# How do you simplify 5i^32 + 7i^11 + 3i^2 - i^17  in a + bi form?

Dec 13, 2017

$5 {i}^{32} + 7 {i}^{11} + 3 {i}^{2} - {i}^{17} = 2 - 8 i$

#### Explanation:

First, you need to know that the imaginary number $i = \sqrt{- 1}$.
Let's write out some powers of $i$:
$i = \sqrt{- 1}$
${i}^{2} = i \cdot i = \sqrt{- 1} \cdot \sqrt{- 1} = - 1$
${i}^{3} = {i}^{2} \cdot i = - 1 i = - i$
${i}^{4} = {i}^{2} \cdot {i}^{2} = - 1 \cdot - 1 = 1$
${i}^{5} = {i}^{4} \cdot i = 1 i = i$
This pattern continues...
I encourage you to try it out. It's much easier with a pencil and paper than here with all the codes.
Essentially, for each power you raise $i$ to, find the number closest to that power that divides evenly by 4 and use it to break the exponent down into smaller powers of $i$.

In this problem, we have:
${i}^{32}$, which may be written as ${\left({i}^{4}\right)}^{8} = {\left(1\right)}^{8} = 1$
${i}^{11}$, which is ${\left({i}^{4}\right)}^{2} \cdot {i}^{3} = 1 \cdot {i}^{3} = - i$
${i}^{2}$, which is just $- 1$,
and
${i}^{17}$, which is ${\left({i}^{4}\right)}^{4} \cdot i = 1 \cdot i = i$

Plug these simplified values in your original expression:
$5 \left(1\right) + 7 \left(- i\right) + 3 \left(- 1\right) - i$
and treat $i$ like a variable to simplify:
$5 - 7 i - 3 + i = 2 - 8 i$

If you're confused about those middle steps, watch a video:
https://goo.gl/a3K93B

Hope that helps! Imaginary numbers are cool!