Core Connections Algebra 2, 2013
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Core Connections Algebra 2, 2013 View details
Chapter Closure

Exercise 184 Page 434

a We can simplify the given expression by combining like terms. This involves treating the imaginary unit i as though it was a variable. In this case we need to remove the parentheses first.
(3+4i)+(7-2i)
3+4i+7-2i
Simplify
3+7+4i-2i
10+2i
b We can simplify the given expression by factoring a perfect square trinomial.
(a+b)^2 = a^2+2ab+b^2To simplify the expression further, recall the definition of the imaginary unit i. i^2 = -1 Let's begin by factoring the perfect square trinomial.
(3+5i)^2
3^2+2(3)(5i)+(5i)^2
Simplify
3^2+2(3)(5i)+5^2i^2
9+2(3)(5i)+25i^2
9+30i+25i^2
9+30i+25(-1)
9+30i-25
-16+30i
c We can simplify the given expression by factoring a conjugate pair of binomials.
(a+b)(a-b) = a^2-b^2To simplify the expression further, let's first recall the definition of the imaginary unit i. i^2 = -1 Let's begin by factoring the conjugate pair of binomials.
(7+i)(7-i)
7^2-i^2
Simplify
49-i^2
49-(-1)
49+1
50
d To simplify the given expression we will use the Properties of Exponents. Let's also recall the definition of the imaginary unit i.
i^2 = -1 We can begin by raising the second pair of parentheses to the power of 2.
(3i)(2i)^2
(3i)(2^2i^2)
Simplify
3* i* 2^2* i^2
3* i* 4* i^2
3* i* 4* (-1)
(-1)* 3* 4 * i
-12i
e We can use the Properties of Exponents to simplify the given expression. To do so, let's also recall the definition of the imaginary unit i.
i^2 = -1 Begin by rewriting the exponent as a sum. Then, use the Properties of Exponents and the definition of i to simplify the expression.
i^3
i^(2+1)
Simplify
i^2* i^1
(-1)* i^1
(-1)* i
- i
f Generally, the value of i^n — where n is a whole number — can be calculated by dividing n by 4 and considering the remainder. Let R be the remainder when n is divided by 4.
i^n= i, ifR=1 - 1, ifR=2 - i, ifR=3 1, ifR=0 Now, let's consider i^(32). Note that 32 is divisible by 4. 32 ÷ 4 = 8 Since the remainder is 0, we obtain that i^(32)=1.

Extra

Powers of i

The Commutative and Associative Properties of Multiplication hold true for imaginary numbers. We can use them to find the powers of i. Let's consider the first four powers. Recall that any number raised to the power of one equals itself, so i^1=i. Moreover, we know that i^2= - 1. ccccccc i^1&=& i i^2&=& - 1 i^3&=& i^2 * i &=& - 1 * i&=& - i i^4&=& i^2 * i^2&=& ( - 1) * ( - 1)&=& 1 Let's now calculate the following four powers. To do so, we will use the results obtained above. ccccccc i^5 &=& i^4 * i &=& 1 * i &=& i i^6 &=& i^4 * i^2 &=& 1 * ( - 1) &=& - 1 i^7 &=& i^4 * i^3 &=& 1 * ( - i) &=& - i i^8 &=& i^4 * i^4 &=& 1 * 1 &=& 1 Notice that the pattern i, - 1, - i, 1, ... repeats in that order continuously after the first four results.