- Addition
- Conjugate Complex Numbers
- Subtraction
- Converting from one form to another
- Multiplication
- Division
- Reciprocals
- De Moivre's Theorem
- Powers of Complex Numbers
- Roots of Complex Numbers
- Example 1
- Example 2 Cube Roots of Unity

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)

re

z=a+ib, Re(z)=a, Im(z)=b (Real and Imaginary parts)

Other Pages:

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

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*=a

That is, the result has a real part only, the imaginary part vanishes.

That is the real and imaginary parts are subtracted.

You can convert from algebraic form to trigonometric form
as follows:

Let z=a+i·b

r=|√(a^{2}+b^{2})|, 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:

e^{iφ}

(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■

If there are two complex numbers, z_{1}=a+ib and z_{2}=c+id
and you want to find:

z_{1}/z_{2}

You can multiply top and bottom by z_{2}* 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:

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!

cos(nφ) +i sin (nφ)

For instance:

(2+3i)

Note that
r=√(a^{2}+b^{2}),
√(2^{2} + 3^{2})=√(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} =2^{3}+3.2^{2}.3i+3.2.(3i)^{2}+(3i)^{3} =

8+36i+54i^{2}+27i^{3}=

8+36i-54-27i (because i^{2}=-1)=

-46+9i

Suppose z=r(cos α +i sin α)

We wish to find z

z

Where φ

We wish to find the 3
roots of (1+1i)^{1/3}

r=1.4142

r^{1/3}=1.1225

phi=atan(1/1)=0.7854

The 3 roots of 1+1i are:

________________________

φ_{0}=0.7854+2*3.1416*0=0.2618

Root_{0}=1.1225*cos(0.2618)+1.1225* i sin(0.2618)

Root_{0}=1.0842+0.2905i

(As a check) Root_{0}^{3}=1+1i

________________________________

φ_{1}=0.7854+2*3.1416*1=2.3562

Root_{1}=1.1225*cos(2.3562)+1.1225* i sin(2.3562)

Root_{1}= – 0.7937+0.7937i

(As a check) Root_{1}^{3}=1+1i

________________________________

φ_{2}=0.7854+2*3.1416*2=4.4506

Root_{2}=1.1225*cos(4.4506)+1.1225* i sin(4.4506)

Root_{2}= – 0.2905 – 1.0842i

(As a check) Root_{2}^{3}=1+1i

________________________________

In the above example, the checks worked out exactly, perhaps by chance.
r=1.4142

r

phi=atan(1/1)=0.7854

The 3 roots of 1+1i are:

________________________

φ

Root

Root

(As a check) Root

________________________________

φ

Root

Root

(As a check) Root

________________________________

φ

Root

Root

(As a check) Root

________________________________

Input data:

We wish to find the 3 roots of (1)^{1/3}

r=1

r^{1/3}=1

phi=atan(0/1)=0

The 3 roots of 1 are:

________________________

φ_{0}=0+2·π·0=0

Root_{0}=1·cos(0)+1· i sin(0)

Root_{0}=1

(As a check) Root_{0}^{3}=1

________________________________

φ_{1}=0+2·π·1

Root_{1}=1*cos(2.094395)+1* i sin(2.094395)

Root_{1}= – 0.5+0.866025i

(As a check) Root_{1}^{3}=1

________________________________

φ_{2}=0+2*3.141593*2=4.18879

Root_{2}=1*cos(4.18879)+1* i sin(4.18879)

Root_{2}= – 0.5 – 0.866025i

(As a check) Root_{2}^{3}=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:We wish to find the 3 roots of (1)

r=1

r

phi=atan(0/1)=0

The 3 roots of 1 are:

________________________

φ

Root

Root

(As a check) Root

________________________________

φ

Root

Root

(As a check) Root

________________________________

φ

Root

Root

(As a check) Root

________________________________

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:

x

Or

(x-1)(x

Ken's book is packed with examples and explanations that enable you to discover more than 150 techniques to speed up your arithmetic and increase your understanding of numbers. Paperback and Kindle: