Now, you may think you can do this:
But this doesn't make any sense! You already have two numbers that square to 1; namely –1 and +1. And i already squares to –1. So it's not reasonable that i would also square to 1. This points out an important detail: When dealing with imaginaries, you gain something (the ability to deal with negatives inside square roots), but you also lose something (some of the flexibility and convenient rules you used to have when dealing with square roots). In particular, YOU MUST ALWAYS DO THE i-PART FIRST!
Simplify sqrt(–9). Copyright © Elizabeth Stapel 2000-2011 All Rights Reserved
(Warning: The step that goes through the third "equals" sign is " ", not "". The i is outside the radical.)
In your computations, you will deal with i just as you would with x, except for the fact that x2 is just x2, but i2 is –1:
Simplify 2i + 3i.
2i + 3i = (2 + 3)i = 5i
Simplify 16i – 5i.
16i – 5i = (16 – 5)i = 11i
Multiply and simplify (3i)(4i).
(3i)(4i) = (3·4)(i·i) = (12)(i2) = (12)(–1) = –12
Multiply and simplify (i)(2i)(–3i).
(i)(2i)(–3i) = (2 · –3)(i · i · i) = (–6)(i2 · i)
=(–6)(–1 · i) = (–6)(–i) = 6i
Note this last problem. Within it, you can see that , because i2 = –1. Continuing, we get:
This pattern of powers, signs, 1's, and i's is a cycle:
In other words, to calculate any high power of i, you can convert it to a lower power by taking the closest multiple of 4 that's no bigger than the exponent and subtracting this multiple from the exponent. For example, a common trick question on tests is something along the lines of "Simplify i99", the idea being that you'll try to multiply i ninety-nine times and you'll run out of time, and the teachers will get a good giggle at your expense in the faculty lounge. Here's how the shortcut works:
i99 = i96+3 = i(4×24)+3 = i3 = –i
That is, i99 = i3, because you can just lop off the i96....