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Question 3

Simplify [tex]7i^{22}[/tex].

A. [tex]-7i[/tex]
B. [tex]7i[/tex]
C. [tex]-7[/tex]
D. [tex]7[/tex]

Sagot :

To simplify [tex]\( 7i^{22} \)[/tex], let's begin by understanding the properties of the imaginary unit [tex]\( i \)[/tex].

The imaginary unit [tex]\( i \)[/tex] is defined by the property [tex]\( i^2 = -1 \)[/tex]. From this, we can derive the higher powers of [tex]\( i \)[/tex]:

[tex]\[ i^1 = i \\ i^2 = -1 \\ i^3 = i \cdot i^2 = i \cdot (-1) = -i \\ i^4 = (i^2)^2 = (-1)^2 = 1 \][/tex]

Notice that [tex]\( i^4 = 1 \)[/tex], and this pattern repeats every four powers:

[tex]\[ i^5 = i^1 = i \\ i^6 = i^2 = -1 \\ i^7 = i^3 = -i \\ i^8 = i^4 = 1 \][/tex]

This cyclical property of [tex]\( i \)[/tex] tells us that any integer power of [tex]\( i \)[/tex] can be reduced by considering the remainder when the exponent is divided by 4.

Next, we simplify [tex]\( i^{22} \)[/tex].

Find the remainder when 22 is divided by 4:

[tex]\[ 22 \div 4 = 5 \text{ remainder } 2 \][/tex]

Therefore, [tex]\( i^{22} \)[/tex] can be simplified as:

[tex]\[ i^{22} = i^{4 \cdot 5 + 2} = (i^4)^5 \cdot i^2 \][/tex]

Since [tex]\( i^4 = 1 \)[/tex], we have:

[tex]\[ (i^4)^5 = 1^5 = 1 \][/tex]

Thus:

[tex]\[ i^{22} = 1 \cdot i^2 = i^2 \][/tex]

And since [tex]\( i^2 = -1 \)[/tex], we get:

[tex]\[ i^{22} = -1 \][/tex]

Now we need to multiply this result by 7:

[tex]\[ 7i^{22} = 7 \cdot (-1) = -7 \][/tex]

Therefore, the simplified form of [tex]\( 7i^{22} \)[/tex] is [tex]\( -7 \)[/tex].

So the correct answer is:

[tex]\[ -7 \][/tex]