In a complex number, a+ib, a is the real part and b is the imaginary part, although, of course, both a and b are real numbers.
The complex conjugate of z=a+ib is z*=a-ib.
You can express complex numbers in
various forms, including algebraic,
trigonometric and
exponential form. For instance:
a+ib (algebraic)
r(cosф +sinф) (trigonometric)
reiφ
(exponential)
z=a+ib, Re(z)=a, Im(z)=b (Real and Imaginary parts)
Complex numbers are added as follows.
Let a+ib, and c+id be a complex number, where a, b, c, and d are real
numbers.
i², or j² = – 1
(a+i·b)+(c+i·d)=(a+b)
+
i(b+d)
For instance: (2+3·i)+(4+5·i)=
(2+4)+(3+5)i = 6+8·i
Conjugate
Complex Numbers
Complex numbers often come in pairs, such as a+b·i,
a–b·i and are called
conjugate. If we use z to refer to a complex number, then the conjugate
is written: z*, or so, if:
z=a+i·b, then z*=a–i·b
Conjugate complex numbers have these properties in arithmetic:
z+z* =2a (2·Re(z) )
z-z*=2b·i (2·Im(z) )
zz*=a2 + b2 (Re(z)2
+ Im(z)2 )
That is, the result has a real part only, the imaginary part vanishes.
Subtraction
(a+b·i) – (c+d·i) = (a-c)+ (b-d)·i
That is the real and imaginary parts are subtracted.
Converting
from one form to another
You can convert from algebraic form to trigonometric form
as follows:
Let z=a+i·b r=|√(a2+b2)|, r
or |z|
is called, variously, the absolute value, phasor, modulus, or radial coordinate, and must be positive
φ, or arg(z) = tan-1(b/a)
z=r(cosφ +i·sinφ )
The exponential
form is therefore: eiφ
Multiplication
In algebraic form,
you can multiply two complex numbers as you would two factors in ordinary
algebra, in this case, keeping
the real and imaginary parts together. For instance
(a+b·i)(c+d·i) =a·c+(b·c+d·a)i +b·d·i²
=(a·c−b·d) + (b·c+d·a)·i [because i²=−1]
For instance, (2+3·i)(4+i)=
(8-3)+(14)·i
=5 +14·i■
Division
If there are two complex numbers, z1=a+ib and z2=c+id
and you want to find:
z1/z2
You can multiply top and bottom by z2* to remove the imaginary parts,
where the conjugate z2* is (c−id),:
You do this because z·z* is a real number.
For instance: ,
multiplied throughout by (5+2i) becomes
which, simplifying the denominator, gives:
And simplifying the numerator, gives, finally:
Reciprocals
Reciprocals are simply division, where the numerator is 1, we proceed
as with division, above.
De
Moivre's Theorem
De Moivre's Theorem states: (n=0,1,2...n-1). Usually φ is
restricted to ±π, or [0,2π]
We can genuinely thank Abraham de Moivre (1667-1754) for this, as will
become obvious later!
Powers
of Complex Numbers
If we have a complex number in trigonometric form, [r(cosφ +i
sin φ)]n
we can use Abe De Moivre's
Theorem, above, to obtain: cos(nφ) +i sin (nφ)
For instance:
(2+3i)3
Note that
r=√(a2+b2),
√(22 + 32)=√(13)=3.6056
φ=tan-1(3/2)=0.9828
3φ=2.9484
[ 3.6056(cos3φ + i sin3φ ) ]3 =
46.8722(cos(2.9484)-i sin (2.9484) )= -46 + 9 i
(all
rounded to 4 decimals)
Alternatively, we can stick with algebra, and use the Binomial Theorem:
(2+3i)3 =23+3.22.3i+3.2.(3i)2+(3i)3 =
8+36i+54i2+27i3=
8+36i-54-27i (because i2=-1)= -46+9i
Roots
of Complex Numbers
The roots of complex numbers can be computed on the same principle as the
powers. using the trigonometric method.
Suppose z=r(cos α +i sin α)
We wish to find z1/n where n is a positive
integer.
zk1/n= [r(cos
φk + i sin φk)
]1/n
Where φk=(α+2*π*k)/n,
and k=0, 1,.... n-1 (n roots required!)
Example
1
We wish to find the 3
roots of (1+1i)1/3
r=1.4142
r1/3=1.1225
phi=atan(1/1)=0.7854
The 3 roots of 1+1i are:
________________________
φ0=0.7854+2*3.1416*0=0.2618
Root0=1.1225*cos(0.2618)+1.1225* i sin(0.2618)
Root0=1.0842+0.2905i
(As a check) Root03=1+1i
________________________________
φ1=0.7854+2*3.1416*1=2.3562
Root1=1.1225*cos(2.3562)+1.1225* i sin(2.3562)
Root1= – 0.7937+0.7937i
(As a check) Root13=1+1i
________________________________
φ2=0.7854+2*3.1416*2=4.4506
Root2=1.1225*cos(4.4506)+1.1225* i sin(4.4506)
Root2= – 0.2905 – 1.0842i
(As a check) Root23=1+1i
________________________________
In the above example, the checks worked out exactly, perhaps by chance.
Example
2 Cube Roots of Unity
Input data:
We wish to find the 3 roots of (1)1/3
r=1
r1/3=1
phi=atan(0/1)=0
The 3 roots of 1 are:
________________________
φ0=0+2·π·0=0
Root0=1·cos(0)+1· i sin(0)
Root0=1
(As a check) Root03=1
________________________________
φ1=0+2·π·1
Root1=1*cos(2.094395)+1* i sin(2.094395)
Root1= – 0.5+0.866025i
(As a check) Root13=1
________________________________
φ2=0+2*3.141593*2=4.18879
Root2=1*cos(4.18879)+1* i sin(4.18879)
Root2= – 0.5 – 0.866025i
(As a check) Root23=1
– 0.000001i
________________________________
Even using more decimal places in the above, there is a rounding error
on the last check. The cube roots of unity are, exactly:
x=1
x2=-0.5 + √(3)/2i
x3=-0.5 - √(3)/2i
When taking a cube root, there are times when we need all three roots.
These roots are, by the way, the solutions to the following equation:
x3-1=0
Or
(x-1)(x2+x+1)=0
Faster Arithmetic - by Ken Ward
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r=|√(a2+b2)|, r
or |z|
is called, variously, the absolute value, phasor, modulus, or radial coordinate, and must be positive
so, if:
, 
(n=0,1,2...n-1). Usually φ is
restricted to ±π, or [0,2π]