How to show $i^{-1} = -i$? How can I show $i^{-1} = -i$, where $i$ is the imaginary unit?
Here's what I've tried:
$i^{-1} = (-1)^{-1/2} = \dots ?$
 A: $$i^{-1} = \frac1i=\frac ii\cdot \frac 1i = \frac{i}{i\cdot i} = \frac{i}{-1} = -i$$
A: $i^{-1}=i^{1-2}=i/i^2=i/(-1)=-i$
A: Write as a fraction and expand with $i$ to get
$$
i^{-1} = \frac{1}{i} = \frac{i}{i\cdot i} = \frac{i}{-1} = -i.
$$
A: $$i^{-1} = \frac{1}{i}$$
$$\frac{1}{i}\cdot i = 1$$
$$-i \cdot i = 1$$
Therefore,
$$-i = \frac{1}{i} = i^{-1}.$$
A: 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 $$\frac{1}{i}=-i$$
A: 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 $\ldots$
A: 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 $i^{-1}=-i$.
A: You can do this:
$$
\frac{1}{i}=\frac{1}{i}\cdot\frac{i}{i}=\frac{i}{i^2}=\frac{i}{-1}=-i.
$$
A: $i^{-1}=i^{3-4}=\frac{i^3}{i^4}=i^3=-i$
A: $i^2 = -1$, and $i^4 = 1$, so $i \cdot i^3 = 1$ and $i^3 = \dfrac1i$.
