How to show $i^{-1} = -i$?

Question

How can I show $i^{-1} = -i$, where $i$ is the imaginary unit?

Here's what I've tried: $i^{-1} = (-1)^{-1/2} = \dots ?$

Answer 1

By definition, $i^{-1}$ is the (unique) complex number $a$ such that $a\cdot i=i\cdot a=1$.

Since $(-i)\cdot i=i\cdot (-i)=1$, we have $i^{-1}=-i$.

Answer 2

You can do this: $$ \frac{1}{i}=\frac{1}{i}\cdot\frac{i}{i}=\frac{i}{i^2}=\frac{i}{-1}=-i. $$

Answer 3

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

Answer 4

$$i^{-1} = \frac1i=\frac ii\cdot \frac 1i = \frac{i}{i\cdot i} = \frac{i}{-1} = -i$$

Answer 5

$i^{-1}=i^{1-2}=i/i^2=i/(-1)=-i$

Answer 6

Write as a fraction and expand with $i$ to get $$ i^{-1} = \frac{1}{i} = \frac{i}{i\cdot i} = \frac{i}{-1} = -i. $$

Answer 7

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

Therefore,

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

Answer 8

Assume that $z=a+bi\neq 0$ is a complex number. You can always do the following in order to express $\frac{1}{z}$ in the form $\alpha+\beta i$. $$\frac{1}{z}=\frac{1}{a+bi}\frac{a-bi}{a-bi}=\frac{a}{a^2+b^2}-\frac{b}{a^2+b^2}i$$

If you apply this to $z=i$, you get that $$\frac{1}{i}=-i$$

Answer 9

Suppose we want to solve the equation

$$x^{-1} = -x$$

Multiplying both sides by the nonzero number $x$ (note that $x=0$ is not possible) gives an equivalent equation. [Two equations are equivalent means that the equations have the same solution set.]

$$x^{-1} \cdot x = -x \cdot x$$ $$1 = -x^2$$

Now multiply both sides by $-1$, which also gives an equivalent equation. $$-1 = x^2$$

Conclusion: The solutions to $x^{-1} = -x$ are the same as the solutions to $x^2 = -1.$ Therefore, the solutions to $x^{-1} = -x$ are $\ldots$

Answer 10

$i^2 = -1$, and $i^4 = 1$, so $i \cdot i^3 = 1$ and $i^3 = \dfrac1i$.