Complex Number Arithmetic (Notes)

On this page:
  1. Addition
  2. Conjugate Complex Numbers
  3. Subtraction
  4. Converting from one form to another
  5. Multiplication
  6. Division
  7. Reciprocals
  8. De Moivre's Theorem
  9. Powers of Complex Numbers
  10. Roots of Complex Numbers
  11. Example 1
  12. Example 2 Cube Roots of Unity
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)

Other Pages:

Roots of Unity



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 complex conjugate symbol 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.


(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
impliesr=|√(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:



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·c+(b·c+d·a)i +b·d·i²
=(a·cb·d) + (b·c+d·a)·i [because i²=1]
For instance, (2+3·i)(4+i)=
=5 +14·i


If there are two complex numbers, z1=a+ib and z2=c+id and you want to find:
You can multiply top and bottom by z2* to remove the imaginary parts, where the conjugate z2* is (cid),:

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 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:

Note that r=√(a2+b2), √(22 + 32)=√(13)=3.6056
[ 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-54-27i (because i2=-1)=

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
The 3 roots of 1+1i are:
Root0=1.1225*cos(0.2618)+1.1225* i sin(0.2618)
(As a check) Root03=1+1i
Root1=1.1225*cos(2.3562)+1.1225* i sin(2.3562)
Root1= – 0.7937+0.7937i
(As a check) Root13=1+1i
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
The 3 roots of 1 are:
Root0=1·cos(0)+1· i sin(0)
(As a check) Root03=1
Root1=1*cos(2.094395)+1* i sin(2.094395)
Root1= – 0.5+0.866025i
(As a check) Root13=1
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:
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:

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