i can"t seem to find a solution to this for the life the me. My sdrta.netematics teacher didn"t understand either.

You are watching: What does i^4 equal

Edit: i asked the teacher that generally teaches my course today, and also she said it was significant that the other teacher didn"t know.

My reasonable goes as follows:

any genuine number: $x$ to the fourth power is same to $(x^2)^2$. Utilizing this logic, $i^4$ would be same to $(i^2)^2$. This would result in $(-1)^2$, and $(-1)^2 = 1$.

Obviously, this logic have the right to be used to any kind of real numbers, but does it likewise apply to complicated numbers?


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edited Feb 5 "17 at 16:05
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Harsh Kumar
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asked jan 5 "17 at 20:47
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TravisTravis
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$egingroup$ Yes, $i^4=1$. But what else could it equal ?? $endgroup$
–user65203
january 7 "17 in ~ 15:00
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Yes. The strength of $i$ room cyclic, repeating themselves ever before time the exponent increases by 4:$$i^0 = 1$$$$i^1=i$$$$i^2 = -1$$$$i^3 = -i$$$$i^4 = 1$$$$i^5 = i$$$$i^6 = -1$$$$i^7 = -i$$$$i^8 = 1$$etc.

Your reasoning is excellent, and you should feel great about the truth that you figured this out on her own. The reality that her sdrta.net teacher didn"t recognize this is, in my professional opinion as a sdrta.net educator, a disgrace.

Edited come add: as Kamil Maciorowski notes in the comments, the sample persists for negative exponents, together well. Specifically,$$i^-1= frac1i = -i$$If $frac1i=-i$ appears odd, an alert that $i(-i) = -i^2 = -(-1) = 1$, for this reason $i$ and also $-i$ space multiplicative inverses; because of this $i^-1 = -i$. When you recognize that, you can expand the pattern:$$i^-1 = -i$$$$i^-2 = -1$$$$i^-3 = i$$$$i^-4 = 1$$and therefore on.

Second update:The OP asks because that some extr discussion of the residential property $left( x^a ight)^b = x^ab$, so below is part background top top that:

First, if $a$ and also $b$ room natural numbers, climate exponentiation is many naturally interpreted in terms of repeated multiplication. In this context, $x^a$ method $(xcdot xcdot cdots cdot x)$ (with $a$ factors of $x$ appearing), and also $left( x^a ight)^b$ way $(xcdot xcdot cdots cdot x)cdot(xcdot xcdot cdots cdot x)cdot cdots cdot (xcdot xcdot cdots cdot x)$, with $b$ set of parentheses, every containing $a$ determinants of $x$. Since multiplication is associative, we can drop the parentheses and also recognize this together a product of $ab$ factors of $x$, i.e. $x^ab$.

Note the this reasoning works for any $x$, even if it is it is positive, negative, or complex. The even uses in setups were multiplication is noncommutative, choose matrix multiplication or quaternions. All we need is that multiplication is associative, and that $a$ and $b$ be natural numbers.

Once us have established that $left( x^a ight)^b = x^ab$ for natural numbers $a,b$ we can prolong the reasonable to essence exponents. If $a$ is a confident number, and if $x$ has actually a multiplicative inverse, climate we specify $x^-a$ to mean the exact same thing as $left(frac1x ight)^a$, or (equivalently) together $frac1x^a$. V this convention in place, it is straightforward to verify the for any mix of signs for $a,b$, the formula $left(x^a ight)^b = x^ab$ holds.

Note however that in prolonging the formula to cover a larger collection of exponents, us have also made it essential to restrict the domain of values $x$ over which this building holds. If $a$ and also $b$ room just organic numbers then $x$ deserve to be virtually any thing in any collection over i beg your pardon an associative multiplication is defined. But if we want to permit $a$ and $b$ to be integers then we need to restrict the formula come the instance where $x$ is one invertible element. In particular, the formula $x^a$ is not really well-defined if $x=0$ and also $a$ is negative.

Now let"s think about the instance where the exponents space not simply integers but arbitrary rational numbers. We begin by specifying $x^1/a$ to typical $sqrtx$. ( check out Why does $x^frac1a = sqrtx$? for a quick explanation the why this convention makes sense.)

In this definition, we space assuming that $a$ is a natural number, and also that $x$ is positive. Why do we require $x$ to be positive? Well, consider an expression choose $x^1/2$. If $x$ is positive, this is (by convention) identified to be the hopeful square source of $x$. However if $x$ is negative, climate $x^1/2$ is not a genuine number, and also even if we extend our number system to include complex numbers, that is not totally clear which of the two complicated square root of $x$ this have to be identified with. An ext or much less the same problem arises when you try to extend the residential or commercial property to complicated $x$: while nonzero complex numbers do have actually square roots (and $n$th root in general), over there is no way to pick a "principal" $n$th root.

Things get really crazy once you try to extend the residential or commercial property $left(x^a ight)^b=x^ab$ come irrational exponents. If $x$ is a hopeful real number and $a$ is a genuine number, we can re-define the expression $x^a$ to average $e^aln x$, and also it can be confirmed that this re-definition produce the same outcomes as every one of the conventions above, but it just works due to the fact that $ln x$ is well-defined for confident $x$. As soon as you try to allow negative $x$, friend run right into trouble, due to the fact that $ln x$ isn"t well-defined in the case. One can specify logarithms of negative and complicated numbers, however they space not single-valued, and also there room all type of technicalities around choosing a "branch" of the logarithm function.

In particular -- and this is really important because that the question available -- the identification $left(x^a ight)^b=x^ab$ does not hold in general if $x$ is not a optimistic real number or if $a,b$ space not both integers. A many of civilization misunderstand this, and also indeed there are many, many, many, numerous questions ~ above this site that are rooted in this misunderstanding.

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But with respect to the inquiry in the OP: that is perfect reasonable come argue that $i^4 = left(i^2 ight)^2$, since even though $i$ is a facility number, the exponents space integers, for this reason the an easy notion of exponentiation as repeated multiplication is reliable.