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$$i^6=i^{\frac{4\times3}{2}}=(i^4)^{3/2}=1^{3/2}=1.$$

Where is the mistake?

Md Abidur Rahman
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    You can use a similar argument to show that $(-1)^1=1$. We have $(-1)^1=(-1)^{\frac{2}{2}}=((-1)^2)^{1/2}=1^{1/2}=1$.

    This is obviously not true, because a number can have multiple square roots in the real and complex numbers. So we can't just take one square root as the answer.

     – MathMinded Dec 31 '24 at 15:45
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    there are many versions of this question on the site, e.g. https://math.stackexchange.com/questions/49169/why-sqrt-1-times-1-neq-sqrt-12 or https://math.stackexchange.com/questions/3219025/why-22-5-isnt-equal-to-2251-10-fractional-powers-of-ne or https://math.stackexchange.com/questions/281528/what-is-wrong-with-this-fake-proof-ei-1?noredirect=1&lq=1 – Matthew Towers Dec 31 '24 at 15:49
  • When you take $(i^4)^{3/2}$ , you are effectively raising $i^4=1$ to the $3/2$-th power . This works fine in real numbers, but you are implicitly assuming that fractional exponents for complex numbers behave the same way, which is not the case. Specifically, going by the "best definition" $$a^b=\exp(b\times \log a)$$ – Antony Theo. Dec 31 '24 at 15:49
  • See this answer to a related question – MPW Dec 31 '24 at 15:50
  • Please only one question per post. Moreover, your second question is too vague and (as shown in MathMinded's example) unrelated to the first one. Better remove it. I disagree with @AntonyTheo.'s "This works fine in real numbers". Your problem is not about real versus complex numbers but about integer versus fractional exponents. For this reason, imo the linked post to close as a duplicate is not very well chosen. A better one would be https://math.stackexchange.com/questions/547181 – Anne Bauval Dec 31 '24 at 16:42

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