# What does it mean to evaluate $i^{786}$?

In my book ,

To evaluate $$i^{786}$$.

They did. $$i^{4*196 + 2}$$

Why do we have to do it this way? I have to write in the exam in a subjective format. So want to know is this the only correct way to write it. Since if the exponent is sometimes bigger, why would anyone even think of doing it like this?

I would have simply said that

$$-1^{786/2}$$ since the power is 393. Therefore the answer is $$-1$$.

I understand it is more of a homework question but I just couldn’t find the tag for it.

• Welcome to MSE! The title supports MathJax as well. Dec 21, 2020 at 7:08
• Thank you @Adola and e2-e4
– user864449
Dec 21, 2020 at 7:17
– MJD
Dec 21, 2020 at 15:43
• @user282657: note that, according to your reasoning, $$i=(i^4)^{1/4}=1^{1/4}=1.$$ Dec 21, 2020 at 15:50

$$i^n$$ is periodic with period of $$4$$

$$i^1=i$$ $$i^2=-1$$ $$i^3=-i$$ $$i^4=1$$

So we only need to know the remainder when $$786$$ is divided by $$4$$:

So $$i^{786}=i^{4*196+2}=i^2=-1$$

Your answer raises a new question: How do you know that $$(-1)^{393}=-1$$? Well, you could argue: $$(-1)^{393} = (-1)^{2\times 196+1} = 1^{196}\times(-1)^1=-1$$ This is basically doing with $$-1$$ what the book did with $$i$$. Since you're making a similar argument either way, the book's author probably decided that it's simpler to skip past $$(-1)^{393}$$ and operate directly on the expression $$i^{786}$$.

Notice that $$i = i$$ $$i^2 = i * i= -1$$ $$i^3 = i^2*i = -1*i = -i$$ $$i^4 = i^3*i = -i*i = -(-1) = 1$$ $$i^5 = i^4 * i = 1 * i = i$$

You can notice that for any integer k you'll have that: $$i^{4k} = 1$$ $$i^{4k+1} = i$$ $$i^{4k+2} = -1$$ $$i^{4k+3} = -i$$

This is commonly written as $$i^k = i^{k \mod 4}$$

It's probably just a way to emphasize that the result will be $$-1$$, as $$i^{4+2}=i^4\cdot i^2 = 1\cdot-1$$That's easier to comprehend.