The imaginary unit can be multiplied by itself as any real number, obtaining the different exponentiations of the imaginary unit.
We will work in the following way:
- by agreement it is established that $$i^0=1$$, as happens with any other real number.
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for the first four degrees we have:
$$i^1=i$$
$$i^2=i\cdot i= \sqrt{-1}\cdot\sqrt{-1}=(\sqrt{-1})^2=-1$$
$$i^3=i^2\cdot i= (-1)\cdot i= -i$$
$$i^4=i^3\cdot i= (-i)\cdot i= -(i^2) =-(-1)=1$$
Where each of the powers is obtained by multiplying the previous one $$i$$ times.
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the following grades we will proceed similarly. Let's see the continuation:
$$i^5=i^4\cdot i=1\cdot i=i=i^1$$
$$i^6=i^5\cdot i=i\cdot i=i^2$$
$$i^7=i^6\cdot i=i^2\cdot i=i^3$$
$$i^8=i^7\cdot i=i^3\cdot i=i^4$$
So they form a periodic succession, since the values of the first four powers, which are $$ \ i,\; -1,\; -i,\; 1 \ $$ appear again and again. This is because if one wants the $$n$$-th power of the imaginary unit (to calculate $$i^n$$), this coincides with the result of raising $$i$$ to the remainder of dividing $$n$$ over $$4$$.
Namely $$ i^n=i^{4q+r} \ $$ where $$n=4q+r$$ is the usual euclidean division. Once we have this, by using of the properties of the exponentiation we can write: $$$i^n=i^{4q+r}=i^{4q}\cdot i^r= (i^4)^q\cdot i^r$$$ but since we have seen that $$i^4=1$$ then we are left with: $$$ i^n=(i^4)^q\cdot i^r=(1)^q \cdot i^r=i^r$$$
So it is enough to calculate $$i^r$$ where $$r$$ corresponds to the remainder of the division of $$n$$ over $$4$$.
It is easy to anticipate that $$i^n=i^r$$ will always give previously given results, since $$r$$ can only be $$0$$, $$1$$, $$2$$ or $$3$$.
Let's see some examples:
$$i^{347}$$ looks like a very difficult problem, but if we do the division of $$347$$ over $$4$$ we obtain $$347=4\cdot 86+3$$ so the remainder is $$3$$. That's why we can write:
$$i^{347}=i^{4\cdot 86}\cdot i^3= i^3$$
Then we look at the table that we have written with the first four powers of $$i$$ and observe that $$i^3=-i$$. Therefore we will have:
$$i^{347}=i^3=-i$$
What if we have a negative exponent?
If we want to calculate $$i^{-n}$$ we only must write it in the following way: $$i^{-n}= \dfrac{1}{i^n}$$. Then we solve the denominator as has been explained for positive powers of $$i$$ and then write the result in the denominator.
For example: $$$i^{-89}=\dfrac{1}{i^{89}}$$$
We will first compute $$i^{89}$$. By proceeding as we learned, we need to calculate the division of $$89$$ over $$4$$ and we obtain $$89=4\cdot 22+1$$, the remainder is $$1$$. So we have: $$$i^{89}=i^{4\cdot 22}\cdot i^1=i$$$
Once we have this, we re-write the result of what is required: $$$i^{-89}=\dfrac{1}{i^{89}}=\dfrac{1}{i}$$$
If we prefer that the imaginary unit stays in the numerator, we can do as we do when we want to eliminate it from the denominator, so we have to multiply and divide by its conjugate. This is: $$$\dfrac{1}{i}\cdot\dfrac{-i}{-i} = \dfrac{-i}{i\cdot(-i)}= \dfrac{-i}{-i^2}= \dfrac{-i}{-(-1)}=-i $$$