
complex numbers are viewed as position vectors.
Addition corresponds to vector addition
Multiplication corresponds to multiplying their magnitudes and adding their arguments (i.e. the
angles they make with the x or R axis).
Viewed in this way the multiplication of a complex number by i corresponds to rotating a
complex number anticlockwise through 90-degrees about the origin.
Addition and subtraction of complex numbers:
Complex numbers are added by adding the real and imaginary parts of the summands.
(a+bi)+(c+di)=(a+c)+(bi+di)
Ex: (2+2i)+(3+2i)=(2+3)+(2i+2i)=5+4i
Complex numbers are subtracted similarly:
(a+bi)-(c+di)=(a-c)+(bi-di)
Ex: (4+5i)-(3+i)=(4-3)+(5i-i)=1+4i
Multiplication of complex numbers:
Multiplication of two complex numbers is defined by the following formula:
(a+bi)*(c+di)=(ac-bd)+(bc+ad)i
Ex: (2+i)*(3+i)=6+2i+3i+i^2 (the i^2 will become -1, because i^2 is equal to -1…which is where
the –bd from the formula comes from) so the result will be: 5+5i.
Trigonometric functions:
Let z=a+bi, with the real part being a, and the imaginary part b. Then |z| is the distance from the z to zero
in the complex plane and is computed by |z|=sqrt(a^2+b^2).
The line containing zero and z makes an angle (arg(z)) with the real axis satisfying:
Tan(arg(z))=b/a
Cos(arg(z))=a/|z|
Sin(arg(z))=b/|z|
So: a=|z|cos(arg(z)), and b=|z|sin(arg(z)),
So: z=|z|cos(arg(z))+ |z|sin(arg(z))i
Thus: z=|z|[ cos(arg(z))+ isin(arg(z))]
We did the proof of Euler’s formula by doing the Taylor Series of e^(ix):
Since i^2 is equal to -1 then the ((ix)^2)/(2!) becomes (- x^2)/(2!) and i^4 is equal to +1 resulting
in (+ x^4)/(4!). This sequence will alternate for i^6, i^8, i^10,i^12... By separating the terms
without an i and the terms containing an i results in:
(1-(x^2)/(2!)+(x^4)/(4!)-(x^6)/(6!)+(x^8)/(8!)-…)+(ix-(ix^3)/(3!)+(ix^5)/(5!)-(ix^7)/(7!)+…)
with the first part of the series being equal to the Taylor Series of cos(x). Then by factoring out a
common factor (i) from (ix-(ix^3)/(3!)+(ix^5)/(5!)-(ix^7)/(7!)+…) gives the Taylor Series for
i*sin(x).
Because of the formula e^(ix)=cos x+i sin x , we can immediately obtain some formulas for the sine and
cosine functions, in terms of complex exponentials. These formulas are often used as definitions of the
trigonometric functions for complex numbers. So you get,
 cos(x)= (e^(ix)+e^(-ix)) / 2
 sin(x) = (e^(ix)- e^(-ix)) / 2
Substituting the quantity ix in place of the variable x , we can produce formulas for pure imaginary
numbers. Notice that the results are related to the hyperbolic functions.
 cos(ix) = (e^(ix)+e^(-ix)) / 2 = cosh x
 sin(ix) = (e^(ix)- e^(-ix)) / 2 = i sinh x
Then, using the formulas for the sine and cosine of the sum of two angles, we can obtain formulas for the
sine and cosine of complex numbers
 cos(a+bi)=(cos a) (cosh b) −(i)(sin a)(sinh b)
 sin(a+bi)=(sin a)(cosh b)+(i)(cos a)(sinh b)
Also, by taking Euler’s Identity: e^(i*θ)=cos(θ)+isin(θ)
and setting theta equal to Pi gives:
e^(i* π)=cos(π)+isin(π)
Since the cos(π) is -1 and the sin(π) is 0, then:
e^(i* π)=-1+0
e^(i* π)+1=0
which gives the identity:
Roots of complex numbers:
To find the cubed root of -8:
z=(-8)^(1/3)
so:
z^3=-8
then convert z into polar form (p*e^(i* θ)); where p is the symbol for rho.
This gives us:
z^3=p^3 e^(3*i*θ)=-8
-8*1=8*e^(π*i)
By taking out the real numbers p^3=8 we find that p=2.
Then looking at our exponents that are left we have: 3*θ= π +2 π *k
By solving for theta we get:
θ=(π/3)+((2*π*k)/3)
(since we are taking the cubed root, k=0,1,2)
When k=0 then:
θ=(π/3)+((2 π*0)/3)) which is equal to (π/3)
The polar form when k=0:
z=2 e^(π*i/3)
By using Euler’s formula e^(i*(π/3))= cos(π/3) + i*sin(π/3), we get:
z=2*cos(π/3)+2*i*sin(π/3)= 1 + i sqrt(3)
When k=1:
θ=(π/3)+((2π*1)/3) which is equal to π
The polar form when k=1:
When k=2:
z=2*e^(i*π) = 2*(-1) = -2
θ=(π/3)+((2π*2)/3) which is (π/3)+((4π)/3) and when added up is ((5*π)/3)
The polar form when k=2:
z=2*e^(i*5*π)/3)
By using Euler’s formula e^(i*((5*π)/3))=cos((5*π)/3)+i*sin((5*π)/3), we get:
z=2*cos((5*π)/3)+2*i*sin((5*π)/3)
z=2*(1/2)+2*i*(-sqrt(3))
z=1- i sqrt (3)
Resulting in the answers: -2, 1+i sqrt (3), and 1- i sqrt (3)
The graph of the cubed root of -8 will look similar to this sketch showing the three points we just found
(π)/3, π, and (5π)/3, which in a circle would have a radius 2: