ns can"t seem to uncover a systems to this for the life that me. My jamesmerse.com teacher didn"t recognize either.

You are watching: What is the value of the expression i 0 × i 1 × i 2 × i 3 × i 4?

Edit: ns asked the teacher that generally teaches my food today, and she claimed it was incredible that the various other teacher didn"t know.

My logic goes together follows:

any actual number: $x$ come the fourth power is same to $(x^2)^2$. Using this logic, $i^4$ would certainly be equal to $(i^2)^2$. This would result in $(-1)^2$, and also $(-1)^2 = 1$.

Obviously, this logic have the right to be applied to any type of real numbers, but does it also apply to facility numbers?


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Harsh Kumar
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asked january 5 "17 in ~ 20:47
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TravisTravis
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$egingroup$ Yes, $i^4=1$. However what else might it same ?? $endgroup$
–user65203
jan 7 "17 in ~ 15:00
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Yes. The strength of $i$ room cyclic, repeating themselves ever time the exponent boosts 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 thinking is excellent, and also you have to feel great about the truth that friend figured this the end on your own. The fact that your jamesmerse.com teacher didn"t understand this is, in my experienced opinion as a jamesmerse.com educator, a disgrace.

Edited to add: together Kamil Maciorowski notes in the comments, the pattern persists for an adverse exponents, as well. Specifically,$$i^-1= frac1i = -i$$If $frac1i=-i$ appears odd, notification that $i(-i) = -i^2 = -(-1) = 1$, so $i$ and also $-i$ room multiplicative inverses; thus $i^-1 = -i$. When you understand that, you can expand the pattern:$$i^-1 = -i$$$$i^-2 = -1$$$$i^-3 = i$$$$i^-4 = 1$$and so on.

Second update:The OP asks for some added discussion the the residential property $left( x^a ight)^b = x^ab$, so below is part background ~ above that:

First, if $a$ and $b$ are natural numbers, then exponentiation is most naturally taken in terms of repeated multiplication. In this context, $x^a$ means $(xcdot xcdot cdots cdot x)$ (with $a$ components of $x$ appearing), and $left( x^a ight)^b$ means $(xcdot xcdot cdots cdot x)cdot(xcdot xcdot cdots cdot x)cdot cdots cdot (xcdot xcdot cdots cdot x)$, v $b$ sets of parentheses, each containing $a$ factors of $x$. Due to the fact that multiplication is associative, we deserve to drop the parentheses and recognize this as a product the $ab$ components of $x$, i.e. $x^ab$.

Note the this thinking works for any $x$, even if it is it is positive, negative, or complex. The even uses in settings were multiplication is noncommutative, prefer matrix multiplication or quaternions. Every we need is the multiplication is associative, and also that $a$ and also $b$ be organic numbers.

Once us have developed that $left( x^a ight)^b = x^ab$ for organic numbers $a,b$ us can extend the logic to integer exponents. If $a$ is a confident number, and if $x$ has actually a multiplicative inverse, then we specify $x^-a$ to average the same thing as $left(frac1x ight)^a$, or (equivalently) as $frac1x^a$. Through this convention in place, that is simple to verify the for any combination of indicators for $a,b$, the formula $left(x^a ight)^b = x^ab$ holds.

Note but that in expanding the formula come cover a larger collection of exponents, we have also made it important to restrict the domain of values $x$ over which this residential property holds. If $a$ and also $b$ space just natural numbers then $x$ have the right to be practically any object in any set over i m sorry an associative multiplication is defined. But if we desire to enable $a$ and $b$ to it is in integers then we need to restrict the formula to the situation where $x$ is one invertible element. In particular, the formula $x^a$ is no really well-defined if $x=0$ and also $a$ is negative.

Now let"s consider the situation where the exponents are not just integers yet arbitrary rational numbers. We start by defining $x^1/a$ to median $sqrtx$. ( view Why does $x^frac1a = sqrtx$? for a short explanation the why this convention provides sense.)

In this definition, we room assuming the $a$ is a herbal number, and also that $x$ is positive. Why execute we require $x$ to be positive? Well, take into consideration an expression like $x^1/2$. If $x$ is positive, this is (by convention) characterized to it is in the hopeful square source of $x$. But if $x$ is negative, then $x^1/2$ is not a real number, and even if we expand our number mechanism to include complicated numbers, the is not completely clear i m sorry of the two facility square root of $x$ this must be determined with. An ext or much less the same difficulty arises as soon as you try to extend the home to facility $x$: while nonzero facility numbers do have square roots (and $n$th roots in general), over there is no means to pick a "principal" $n$th root.

Things get really crazy as soon as you shot to prolong the building $left(x^a ight)^b=x^ab$ come irrational exponents. If $x$ is a confident real number and $a$ is a genuine number, we have the right to re-define the expression $x^a$ to mean $e^aln x$, and it have the right to be verified that this re-definition to produce the same results as all of the conventions above, however it only works since $ln x$ is well-defined for positive $x$. As shortly as you shot to allow an adverse $x$, friend run right into trouble, due to the fact that $ln x$ isn"t well-defined in that case. One can define logarithms of an unfavorable and complicated numbers, but they are not single-valued, and also there room all type of technicalities about choosing a "branch" the the logarithm function.

In particular -- and this is really important for the question available -- the identification $left(x^a ight)^b=x^ab$ does not host in general if $x$ is no a confident real number or if $a,b$ are not both integers. A lot of of people misunderstand this, and also indeed there are many, many, many, numerous questions top top this site that are rooted in this misunderstanding.

See more:
Kyouju To Shitsuji No Shizuka Na Seikatsu Vol 1 Chapter 3, Just A Moment

But v respect to the inquiry in the OP: the is perfectly reasonable come argue the $i^4 = left(i^2 ight)^2$, due to the fact that even though $i$ is a complicated number, the exponents space integers, so the basic notion of indexes as repetitive multiplication is reliable.