[highschool math] algebra with imaginary A debate with my teacher about [(-1)^(2÷4)]

The answer that would typically be given would be i. Explaining why is kind of technical. The short answer is that you run into issues when trying to define roots of negative numbers, because you end up having to deal with multivalued functions.

The first method interprets the expression as "The square root of -1." There are two square roots of -1. Theres i and -i.

The second method interprets the expression as "The fourth root of 1." There are 4 fourth roots of 1. There's, 1, i, -1, and -i.

So, you have to decide what you want the expression (-1)\^(2/4) to mean. Depending on the interpretation, it can mean different things. It's generally safer to spell out exactly what you mean rather than relying on typical exponent rules in these cases.

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If you want to get technical, typically what we do is define exponential expressions via the logarithm. We say a\^b = e\^(b \* ln(a)). So your expression would become:

(-1)\^(2/4) = e\^(2/4 \* ln(-1)).

To find ln(-1) you are looking for x in the equation e\^x = -1. There are an infinite number of such x values. They are given by x = i(pi + 2pi \* n) where n is any integer. So, there are many answers to ln(-1). It is a multivalued function. So, we have

(-1)\^(2/4) = e\^(2/4 \* i \* (pi + 2pi \* n)).

By plugging in different values of n you get the answers i and -i. To rule out one of these we will typically begin by choosing a preferred "branch" of the logarithm (called the principal branch), in which we restrict the domain so there is only one answer.
In complex numbers, (e^(z1))^z2 ≠ e^(z1\*z2) if z1 and z2 are both non-integers. (-1)^(2/4) = (-1)^(1/2) = ±i where the principle root is +i.
You’re teacher is correct. Every order of operations convention (PEMDAS, BODMAS, etc.) puts priority on the exponents. This means doing any operations in the exponents first. Therefore the 2÷4 would first be reduced to ½ and lead to *i* as the final result.
i

2÷4 is done first.

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