Question

How do you Simplify this

Answer 1

First, recall some rules of exponents.

$a^m * a^n = a^{m + n}$

$\dfrac{a^m}{a^n} = a^{m - n}$

$-a^n = - (a^n)$

$a^{-n} = \dfrac{1}{a^n}$

Here is the solution.

$\dfrac{-i^{-44} \cdot i^{35}}{i^8} =$

$= -\dfrac{i^{-9} }{i^8}$

$= -i^{-17}$

$= -\dfrac{1}{i^{17}}$

$= -\dfrac{1}{i}$

(For an explanation of how $i^{17}$ became $i$,
see below the final answer.)

$= -\dfrac{1}{i} \times \dfrac{i}{i}$

$= - \dfrac{i}{i^2}$

$= \dfrac{-i}{-1}$

$= i$

Final answer: $i$



Integer powers of i.

$i^0 = 1$

$i^1 = i$

$i^2 = -1$

$i^3 = -1 \times i = -i$

$i^4 = i^3 \times i = -i \times i = 1$ Notice that $i^4 = i^0$

Now the pattern is complete, and for each consecutive integer power, you get the same pattern of results. Let n be a positive integer multiple of 4.
Then,
i^n = 1
i^(n + 1) = i
i^(n + 2) = -1
i^(n + 3) = -i

After power n + 3, power n + 4 is again an multiple of 4, and you start again with 1. Notice that integer 17 is 16 + 1, so it is 1 more than a multiple of 4, so i^17 is the same as i^(n + 1) = i.