40
$\begingroup$

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?

$\endgroup$
  • 7
    $\begingroup$ Hint $i^2 = -1$ $\endgroup$ – Mann May 11 '15 at 12:14
  • 7
    $\begingroup$ Multiply by $i/i$. $\endgroup$ – David Mitra May 11 '15 at 12:14
  • 1
    $\begingroup$ Hint $$z=\frac{1}{i}\iff zi=1\implies \dots$$ $\endgroup$ – John Joy May 11 '15 at 12:56
  • 31
    $\begingroup$ 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
  • 4
    $\begingroup$ 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
39
$\begingroup$

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

$\endgroup$
39
$\begingroup$

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

$\endgroup$
18
$\begingroup$

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 $2$.

$\endgroup$
  • 1
    $\begingroup$ I appreciate that this answer gives context to the calculation. +1 ! $\endgroup$ – pjs36 May 11 '15 at 15:04
8
$\begingroup$

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 $-i$.

$\endgroup$
5
$\begingroup$

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

$\endgroup$
4
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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

This agrees with

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

$\endgroup$
0
$\begingroup$

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

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

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

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

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

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

$\endgroup$
  • 1
    $\begingroup$ How do you know that $|1/i|=1$??? $\endgroup$ – JP McCarthy May 11 '15 at 17:24
  • $\begingroup$ @JpMcCarthy look in the edit! $\endgroup$ – Jan May 11 '15 at 17:30
  • 5
    $\begingroup$ 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! $\endgroup$ – Jan May 11 '15 at 17:36
  • 3
    $\begingroup$ How did you know that the imaginary part of $1/i$ is $-1$??? $\endgroup$ – JP McCarthy May 12 '15 at 6:50
0
$\begingroup$

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 $$\frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$$ $$\frac{1}{\sqrt{17}} = \frac{\sqrt{17}}{17}$$ $$\frac{1}{i} = \frac{i}{-1}.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.