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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?

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    $\begingroup$ $1/i=-i{{{{}}}}$ $\endgroup$ – Wojowu Oct 8 '15 at 20:18
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    $\begingroup$ 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. $\endgroup$ – user137731 Oct 8 '15 at 20:18
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    $\begingroup$ 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 $-i$. $\endgroup$ – Ramiro Oct 8 '15 at 21:14
  • $\begingroup$ 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). $\endgroup$ – Ramiro Oct 9 '15 at 0:17
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Hint:

$(-i) \cdot i=1$, so the inverse of $i$ is: $\dfrac{1}{i}=-i$.

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$\cfrac{1}{\sqrt{-1}}=\color{blue}{\cfrac{1}{i} = \cfrac{1}{i} \times \cfrac{i}{i}} = \color{#F80}{\cfrac{i}{i^2}=\cfrac{i}{-1}}=-i$

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 $$i^2=-1$$

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  • $\begingroup$ @marc Is there something else? $\endgroup$ – BLAZE Oct 8 '15 at 20:54
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    $\begingroup$ 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). $\endgroup$ – Ramiro Oct 8 '15 at 23:25
  • $\begingroup$ @Ramiro Okay, thanks again Ramio $\endgroup$ – BLAZE Oct 8 '15 at 23:28
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Hint:

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

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