2

For example, what is the value of $i^{83}~$? To my understanding, I can solve this problem in 2 different ways that give me different answers:

Solution 1: $$ i^{83} = i^{82}i=(i^2)^{41}i=(-1)^{41}i=-i $$

Solution 2: $$ i^{83} = ((-1)^{1/2})^{83}=((-1)^{83})^{1/2}=(-1)^{1/2}=i $$

Both solutions seem to be correct to me but they yield different results. Could someone please help me understand which one of these solutions is wrong and why?

strxwberri
  • 29
  • 1
    For $n \equiv 3 \pmod{4}$, $i^n=-i$. In particular, $i^{83}=-i$ since $83 \equiv 3 \pmod 4$. – Geoffrey Trang Aug 16 '22 at 04:47
  • 1
    The index law you used in the second method is not valid when using the principal value of the square root (i.e. that $\sqrt{-1} = i$ and not $-i$) – messenger Aug 16 '22 at 04:50
  • 1
    I think this is quite a good question, but I think you should generalize it a bit more to get it more attention. As it is now, the title seems to be asking for the solution to a homework question, whereas the real question is one about applying specific exponent laws to $i$, which is a far more general and on-topic question. Your generalized title could be something of the ilk "Is it valid to turn $i^n$ into $((-1)^n)^{1/2}$?" And then you could keep the body of your question as an example of that question. – user110391 Aug 16 '22 at 05:10
  • 1
    Thank you for your suggestion @user110391 . I updated the title per your suggestion. – strxwberri Aug 16 '22 at 05:25
  • 1
    Thanks @messenger. I think your answer is helpful, it would help me a lot if you could please elaborate with some details on why it is not valid to use the principal value of the square root. Thanks! – strxwberri Aug 16 '22 at 05:27

1 Answers1

0

Your solution.1 is correct.

$i=i,~i^2=-1,~i^3=-i,~i^4=1$

$$i^{83}=i^3\cdot i^{80}=i^3=-i$$

Your solution.2 is wrong, because you need to specify the branch. If you take $-1=e^{i\pi}$, then $\sqrt{-1}=e^{i\pi /2}=i$ $$(i)^{83}=(-1)^{83/2}=((-1)^{83})^{1/2}=((e^{i\pi})^{83})^{1/2}=(e^{i\pi3})^{1/2}=e^{i\pi3/2}=-i$$

MathFail
  • 21,128
  • 2
    I think a good question to this ought to resolve the contradiction as well, instead of just asserting that one or the other is correct. Why was their second solution incorrect? – user110391 Aug 16 '22 at 05:13
  • You need to specify the branch @user110391 – MathFail Aug 16 '22 at 05:16
  • 1
    Thank you @MathFail. I can understand the substitution with Euler's formula and how you got $-i$ as the result. Could you please elaborate on what "you need to specify the branch" means in this context? Thank you! – strxwberri Aug 16 '22 at 05:32
  • 1
    @strxwberri We need to specify a certain branch so that we don't have a function taking a single input to multiple outputs. We represent complex numbers as $z = re^{i(\theta + 2n\pi)}$ for any $n \in \mathbb{Z}$. For instance, $-1$ can also equal $e^{i3\pi}$ (so $n = 1$ in this case), which would imply $\sqrt{-1} = e^{i3\pi/2} = \left(e^{i\pi/2}\right)^3 = i^3 = -i$. In fact, you can do this with any integer $n$ and see that you'd either get $i$ or $-i$ as a solution. – Accelerator Aug 16 '22 at 05:58