## Friday, December 19, 2014

### Unizor - Trigonometry - Exponentiation of Complex Numbers

Continuing on trigonometric representation of complex numbers, let's research how to raise a complex number in trigonometric form to some power (the process called exponentiation). It's important to understand that the rules we will be dealing with are not theorems that we prove, but definitions of new operations. We just prove that these definitions are reasonable in a sense that they preserve important properties of new operations, similar to properties we already know for real numbers, like

a^0 = 1 and

a^(P+Q) = a^P·a^Q and

a^(-P) = 1/(a^P) and

a^(P·Q) = (a^P)^Q

Since we know how to multiply two complex numbers in trigonometric form, we can easily derive a formula for raising a complex number to a positive integer power since this process is just a multiplication by itself certain number of times.

Indeed, using the property of multiplication addressed in the previous lecture,

[r·cos(φ)+i·r·sin(φ)]^2 =

= [r·cos(φ)+i·r·sin(φ)] ·

· [r·cos(φ)+i·r·sin(φ)] =

=r·r·cos(φ+φ)+i·r·r·sin(φ+φ)=

= r^2·cos(2·φ)+i·r^2·sin(2·φ) =

= r^2·[cos(2·φ)+i·sin(2·φ)]

Similarly, by induction, for any natural N,

[r·cos(φ)+i·r·sin(φ)]^N =

= r^N·cos(N·φ)+i·r^N·sin(N·φ) =

= r^N·[cos(N·φ)+i·sin(N·φ)]

Expanding this to negative integer numbers, we will use the main property of exponentiation we would certainly want to preserve:

a^(M+N) = a^M · a^N

and, derived from it for N=−M

1 = a^0 = a^(M−M) = a^[M+(−M)] = a^M · a^(−M)

from which follows:

a^(−M) = 1 / (a^M)

Therefore, it's reasonable to define an exponentiation of complex numbers with negative integer power −N (where N is a positive integer) as

[r·cos(φ)+i·r·sin(φ)]^(−N) =

= 1/[r^N·cos(N·φ)+i·r^N·sin(N·φ)]

Now it's easy to notice that for any argument (phase) ψ of a complex number in polar form

[cos(ψ)+i·sin(ψ)]·[cos(−ψ)+i·sin(−ψ)] =

= cos(ψ−ψ)+i·sin(ψ−ψ) =

= cos(0) + i·sin(0) = 1

Therefore,

1 / [cos(ψ)+i·sin(ψ)] =

= cos(−ψ)+i·sin(−ψ)

Using the above equality for exponentiation to a negative integer power and similar equality for real numbers 1/(r^N)=r^(−N), we derive

[r·cos(φ)+i·r·sin(φ)]^(−N) =

= r^(−N)·[cos(−N·φ)+i·sin(−N·φ)]

Notice the universality of the formula for natural exponent (a consequence of plain multiplication)

{r·[cos(φ)+i·sin(φ)]}^N =

= r^N·[cos(N·φ)+i·sin(N·φ)]

which has exactly the same form if N is a negative integer.

Let's proceed to rational exponent of a complex number in trigonometric form.

The simple approach to define exponentiation with rational exponent is to use the above formula for integer N.

Let's define a new absolute value (modulus) q=r^N and a new argument ψ=N·φ.

Then r=q^(1/N), φ=ψ/N and the formula would look like this:

{q^(1/N)·[cos(ψ/N)+i·sin(ψ/N)]}^N =

= q·[cos(ψ)+i·sin(ψ)]

This is a justification for the general definition of raising a complex number to a power equal to a rational number 1/N (where N is an integer):

{q·[cos(ψ)+i·sin(ψ)]}1/N =

= q^(1/N)·[cos(ψ/N)+i·sin(ψ/N)]

Notice, again, the general format of the formula is exactly the same as for positive integer exponents, that is the absolute value is raised to a power and an argument is multiplied by this power.

From here it's just one little step to derive a formula for any rational exponent. We just combine the rules for z^M and z^(1/N) into one rule for z^(M/N):

{r·[cos(φ)+i·sin(φ)]}^(M/N) =

= r^(M/N)·[cos((M/N)·φ)+i·sin((M/N)·φ)]

Imagine how cumbersome the formula for raising a complex number into a rational power would look in the traditional representation of a complex number as z=a+b·i.

To make this lecture complete, it's necessary to say a few words about irrational exponents. Here, as with raising real numbers to irrational power, exact rigorous mathematics is quite involved. Sufficient to say that the theory of limits is used and, as irrational numbers can be considered as limits of sequences of rational numbers, the definition of irrational exponentiation is based on the same limit principle and is defined as a limit of corresponding rational exponentiation.

Example

Let's see what (−1)^1/2 is if we use the trigonometry.

(−1)^1/2 = [cos(π)+i·sin(π)]^1/2 = cos(π/2)+i·sin(π/2) = 0+i·1 = i

as expected, because this is a definition of a complex i.

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