# Why is $1/i$ equal to $-i$?

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?

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

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

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

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 $2$.

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

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 $-i$.

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

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

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

This agrees with

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

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

I want to add the method that I like.

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

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

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

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

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

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

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

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

• How do you know that $|1/i|=1$??? – JP McCarthy May 11 '15 at 17:24
• @JpMcCarthy look in the edit! – Jan Eerland May 11 '15 at 17:30
• 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. – Laertes May 11 '15 at 17:33
• I think it's the easiest way! – Jan Eerland May 11 '15 at 17:36
• How did you know that the imaginary part of $1/i$ is $-1$??? – JP McCarthy May 12 '15 at 6:50