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When I entered the value $$\frac{1}{i}$$ in my calculator, I received the answer as $-i$ whereas I was expecting the answer as $i^{-1}$. Even google calculator shows the same answer (Click here to check it out).

Is there a fault in my calculator or $\frac{1}{i}$ really equals $-i$? If it does then how?

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    $\begingroup$ Hint $i^2 = -1$ $\endgroup$
    – Mann
    May 11 '15 at 12:14
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    $\begingroup$ Multiply by $i/i$. $\endgroup$ May 11 '15 at 12:14
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    $\begingroup$ Hint $$z=\frac{1}{i}\iff zi=1\implies \dots$$ $\endgroup$
    – John Joy
    May 11 '15 at 12:56
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    $\begingroup$ Three down votes for someone exhibiting natural mathematical curiosity and having the wherewithal to ask about it is shameful. $\endgroup$
    – Emily
    May 11 '15 at 14:50
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    $\begingroup$ Excellent question I wondered that myself when I read it. I could say $+1$ but given the context of the question I should say $+i$! $\endgroup$
    – Math Man
    May 13 '15 at 1:04

11 Answers 11

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$$\frac{1}{i}=\frac{i}{i^2}=\frac{i}{-1}=-i$$

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Note that $i(-i)=1$. By definition, this means that $(1/i)=-i$.

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The notation "$i$ raised to the power $-1$" denotes the element that multiplied by $i$ gives the multiplicative identity: $1$.

In fact, $-i$ satisfies that since

$$(-i)\cdot i= -(i\cdot i)= -(-1) =1$$

That notation holds in general. For example, $2^{-1}=\frac{1}{2}$ since $\frac{1}{2}$ is the number that gives $1$ when multiplied by $2$.

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    $\begingroup$ I appreciate that this answer gives context to the calculation. +1 ! $\endgroup$
    – pjs36
    May 11 '15 at 15:04
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There are multiple ways of writing out a given complex number, or a number in general. Usually we reduce things to the "simplest" terms for display -- saying $0$ is a lot cleaner than saying $1-1$ for example.

The complex numbers are a field. This means that every non-$0$ element has a multiplicative inverse, and that inverse is unique.

While $1/i = i^{-1}$ is true (pretty much by definition), if we have a value $c$ such that $c * i = 1$ then $c = i^{-1}$.

This is because we know that inverses in the complex numbers are unique.

As it happens, $(-i) * i = -(i*i) = -(-1) = 1$. So $-i = i^{-1}$.

As fractions (or powers) are usually considered "less simple" than simple negation, when the calculator displays $i^{-1}$ it simplifies it to $-i$.

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$-i$ is the multiplicative inverse of $i$ in the field of complex numbers, i.e. $-i * i = 1$, or $i^{-1} = -i$.

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$$\frac{1}{i}=\frac{i^4}{i}=i^3=i^2\cdot i = -i$$

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I always like to point out that this fits well into a pattern you see when "rationalising the denominator", if the denominator is a root: $$\frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}}\cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{1}{2}\sqrt{2}$$ $$\frac{1}{\sqrt{17}} = \frac{1}{\sqrt{17}}\cdot \frac{\sqrt{17}}{\sqrt{17}} = \frac{1}{17}\sqrt{17}$$ $$\frac{1}{\sqrt{a}} = \frac{1}{\sqrt{a}}\cdot \frac{\sqrt{a}}{\sqrt{a}} = \frac{1}{a}\sqrt{a}$$ $$\frac{1}{i} = \frac{1}{\sqrt{-1}} = \frac{1}{\sqrt{-1}}\cdot \frac{\sqrt{-1}}{\sqrt{-1}} = \frac{1}{-1}\sqrt{-1} = - i.$$ In this vein, it is almost more suggestive to write $$\frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$$ $$\frac{1}{\sqrt{17}} = \frac{\sqrt{17}}{17}$$ $$\frac{1}{i} = \frac{i}{-1}.$$

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By the definition of the inverse $$\frac1i\cdot i=1.$$

This agrees with

$$(-i)\cdot i=1.$$

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I want to add the method that I like.

$\frac{1}{i}$ $=\frac{1}{cis(\frac{\pi}{2})}$ $= cis(- \frac{\pi}{2})$ $=-i $

Where $cis(x)= \cos(x)+i \sin(x)$

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Any complex number is fully described by its magnitude and phase (argument) via the complex exponential.

$$ X = |X|e^{i\arg{X}}$$

It is useful to write complex numbers in this form when multiplying and dividing as we can make use of exponent rules. Division in this instance simplifies to dividing the magnitudes and subtracting the phases.

Before we compute this division, lets calculate the magnitude and phase of $1$ and $i$. It is quite obvious that the magnitudes of both numbers are $1$ (i.e. $|1|=|i|=1$). And by definition the phases are:

$$\arg{1} = 0$$ $$\arg{i} = \frac{\pi}{2}$$

Our two complex exponentials are therefore:

$$1 = e^{i0}$$ $$i = e^{i\frac{\pi}{2}}$$

Now we perform the division making use of the exponent rules:

$$\frac{1}{i}=\frac{e^{i0}}{e^{i\frac{\pi}{2}}}=e^{-i\frac{\pi}{2}}$$

If you consult the unit circle (since the magnitude is 1), you will find that a phase of $-\frac{\pi}{2}$ corresponds to $−i$. Alternatively you can apply Euler's formula:

$$e^{-i\frac{\pi}{2}} = \cos\left(-\frac{\pi}{2}\right) +i\sin\left(-\frac{\pi}{2}\right) =-i$$

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$$\frac{1}{i}=\left|\frac{1}{i}\right|e^{\arg\left(\frac{1}{i}\right)i}=$$

$$1e^{\left(-\frac{1}{2}\pi\right) i}=e^{\left(-\frac{1}{2}\pi\right) i}=$$

$$1\left(\cos\left(-\frac{1}{2}\pi\right)+\sin\left(-\frac{1}{2}\pi\right)i\right)=\cos\left(-\frac{1}{2}\pi\right)+\sin\left(-\frac{1}{2}\pi\right)i=$$

$$0+(-1)i=0-1i=-i$$

So:

$$\frac{1}{i}=-i$$

Why is $\left|\frac{1}{i}\right|=1$:

$$\left|\frac{1}{i}\right|=\sqrt{\Re\left(\frac{1}{i}\right)^2+\Im\left(\frac{1}{i}\right)^2}=\sqrt{0^2+(-1)^2}=\sqrt{(-1)^2}=\sqrt{1}=1$$

Second way to show $\left|\frac{1}{i}\right|=1$:

$$\left|\frac{1}{i}\right|=\frac{|1|}{|i|}=\frac{\sqrt{1^2}}{\sqrt{1^2}}=\frac{\sqrt{1}}{\sqrt{1}}=\sqrt{\frac{1}{1}}=\sqrt{1}=1$$

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    $\begingroup$ How do you know that $|1/i|=1$??? $\endgroup$ May 11 '15 at 17:24
  • $\begingroup$ @JpMcCarthy look in the edit! $\endgroup$ May 11 '15 at 17:30
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    $\begingroup$ This is the most convoluted way of explaining this that I have ever seen, albeit correct and in some ways more mathematically interesting than the conventional ones. $\endgroup$
    – Laertes
    May 11 '15 at 17:33
  • $\begingroup$ I think it's the easiest way! $\endgroup$ May 11 '15 at 17:36
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    $\begingroup$ How did you know that the imaginary part of $1/i$ is $-1$??? $\endgroup$ May 12 '15 at 6:50

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