If there is one prayer that you should pray/sing every day and every hour, it is the LORD's prayer (Our FATHER in Heaven prayer)
It is the most powerful prayer. A pure heart, a clean mind, and a clear conscience is necessary for it. - Samuel Dominic Chukwuemeka

For in GOD we live, and move, and have our being. - Acts 17:28

The Joy of a Teacher is the Success of his Students. - Samuel Dominic Chukwuemeka

Solved Examples: Simplify the Powers of i

Samuel Dominic Chukwuemeka (SamDom For Peace) Calculators: Calculators for Complex Numbers
Prerequisites:
(1.) Laws of Exponents
(2.) Polynomials

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 a 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 \\[3ex] B.\:\: -1 \\[3ex] C.\:\: 0 \\[3ex] D.\:\: 1 \\[3ex] E.\:\: i \\[3ex] $

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.) ACT Given a positive integer $n$ such that $i^n = 1$, which of the following statements about $n$ must be true?
(Note: $i^2 = -1$)

F. When $n$ is divided by $4$, the remainder is $0$
G. When $n$ is divided by $4$, the remainder is $1$
H. When $n$ is divided by $4$, the remainder is $2$
J. When $n$ is divided by $4$, the remainder is $3$
K. Cannot be determined from the given information


We know that if:

$ i^n = 1 \\[3ex] Then:\:\:n = 4 \:\:OR\:\:multiple\:\:of\:\:4 \\[3ex] $ When $4$ or any multiple of $4$ is divided by $4$, the remainder is $0$
(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 $
(17.) ACT If $i = \sqrt{-1}$, then $\dfrac{i + i^2 + i^3}{i^3 + i^4 + i^5} = ?$

$ A.\;\; -3 \\[3ex] B.\;\; -1 \\[3ex] C.\;\; \dfrac{1}{2} \\[5ex] D.\;\; 1 \\[3ex] E.\;\; 3 \\[3ex] $

$ i = \sqrt{-1} \\[3ex] i^2 = (\sqrt{-1})^2 = -1 \\[3ex] i^3 = i^2 \cdot i = -1 \cdot i = -i \\[3ex] i^4 = i^2 \cdot i^2 = -1 \cdot -1 = 1 \\[3ex] i^5 = i^2 \cdot i^3 = -1 \cdot -i = i \\[3ex] \implies \\[3ex] \dfrac{i + i^2 + i^3}{i^3 + i^4 + i^5} \\[5ex] = \dfrac{i + -1 + -i}{-i + 1 + i} \\[5ex] = \dfrac{i - 1 - i}{1} \\[5ex] = \dfrac{-1}{1} \\[5ex] = -1 $
(18.)


$ 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 $
(19.)

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