We usually represent sqrt(-1) as i

[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 don't 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??

 

[matt grime]

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.

 

[HallsofIvy]

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.

 

[svigneshkumars]

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

 

[matt grime]

"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.

 

[HallsofIvy]

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).

 

[HallsofIvy]

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

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.

 

[svigneshkumars]

Thanks to all ur replies



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

 

[matt grime]

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).

 

[Crosson]

Thanks to all ur replies



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

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

 

[nicksauce]

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

 

[HallsofIvy]

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

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

 

[nicksauce]

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

True

 

[uman]

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

 

[nicksauce]

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

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

 

[HallsofIvy]

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 $i^i$.

 

[uman]

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"

 

[nicksauce]

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.