QuestionSimplify the expression to a + bi form: $-2 i^{74}+11 i^{36}+5 i^{79}-6 i^{18}$

Studdy Solution

STEP 1

What is this asking? We're dealing with powers of the imaginary unit $i$ and want to simplify a bunch of these terms into the standard complex number form $a + bi$ , where $a$ is the real part and $b$ is the imaginary part. Watch out! Remember the cyclic pattern of $i$ : $i^1 = i$ , $i^2 = -1$ , $i^3 = -i$ , and $i^4 = 1$ .
This pattern repeats every four powers, so we'll use that to our advantage!

STEP 2

1. Simplify each term
2. Combine like terms

STEP 3

Let's tackle the first term, $-2i^{74}$ .
We can rewrite the exponent as a multiple of 4 plus a remainder: $74 = 4 \cdot 18 + 2$ .
So, $i^{74} = i^{4 \cdot 18 + 2} = (i^4)^{18} \cdot i^2$ .
Since $i^4 = 1$ , we have $(i^4)^{18} = 1^{18} = 1$ .
Therefore, $i^{74} = 1 \cdot i^2 = i^2 = -1$ .
Multiplying by the coefficient, we get $-2 \cdot (-1) = \textbf{2}$ .

STEP 4

Now, let's simplify $11i^{36}$ .
We have $36 = 4 \cdot 9$ , so $i^{36} = i^{4 \cdot 9} = (i^4)^9 = 1^9 = 1$ .
Thus, $11i^{36} = 11 \cdot 1 = \textbf{11}$ .

STEP 5

Next up is $5i^{79}$ .
We have $79 = 4 \cdot 19 + 3$ , so $i^{79} = i^{4 \cdot 19 + 3} = (i^4)^{19} \cdot i^3 = 1^{19} \cdot i^3 = 1 \cdot (-i) = -i$ .
Multiplying by the coefficient gives us $5 \cdot (-i) = \textbf{-5i}$ .

STEP 6

Finally, let's simplify $-6i^{18}$ .
We have $18 = 4 \cdot 4 + 2$ , so $i^{18} = i^{4 \cdot 4 + 2} = (i^4)^4 \cdot i^2 = 1^4 \cdot i^2 = 1 \cdot (-1) = -1$ .
Multiplying by the coefficient gives us $-6 \cdot (-1) = \textbf{6}$ .

STEP 7

Putting it all together, we have $2 + 11 - 5i + 6$ .
Combining the real parts, $2 + 11 + 6 = 19$ , and the imaginary part is $-5i$ .
So, our final answer is $19 - 5i$ .

STEP 8

The simplified expression in $a + bi$ form is $19 - 5i$ .

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