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?

Solutions Collecting From Web of "Why is $1/i$ equal to $-i$?"

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

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 $2$.

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

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

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

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

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

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

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

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