# Which algebraic rules altered for complex numbers?

1. Nov 4, 2008

### Gerenuk

What changes to the algebraic rules when imaginary "i" is introduced in addition to real numbers?

I know that complex numbers do not have a "<" or ">" relation.
Some functions become multivalued.

What about school algebra? Maybe exponentiation rules are slightly modified?

2. Nov 4, 2008

### Office_Shredder

Staff Emeritus
Exponentiation is hellofalot different. You have the relationship eix = cos(x) + isin(x) if x is a real number (This is seen from the Taylor series of the three functions if you know what those are, otherwise just accept it on faith). Notice the RHS of that equation is periodic. So you get that exponentiation is not 1-1, and often isn't even single valued. If you have a, b complex numbers, we define

$$a^b:= e^{bln(a)}$$

where the natural log of a complex number is defined by its polar coordinates, as if a has polar coordinates $$(r,\theta)$$ (notice that this isn't unique, you can choose $$(r,\theta + 2\pi$$ instead for example) then

$$a=re^{i\theta}$$

So

$$ln(a) = ln(r) + i\theta$$ (this is one of those multivalued functions)

Going back to the original point
$$a^b = e^{bln(a)} = e^{b(ln|a| + iarg(a)}$$ which can be calculated by multiplying the real and complex parts of b and ln(a), and then using the identity for eix above.

The effect of this (and you can investigate this yourself) is that while ab is single valued when be is an integer , if b isn't it becomes multivalued. Similar to how square root returns two values, a positive and a negative one, over the real numbers, taking a cube root returns three values, a fourth root returns four values, etc. Taking irrational roots will return infinite values

3. Nov 4, 2008

### Gerenuk

So would rules like
$$x^{a+b}=x^ax^b$$
and
$$(xy)^a=x^ay^a$$
and
$$(x^a)^b=x^{ab}$$
change?

Last edited: Nov 4, 2008
4. Nov 6, 2008

### Office_Shredder

Staff Emeritus
For the most part, but you have to be careful about the multivalued aspect of exponentiation (particularly for the last one). For example

$$\sqrt{(-1)^2} = \sqrt{-1}^2$$

naively gives

$$\sqrt{1} = i^2$$ hence $$1=-1$$ It IS true that ONE OF the square roots of 1 is -1, but not all of them. This can also cause trouble if, for example, you have

$$a^3 = b^3$$ and take the third root of each side to conclude

$$a = b$$

True in the reals, but not in the complex numbers as 1 actually has three distinct cube roots $$e^{\pi 2i/3}$$ $$e^{\pi 4i/3}$$ and 1 itself...you can see how cubing the first one gives 1 from using the formula in my first post, and then the second one is just the square of the third, and the third one is the cube of the third... but if you take the fourth power of $$[e^{\pi 2i/3}]^4 = e^{\pi 8i/3} = e^{\pi 2i/3}$$ as $$e^{\pi 6i/3} = 1$$. Exercise for the reader: How do I know that I have the 'right' value of $$[e^{\pi 2i/3}]^4$$ so that I can just multiply the values, especially right after giving two examples of how you can't?

5. Nov 6, 2008

### Gerenuk

So if I want to write down some rules to brainlessly apply and still get correct results, what should I do?

I was thinking of something like
$$x^ax^b=x^{a+b}e^{2\pi\mathrm{i}(ma+nb)}$$

I mean how would I define my algebra (in particular exponentiation) in a strict way?

6. Nov 6, 2008

### Tac-Tics

Functions are not multivalued. However, functions which are invertible in R, such as exp (whose inverse is ln) no longer have inverses. Given a complex y and an unknown complex x, if you know that exp(x) = y, you *don't* know x (although you have narrowed it down to a countably infinite subset of R).

This rule is a combination of two rules. You can split it into "x^a x^b = x^(ab)" (with some restrictions I'm not sure of) and "e^(2pi i) = 1" is sufficient to show this relationship. And even that is just a specification of Euler's formula.