# Why doesn't simplifying $i$ using division not seem to work?

I was wondering why simplifying $i$ doesn't seem to work with division the same way it does multiplication.

For example, the following works:

$i^4 = 1$

$i^{31} = (i^4)^7 \cdot i^3 = 1 \cdot (-1) \cdot i = -i$

But not when applying the rules of exponents to this concept with division:

$i^{31} = \cfrac{(i^4)^8}{i^1} = \cfrac{1}{i} = \cfrac{1}{\sqrt{-1}} \ne -i$

Have I messed things up with the rules?

• $1/i=-i{{{{}}}}$ – Wojowu Oct 8 '15 at 20:18
• How to format your questions: type $i^{-31i}$ to get $i^{-31i}$ and $\sqrt{-1}$ to get $\sqrt{-1}$. To get an expression all on one line, like $$(i^4)^7\cdot i^3$$ you need to type double $'s like $$(i^4)^7 \cdot i^3$$. Doing so makes your questions/ answers much more readable. For more info, see here. – user137731 Oct 8 '15 at 20:18 • The number$i$works perfectly fine with division:$1/i = -i$and the simplest way to prove it is to note that$i(-i)=-i^2=1$. The issue you are having has another source. It come from the equality$\sqrt{-1} =i$. Then you have $$1/i = 1/\sqrt{-1} = \sqrt{1/-1}= \sqrt{-1}= i \neq -i$$ This is wrong! But, where is the error? The error is that IF$a\geq 0$, then$\sqrt{a}$is the positive square root of$a$and$-\sqrt{a}$is the negative square root of$a$, BUT this does not apply if$a<0$or$a$is complex. In fact,$\sqrt{-1}$is multi-valued, it has two values:$i$and$-i$. – Ramiro Oct 8 '15 at 21:14 • Maybe it is worth to mention that we can produce incorrect results from$i=\sqrt{-1}$just using multiplication. For instance: $$-i=(-1).i= i.i.i = \sqrt{-1}.\sqrt{-1}.\sqrt{-1}= \sqrt{(-1).(-1).(-1)}= \sqrt{-1}=i$$ The issue is the same I explained in my comment above:$\sqrt{-1}$has actually two values$i$and$-i$(and we can not distinguish between them as "positive" square root and "negative" square root). – Ramiro Oct 9 '15 at 0:17 ## 3 Answers Hint:$(-i) \cdot i=1$, so the inverse of$i$is:$\dfrac{1}{i}=-i$.$\cfrac{1}{\sqrt{-1}}=\color{blue}{\cfrac{1}{i} = \cfrac{1}{i} \times \cfrac{i}{i}} = \color{#F80}{\cfrac{i}{i^2}=\cfrac{i}{-1}}=-i$In the$\color{blue}{\mathrm{blue}}$step we have multiplied by$1$; since$\cfrac{i}{i}=1$. In the$\color{#F80}{\mathrm{orange}}$step we have used the definition of$i$such that $$i=\sqrt{-1}$$ so $$i^2=-1$$ • @marc Is there something else? – BLAZE Oct 8 '15 at 20:54 • In fact,$\sqrt{-1}$is multivalued. It has two values$i$and$-i$. Your solution is OK because you actually used only$i^2=-1\$. (See my comment below the question). – Ramiro Oct 8 '15 at 23:25
• @Ramiro Okay, thanks again Ramio – BLAZE Oct 8 '15 at 23:28

Hint:

$$\frac 1 i=\frac 1 i\cdot \frac i i$$