Rational power of the imaginary unit

[coreanphysicsstudent]

Homework Statement

solve i^2/3 using complex number.

Relevant Equations

below is my solution.

i^2/3=e^(2/3(lni)), ※(lni = i(pi/2+-2npi))
=e^2/3(i(pi/2+-2npi))=e^i(pi/3+-4/3npi)

※e^i(+-4/3npi)=+-1 (is tha correct?) .....(*)

e^ipi/3 = (1/2 + root3/2i)..........(**)

(**)(*) =e^i(pi/3+-4/3npi)= i^2/3 (using property of exponents)

So, my answer is +-(1/2+root3/2i)

but the real answer is −1, (1±i√3)/2

What's wrong with my solution?

please help me through.

 

[haruspex]

e^i(+-4/3npi)=+-1 (is tha correct?) ...(*)

No.

 

[BvU]

Hello @coreanphysicsstudent , !

First: you want to learn some $\LaTeX$ to typeset math stuff; it's worth the investment.
$$i^{2/3}$$ or $$i^{2/3}$$looks so much better than i^2/3 !
$$ $$
Or else learn to use the editing buttons to at least get i^2/3

Re your answer: $^e\log$ domain (principal value) is restricted to $\arg \log \in (-\pi,\pi]$

Furthermore: you have two answers ($\pm$ the book answer) and you can check if both yield $i$ when raised to the 3/2^th power !

Finally, it is good to use a simpler method for exponentiating complex numbers:

$${\bf z} = {\bf a}^x \Leftrightarrow \quad |{\bf z}| = |{\bf a}|^x \quad \& \quad \arg {\bf z} = x \arg {\bf a}$$

 

[etotheipi]

I think the quickest way here is to notice that $i = e^{i(\frac{\pi}{2} + 2n\pi)}$

 

[BvU]

I think the quickest way here is to notice that $i = e^{i(\frac{\pi}{2} + 2n\pi)}$

Easier: $i = e^{i\,\pi/2}$. Therefore the modulus of $i^{2/3}$ is 1 and the argument $\pi/3$

We also don't say 1 = 1 * 1ⁿ

 

[etotheipi]

Easier: $i = e^{i\,\pi/2}$. Therefore the modulus of $i^{2/3}$ is 1 and the argument $\pi/3$

We also don't say 1 = 1 * 1ⁿ

Right, but if you use the more general form then you also get other solutions like $e^{\frac{5\pi}{3}i}$.

 

[BvU]

That is not $i^{2/3}$

 

[etotheipi]

That is not $i^{2/3}$

Wolfram alpha seems to give three roots, which correspond to different values of $n$.

 

[haruspex]

Easier: $i = e^{i\,\pi/2}$. Therefore the modulus of $i^{2/3}$ is 1 and the argument $\pi/3$

We also don't say 1 = 1 * 1ⁿ

As @etotheipi notes, $i = e^{i\,\pi/2+2ni\pi}=e^{\frac {i\pi}2(4n+1)}$, so $i^{\frac 23}=e^{\frac {i\pi}3(4n+1)}$

 

[coreanphysicsstudent]

Hello @coreanphysicsstudent , !

First: you want to learn some $\LaTeX$ to typeset math stuff; it's worth the investment.
$$i^{2/3}$$ or $$i^{2/3}$$looks so much better than i^2/3 !
$$ $$
Or else learn to use the editing buttons to at least get i^2/3

Re your answer: $^e\log$ domain (principal value) is restricted to $\arg \log \in (-\pi,\pi]$

Furthermore: you have two answers ($\pm$ the book answer) and you can check if both yield $i$ when raised to the 3/2^th power !

Finally, it is good to use a simpler method for exponentiating complex numbers:

$${\bf z} = {\bf a}^x \Leftrightarrow \quad |{\bf z}| = |{\bf a}|^x \quad \& \quad \arg {\bf z} = x \arg {\bf a}$$

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Well Thank you for your advice. I wil learn LaTex sooner or later.
But I really don't understand your answer, it's quite difficult that using arg. I don't have any ideas of arg method.here's my method below, can you please adjust my method?

and how can i find correct answer that book refer.

 

[coreanphysicsstudent]

No.

Can you inform why (*) is wrong, and the correct method.

 

[BvU]

Wolfram alpha seems to give three roots, which correspond to different values of $n$.

View attachment 262640

You mean here ?

To which I can subscribe. But:

And this site seriously claims $i^{2/3} = -1$ ? and also that $e^z = e^{-z}$ ?

 

[etotheipi]

Down at the bottom it gives you one with all the roots on it:

We can take ${(e^{\frac{5\pi i}{3}})}^{\frac{3}{2}} = e^{\frac{5\pi i}{2}} = i$. I don't think it is claiming $e^z = e^{-z}$, it is just that $e^{-\frac{i \pi}{3}} \equiv e^{\frac{ 5 i \pi}{3}}$, which is another solution.

It's slightly odd behaviour, but I wonder if that is down to the weirdness of rational exponents like in this case... I could well have this all wrong, if I am missing a key rule or something!

 

[BvU]

Perhaps I am the one with egg on my face, stubbornly holding on to demanding a unique value in the complex world.

Here (at 4/5 into the page) is a counter-example ($i^i$) :

Another common pitfall is finding "the solution" to an exponential and assuming that it is the unique solution.

I'm inciined to ask @LCKurtz !PS And that would mean OP has 2/3 of the answers and his book answer only 1/3 !

 

[etotheipi]

It's definitely weird behaviour, and slightly confuzzling!

 

[SammyS]

@coreanphysicsstudent
You changed which expression is labelled (*) and which is labelled (**).

This is correct: $\displaystyle e^{i \pi/3}=\frac{1}{2}+\frac{\sqrt{3~}}{2}i$ .

This is incorrect in general: $\displaystyle e^{\pm i~ 4n\pi/3}=\pm 1$ . Moreover, it doesn't follow from what precedes it in your work.

What you have preceding it is (you do not need the ± ) is: $\displaystyle e^{i (\pi/3+4n\pi/3)}$.

But $\dfrac{\pi}{3}+\dfrac{4n\pi} 3 \ne \dfrac{4n\pi} 3$.

Rather $\dfrac{\pi}{3}+\dfrac{4n\pi} 3 = \dfrac{(4n+1)\pi} 3$ .

Try some values for $n$.

 

[haruspex]

that would mean OP has 2/3 of the answers and his book answer only 1/3 !

No, the book answer presents all three:
"but the real answer is −1, (1±i√3)/2."

It comes down to the exact wording of the question. It is quoted as saying "solve i^2/3" but that doesn’t quite make sense. If it actually says "find the solutions of" then we are clearly not only concerned with the principal solution. If it says "find" then I would consider it ambiguous.

 

[BvU]

 

[SammyS]

I think the quickest way here is to notice that $i = e^{i(\frac{\pi}{2} + 2n\pi)}$

Maybe a quicker way:

$i^{2/3} = \left(i^2\right)^{1/3}=\left(-1\right)^{1/3}$.

Of course, treat this as complex.

 

[WWGD]

In the main branch:
$i^{2/3}=e^{(2/3)Logi}=e^{(2/3)(i(\pi/2+2n \pi))}$. Exponentiation is single-valued within a branch.

 

[WWGD]

Or looking at $i^{2/3}$ as the solutions to $z^3=i^2=-1$ or the 3 roots of ( negative) unity.