3

I feel like there is a pretty critical flaw with my logic here but I cant figure out where it is.

I don't have much knowledge about imaginary and complex numbers , but I've watched Khan academy's video on them earlier and saw this video here , where he demonstrated that very high powers of $i$ can be simplified by taking multiples of 4 of them , since $i^4=1$

say we have $i^{25}$, cant we rewrite this as $i^{4*\frac{25}{4}} = (i^4)^{\frac{25}{4}} = 1^{\frac{25}{4}} = 1$ ? and do this for every number that belongs to R?

Dahen
  • 339
  • 2
  • 14
  • 2
    $z\mapsto z^4$ is not injective over $\mathbb{C}$, hence for any $z\in\mathbb{C}^*$ there are four distinct complex numbers $w$ such that $w^4=z$. You have to be careful in writing $z^{1/4}$. – Jack D'Aurizio Nov 01 '17 at 20:41
  • so basically it isn't as simple as I'm making it out to be? – Dahen Nov 01 '17 at 20:43
  • $(-1)^{2*\frac{25}{2}}=((-1)^2)^{\frac{25}{2}}=1^{\frac{25}{2}}=1$? – Eclipse Sun Nov 01 '17 at 20:46
  • @EclipseSun actually yeah shouldn't that work for that too? – Dahen Nov 01 '17 at 20:50
  • 1
    @Dahen: your argument fails to work over $\mathbb{R}$, too. According to it: $$-2=\sqrt[3]{-8}=(-8)^{1/3}=(-8)^{2/6}=\left[(-8)^2\right]^{1/6}=64^{1/6}=2$$ – Jack D'Aurizio Nov 01 '17 at 20:50
  • @JackD'Aurizio i know my argument is pretty nonsensical, but I don't really understand why I guess when I think of it – Dahen Nov 01 '17 at 20:52
  • 4
    The "obvious" identity $(z^a)^b=z^{ab}$ works for $z\in\mathbb{R}^+$, but not in general. – Barry Cipra Nov 01 '17 at 20:52
  • 1
    @Dahen: the issue relies in defining $z^{1/4}$ or $z^{1/3}$ when $z$ is not a non-negative real number, since you have to deal with the lack of injectivity of some function. – Jack D'Aurizio Nov 01 '17 at 20:53
  • @BarryCipra ohh , okay, well in that case it'd make a lot more sense that way – Dahen Nov 01 '17 at 20:53
  • @Dahen Apparently the first equality cannot hold. You may think $x^{ab}=(x^a)^b$. This does not hold in general. More easier one is $(-1)^{2*\frac12}\ne((-1)^2)^{\frac12}$. – Eclipse Sun Nov 01 '17 at 20:54
  • @EclipseSun I didnt know that this exponent property didn't hold true for most cases. Because that way I could understand why it wouldn't work like that. but in the video I've linked with it he states that $i^{4k} = (i^4)^k = 1$ , but I guess this only works where K is a natural number – Dahen Nov 01 '17 at 20:59
  • It is true for any complex number $z$, $z^{mn}=(z^{m})^n$, for $m,n\in\mathbb{N}$. – Eclipse Sun Nov 01 '17 at 21:02
  • @JackD'Aurizio I see, also just so we're on the same page, what is the function you're talking about? – Dahen Nov 01 '17 at 21:05

0 Answers0