Finding a negative power of $i$

Question

How to find the value of $i$ when it has negative power? When solving for $i$ with positive power, I use something like $$i^{101} = (i^{2})^{50}\times i = (-1)^{50} \times i = 1 \times i = i.$$ But how to solve for negative power of $i$ such as $i^{-10}$?

Can anyone explain what to do in this case?

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Solution attempt:

I will solve for $i^{-3}$ as $$i^{-3} = \frac{1}{i^3} = \frac{1}{i^2 \times i} = \frac{1}{-1 \times i} = \frac {1}{-i},$$ so the answer we get is $\dfrac{1}{-i}$. But my book is saying that I should get $i$.

Answer 1

By your analogy, $$i^{-10}=\frac{1}{i^{10}}=\frac{1}{i^8*i^2}=\frac{1}{(i^{2})^4* i^2}.$$ Since $i^2=-1$, it follows that $$i^{-10}=\frac{1}{(-1)^4*-1}=\frac{1}{-1}=-1.$$

Answer 2

The key point is that $$ i=e^{i\pi/2} $$ from which you obtain immediately $$ -\frac1i=-e^{-i\pi/2}=-(-i)=i. $$

Answer 3

Your answer is correct, just not the simplest form. Note that $i^4=1$, so $$ i^{-3}=i^{-3}i^4=i $$ In your answer $$ \frac1{-i}=\frac{i^4}{-i}=-i^3=-i^2i=i $$

Answer 4

You were on the right track with $$i^{-3} = \frac{1}{-i}$$

but to express $i^{-3}$ in the form $a + bi$ (where $a$ and $b$ are real numbers), multiply $ \frac{1}{-i}$ by the "conjugate" of the denominator $-i$ (that is, $+i$ or simply $i$) like this:

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

The above expression evaluates to

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

Since $i^{2} = -1$, $-i^{2} = -(-1) = 1$ and thus

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

Hence, $i^{-3} = i $.