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# Solved Examples - Simplify the Powers of $i$

For ACT Students
The ACT is a timed exam...$60$ questions for $60$ minutes
This implies that you have to solve each question in one minute.
Some questions will typically take less than a minute a solve.
Some questions will typically take more than a minute to solve.
The goal is to maximize your time. You use the time saved on those questions you solved in less than a minute, to solve the questions that will take more than a minute.
So, you should try to solve each question correctly and timely.
So, it is not just solving a question correctly, but solving it correctly on time.
Please ensure you attempt all ACT questions.
There is no "negative" penalty for any wrong answer.

Simplify these powers of $i$
Show all work.

(1.) $i^{8}$

$i^{8} \\[3ex] = (i^4)^2 \\[3ex] = 1^8 \\[3ex] = 1$
(2.) $i^{19}$

$i^{19} \\[3ex] = i^{16} * i^3 \\[3ex] = (i^4)^4 * -i \\[3ex] = 1^4 * -i \\[3ex] = -i$
(3.) $i^{561}$

$i^{561} \\[3ex] = i^{560} * i \\[3ex] = (i^4)^{140} * i \\[3ex] = 1^{140} * i \\[3ex] = i$
(4.) $i^{1234}$

$i^{1234} \\[3ex] = i^{1232} * i^2 \\[3ex] = (i^4)^{308} * -1 \\[3ex] = 1^{308} * -1 \\[3ex] = -1$
(5.) Given that $i^2 = -1$ and that $k$ is a positive integer, what is the value of $i^{4k}$?

First method: Arithmetic Method/By Testing (Easy way) - Recommended for ACT because it is a timed test.
$k$ is a positive integer.
So, test for $k = 1$

$i^{4k} \\[3ex] Test\:\: k = 1 \\[3ex] i^{4 * 1} \\[3ex] i^{4} \\[3ex] i^4 \\[3ex] = 1 \\[3ex]$ Second method: Algebra Method/By Solving

$i^{4k} \\[3ex] = i^{2(2k)} \\[3ex] = (-1)^{2k} \\[3ex] k \:\:is\:\: positive \\[3ex] means\:\: that\:\: 2k \:\:is\:\: even\:\: (multiples \:\:of\:\: 2) \\[3ex] = (-1)^{even} \\[3ex] = 1$
(6.) $(-i)^{82}$

$(-i)^{82} \\[3ex] = (-1 * i)^{82} \\[3ex] = (-1)^{82} * i^{82} \\[3ex] = 1 * i^{80} * i^{2} \\[3ex] = 1 * (i^4)^{20} * -1 \\[3ex] = 1 * 1^{20} * -1 \\[3ex] = 1 * 1 * -1 \\[3ex] = -1$
(7.) ACT Given that $i^2 = -1$ and that $k$ is a positive integer, what is the value of $i^{4k - 2}$?
A. $-i$
B. $-1$
C. $0$
D. $1$
E. $i$

First method: Arithmetic Method/By Testing (Easy way) - Recommended for ACT because it is a timed test.
$k$ is a positive integer.
So, test for $k = 1$

$i^{4k - 2} \\[3ex] Test\:\: k = 1 \\[3ex] i^{4 * 1 - 2} \\[3ex] i^{4 - 2} \\[3ex] i^2 \\[3ex] = -1 \\[3ex]$ Second method: Algebra Method/By Solving

$i^{4k - 2} \\[3ex] = i^{2(2k - 1)} \\[3ex] = (-1)^{2k - 1} \\[3ex] k \:\:is\:\: positive \\[3ex] means\:\: that\:\: 2k \:\:is\:\: even\:\: (multiples \:\:of\:\: 2) \\[3ex] means\:\: that\:\: 2k - 1 \:\:is\:\: odd\:\: (any\:\: even\:\: number\:\: - 1 \:\:is\:\: odd) \\[3ex] = (-1)^{odd} \\[3ex] = -1$
(8.) $i^{22} - 1$

$i^{22} - 1 \\[3ex] = i^{20} * i^2 - 1 \\[3ex] = ((i^4)^5 * -1) - 1 \\[3ex] = (1^5 * -1) - 1 \\[3ex] = (1 * -1) - 1 \\[3ex] = -1 - 1 \\[3ex] = -2$
(9.) $i^{1372576434830548}$

1372576434830548 is even
It is divisible by $2$
It is also divisible by $4$

$i^{1372576434830548} \\[3ex] = i^{4(some\:\: factor)} \\[3ex] = 1^{(some\:\: factor)} \\[3ex] = 1$
(10.) $(2i)^3$

$(2i)^3 \\[3ex] = 2^3 * i^3 \\[3ex] = 8 * -i \\[3ex] = -8i$
(11.) $i^{1372576434830542}$

We can do this two ways.

1372576434830542 is even
It is divisible by $2$
However, it is not divisible by $4$
So, when it is divided by $2$, the quotient will be an odd number

$i^{1372576434830542} \\[3ex] = i^{2(some\:\:odd\:\: factor)} \\[3ex] = (-1)^{(some\:\:odd\:\: factor)} \\[3ex] = -1 \\[3ex] OR \\[3ex] 42 = 40 + 2 \\[3ex] Apply\:\: the\:\: Law\:\: of\:\: Exponents...Law\:\: 1...Exp \\[3ex] i^{1372576434830542} \\[3ex] = i^{1372576434830540} * i^2 \\[3ex] = i^{4(some\:\: factor)} * -1 \\[3ex] = 1^{(some\:\: factor)} * -1 \\[3ex] = 1 * -1 \\[3ex] = -1$
(12.) $7i^3 - 3i^7$

$7i^3 - 3i^7 \\[3ex] i^7 = i^4 * i^3 = 1 * -i = -i \\[3ex] = 7(-i) - 3(-i) \\[3ex] = -7i + 3i \\[3ex] = -4i$
(13.) $i^{1372576434830545}$

1372526474830545 is odd
It is not divisible by $4$
It is not also divisible by $2$
So, how do we do it?
Well, let us look at the last two digits.
Last two digits is $45$
If it was $44$, it would have been divisible by $4$
$45 = 44 + 1$
Let us apply the Law of Exponents
So, we have:

$i^{1372576434830545} \\[3ex] = i^{1372576434830544} * i^1 \\[3ex] = i^{4(some\:\: factor)} * i \\[3ex] = 1^{(some\:\: factor)} * i \\[3ex] = 1 * i \\[3ex] = i$
(14.) $i^{-9}$
Please review the Division of Complex Numbers first.

$i^{-9} \\[3ex] = \dfrac{1}{i^9} \\[5ex] i^9 = i^8 * i = (i^4)^2 * i = 1^2 * i = 1 * i = i \\[5ex] = \dfrac{1}{i} \\[5ex]$ Conjugate of $i$ is $-i$
However, use $i$ in this case. It does not matter.
Even if you use $-i$, the negatives will cancel out.

$= \dfrac{1}{i} * \dfrac{i}{i} \\[5ex] = \dfrac{1 * i}{i * i} \\[5ex] = \dfrac{i}{i^2} \\[5ex] = \dfrac{i}{-1} \\[5ex] = -i$
(15.) $i^{10} + i^4 + i^{18} + 1$

$i^{10} + i^4 + i^{18} + 1 \\[3ex] i^{10} = i^8 * i^2 = (i^4)^2 * -1 = 1^2 * -1 = 1 * -1 = -1 \\[3ex] i^{18} = i^16 * i^2 = (i^4)^4 * -1 = 1^4 * -1 = 1 * -1 = -1 \\[3ex] = -1 + 1 + (-1) + 1 \\[3ex] = -1 + 1 - 1 + 1 \\[3ex] = 0$
(16.) $i^{1372576434830547}$

1372526474830547 is odd
It is not divisible by $4$
It is not also divisible by $2$
So, how do we do it?
Well, let us look at the last two digits.
Last two digits is $45$
If it was $44$, it would have been divisible by $4$
$47 = 44 + 3$
Let us apply the Law of Exponents
So, we have:

$i^{1372576434830547} \\[3ex] = i^{1372576434830544} * i^3 \\[3ex] = i^{4(some\:\: factor)} * -i \\[3ex] = 1^{(some\:\: factor)} * -i \\[3ex] = 1 * -i \\[3ex] = -i$