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Is the following result true?

If $i^4 = 1$, $(i^4)^{1/4} = 1^{1/4}$, and$i = 1$, then won't it disprove $i = (-1)^{1/2}$?

If no, give a proper reason? Can this be a contradiction to complex numbers?

amWhy
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Sandesh Raj
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    Welcome to MSE. This looks suspiciously like a homework problem. In any case, you are expected to show some effort of your own rather than simply expect us to provide you with a solution. – Ittay Weiss Jan 27 '21 at 14:31
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    There is no contradiction here. You incorrectly used laws of exponents which were only guaranteed to work in a purely positive real setting by trying to use them in a complex setting. It is not true in general that $(a^b)^c = a^{bc}$ when dealing with numbers which are not positive reals. – JMoravitz Jan 27 '21 at 14:32
  • Also related: which step in this process allows me to erroneously conclude that $i=1$? – JMoravitz Jan 27 '21 at 14:36
  • This is one reason why it is misleading to write $i=\sqrt{-1}=(-1)^{1/2}$. $i$ is not the square root of $-1$, it's a square root of $-1$. The only logical notation is to write $i^2 = -1$. Note that there is no natural way in the complex numbers to designate $i$ as the principal square root, in contrast to the real numbers, where we take $\sqrt{x}$ to mean the positive square root of $x$. – Joe Jan 27 '21 at 14:43

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No, because you cannot use the usual rules on exponents when you are working with complex numbers !

The definition $a^b = \underbrace{a \times \dots \times a}_{b \text{ times}}$ is always valid if $b$ is a positive integer, but in order to define what $a^{\frac 1 4}$ means, you need to find something else !

When working over the positive, real numbers, a way to define $a^{\frac 1 b}$ for some positive integer means is to say that it is the only positive real $x$ such that $x^b = a$. However, when working over the complex numbers, this $x$ is no longer unique, because there are non-trivial roots of the unity (numbers that are not $1$, but one of their powers is equal to $1$).

(If you know what the exponential function is) An even more general way to define exponentiation is to say that $x^\alpha = \exp(\alpha \ln(x))$ for any positive, real number $x$. However, this definition also fails terribly when $x$ is a complex number, as there is no "nice" way of extending $\ln$ to the complex numbers.

In short: the rules have changed, you have to be very careful when manipulating complex numbers along with exponents.

Numbra
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