Math Help - Confusion with'i'

1. Confusion with'i'

Ok hey guys,
I have sort of confused myself with the following:
$\frac{1}{i}\cdot\sqrt{\frac{1}{i}} = -i\left(\frac{1}{i}\right)^{\frac{1}{2}}=-\left(\frac{i^2}{i}\right)^{\frac{1}{2}}=-\sqrt{i}$
$\frac{1}{i}\cdot\sqrt{\frac{1}{i}} = \left(\frac{1}{i^3}\right)^{\frac{1}{2}} = \left(\frac{-i}{i^2}\right)^{1/2} = \sqrt{i}$.

So which one is correct?? And which operation is illegal/ill-defined?
Thanks a lot for the help.

2. Re: Confusion with'i'

The symbol:
$$\sqrt{i}$$

is not "well-defined": the complex number i has TWO square roots, and there is no clear choice for making one "the positive" one.

3. Re: Confusion with'i'

They are both correct. i2 has two roots, +i and -i
If you use the negative root in the first line you get:
-(i2/i)1/2=-((-i)(-i)/i)1/2 = (i)1/2

4. Re: Confusion with'i'

Ok, I feared that's what's happening. Thanks for clearing that up!

5. Re: Confusion with'i'

Notice Hartlw saying that " $i^2$ has two roots, i and -i" and Deveno saying "there is no clear choice for the positive one". They are both correct. Since the complex numbers do NOT form an "ordered" field, there is no single choice for "positive" and "negative" complex numbers.

6. Re: Confusion with'i'

Originally Posted by Deveno
The symbol:
$$\sqrt{i}$$

is not "well-defined": the complex number i has TWO square roots, and there is no clear choice for making one "the positive" one.
In the complex number system, c1/n is very well defined. That's why the complex number system was created.

And anyhow, the square roots of i have nothing to do with OP. It's the square roos of i2 that explains the OP.

7. Re: Confusion with'i'

Me also Confuse what are you doing now friends ?

8. Re: Confusion with'i'

Why does an incorrect and irrelevant post get three thanks, and when corrected (#6), a retard suddenly appears for the first time out of nowhere to upstage the correction?

9. Re: Confusion with'i'

Originally Posted by Looknan
Ok hey guys,
I have sort of confused myself with the following:
$\frac{1}{i}\cdot\sqrt{\frac{1}{i}} = -i\left(\frac{1}{i}\right)^{\frac{1}{2}}=-\left(\frac{i^2}{i}\right)^{\frac{1}{2}}=-\sqrt{i}$
$\frac{1}{i}\cdot\sqrt{\frac{1}{i}} = \left(\frac{1}{i^3}\right)^{\frac{1}{2}} = \left(\frac{-i}{i^2}\right)^{1/2} = \sqrt{i}$.
Have a look at this discussion.
There is a growing trend away from using $\sqrt{t}$ unless $t\in\mathbb{R}$.

Here is an early theorem in Marsden/Hoffman: If $z\in\mathbb{C}$ then $\exists w\in\mathbb{C}$ such that $w^2=z$ (notice that $-w$ also works).
We would use $-i\left(\frac{1}{i}\right)^{\dfrac{1}{2}}$ or $-i\left(-i\right)^{\dfrac{1}{2}}$

Because $\mathbb{C}$ is not ordered it is difficult to talk about a principal root.
But some authors do it this way.
If $z\in\mathbb{C}$ If $z^{\dfrac{1}{2}}=\sqrt{|z|}$$\exp\left(\dfrac{i \text{Arg}(z)}{2}\right) If you use that convention then (-i)^{\dfrac{1}{2}}=\exp\left(\dfrac{-i\pi}{4}\right)=$$\left(\dfrac{1}{\sqrt{2}}\right) +\left(\dfrac{-i}{\sqrt{2}}\right)$

10. Re: Confusion with'i'

Last post by Plato totally irrelevant, of course.

In the complex number system, c1/n is very well defined. That's why the complex number system was created.

11. Re: Confusion with'i'

Originally Posted by Hartlw
Last post by Plato totally irrelevant, of course.
In the complex number system, c1/n is very well defined. TRUE
That's why the complex number system was created. Absolutely FALSE
I understand that you think you are able to help with higher mathematics but you just know enough about the subject.

12. Re: Confusion with'i'

Originally Posted by Hartlw
Why does an incorrect and irrelevant post get three thanks,
Maybe that are four people who actually know enough about complex numbers to comment.

13. Re: Confusion with'i'

Originally Posted by Looknan
Ok hey guys,
I have sort of confused myself with the following:
$\frac{1}{i}\cdot\sqrt{\frac{1}{i}} = -i\left(\frac{1}{i}\right)^{\frac{1}{2}}=-\left(\frac{i^2}{i}\right)^{\frac{1}{2}}=-\sqrt{i}$
$\frac{1}{i}\cdot\sqrt{\frac{1}{i}} = \left(\frac{1}{i^3}\right)^{\frac{1}{2}} = \left(\frac{-i}{i^2}\right)^{1/2} = \sqrt{i}$.
As it shows that real operations may(rules) not obey apply to complex numbers.
However, complex operation are compatible with real numbers.

14. Re: Confusion with'i'

Originally Posted by Hartlw
Last post by Plato totally irrelevant, of course.

In the complex number system, c1/n is very well defined. That's why the complex number system was created.
In actual fact, there are "difficulties" with defining c1/n in the complex numbers. To see why, let's look at the analogous definition in the real numbers:

c1/n = cx = elog(c)x for x = 1/n.

The trouble is: "log" isn't definable on ALL the complex numbers because the analogue of ex isn't 1-1 on the complex numbers anymore (so it doesn't have a UNIQUE inverse: log).

To get around this difficulty, one can "restrict" the domain of the complex exponential to a horizontal "swath" of height 2*pi (this is what the "Arg" function does, but there are actually "many" arg functions we could use, so we have to be CLEAR about "which one" we are using).

The two complex numbers:

$$\frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2}$$
$$\frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2}$$

BOTH square to i, and there is no clear-cut choice for calling either one THE square root of i.

In a similar vein, there is no reason to say i OR -i is THE square root of -1. In fact, choosing EITHER ONE leads to a version of the complex numbers that we "can't tell apart algebraically". This is because:

(z + w)* = z* + w*
(zw)* = z*w*

So the mapping z-->z* preserves all the algebraic properties of the complex numbers (it is a field automorphism). The choice of i as the "positive direction of the y-axis" is a completely arbitrary one.

15. Re: Confusion with'i'

Let $a,~b\in\mathbb{C},~~ a\ne 0$ then $a^b=e^{b\log(a)}$

Here is an excellent quote from Jerrold E. Marsden: $a^b$ is singled valued if and only if $b\in\mathbb{Z}$; if $b\in\mathbb{Q}$ where $b=p/q$ in lowest terms then $a^b$ has exactly $q$ distinct values; if $b\in\mathbb{R}\setminus\mathbb{Q}\text{ or if Im}(b)\ne 0$ then $a^b$ has infinitely many values.

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