We usually represent sqrt(-1) as i

1. Jun 30, 2008

svigneshkumars

we usually represent sqrt(-1) as i , but only a quadratic eqn or a eqn of high power would have led to represent the value as i.so for a quadratic eqn we obtain two values ,so why dont we have the two values as +i and -i.

statement 1:
X^2=4 so , X=+2 and -2
Statement 2:
X^2= -1 so X=+i and -i

is statement 2 possible or my question is wrong??

Have i asked a stupid q so that i should brush up my basics??

2. Jun 30, 2008

matt grime

Re: iota

i and -i are both solutions to x^2=-1. And it i isn't an iota; iotas don't have the dot on them.

3. Jun 30, 2008

HallsofIvy

Staff Emeritus
Re: iota

In fact, in the complex number system, any equation of the form xn= a, where a can be any complex number, has n distinct solutions.

4. Jul 1, 2008

svigneshkumars

Re: iota

So if i use -i instead of +i does it make any change ?

5. Jul 1, 2008

matt grime

Re: iota

"Use"? How?

There are 2 roots, they are indistinguishable. If we label one i, the other is -i. But there is no way to say which is which.

6. Jul 1, 2008

HallsofIvy

Staff Emeritus
Re: iota

This is going to get much more technical than the original poster will like! matt grime's point that "they are indistinguishable" is exactly why we CANNOT simply define "i" by "i2= -1" or as "$\sqrt{-1}$". What can be done is to define the complex number system as a set of objects of the form (a, b) where a and b can be any two real numbers, and define addition and multiplication by (a,b)+ (c,d)= (a+ c,b+ d) and (a,b)(c,d)= (ac-bd, ad+ bc). If we identify with complex numbers of the form (a, 0), imaginary numbers with complex numbers of the form (0,, b) and define "i" as (0, 1), then we have (a, b)= (a,0)(1,0)+ (b,0)(0,1)= a+ bi. It then follows that i2= (0,1)(0,1)= (0(0)-1(1),1(0)+ 0(1))= (-1, 0)= -1. It is also true that (0,-1)(0,-1)= (0(0)- (-1)(-1), 0(-1)+ 0(-1))= (-1, 0)= -1 but we have specifically defined i to be (0, 1), not (0, -1).

7. Jul 1, 2008

HallsofIvy

Staff Emeritus
Re: iota

Use how and what kind of change? (-i)2= (-1)2(i)2= (1)(-1)= -1 = i2 but 1+ i is certainly different from 1- i. It makes the same change as if you use -1 instead of +1.

8. Jul 1, 2008

svigneshkumars

Re: iota

Thanks to all ur replies

now i have another question .what would be the value of ii

9. Jul 1, 2008

matt grime

Re: iota

It has infinitely many 'values', since raising to the i'th power is not single-valued. To make it single valued one chooses a so called branch.

Note that a^b is defined to be exp(log(a^b)=exp(b log a) in general, so it is the same as defining log(i). Write i in polar coordinates to do this, and choose Arg (i.e. pi/2).

10. Jul 1, 2008

Crosson

Re: iota

Good question, one answer is ii = 0.20788...

11. Jul 1, 2008

nicksauce

Re: iota

To be exact, i^i = e^(-pi/2).

12. Jul 1, 2008

HallsofIvy

Staff Emeritus
Re: iota

To be exact, that is one of the infinitely many values of $$e^{-pi/2}$$.

13. Jul 1, 2008

nicksauce

Re: iota

True

14. Jul 3, 2008

uman

Re: iota

$$e^{-\frac{\pi}{2}}$$ has multiple values? Since when?

15. Jul 3, 2008

nicksauce

Re: iota

i^i has multiple values. That one is one of them.

16. Jul 3, 2008

HallsofIvy

Staff Emeritus
Re: iota

I was about to retort that no one had said any such thing- then noticed my typo.

I meant to say that $e^{-\pi/2}$ was one of the infinitely many values of $i^i$.

17. Jul 3, 2008

uman

Re: iota

I see.

I suppose it depends on how you define $$i^i$$. My book defined it as $$e^{Log(i)*i}$$ which was in turn defined completely unambiguously...

And by "my book" I mean "the only book I happen to have looked at that talks about complex exponential functions"

18. Jul 3, 2008

nicksauce

Re: iota

I suppose that this is what one might call the principal value of i^i, by taking Log(i) = log(r) + i*theta, where theta is in the range[0,2pi). But you can also have other values by modifiying Log(i) by any multiple of i*2pi.