Question

simplify the expression to a + bi form:
$-2i^{19} - 10i^{116} - 10i^{17} + i^{85}$
answer
attempt 1 out of 2

Explanation:

Step1: Recall the powers of \(i\)

We know that \(i^1 = i\), \(i^2=-1\), \(i^3 = i^2\times i=-i\), \(i^4=(i^2)^2 = (- 1)^2 = 1\), and the powers of \(i\) repeat every 4. So we can find the remainder when the exponent is divided by 4 to simplify \(i^n\).

Step2: Simplify \(i^{19}\)

Divide 19 by 4: \(19\div4 = 4\) with a remainder of 3. So \(i^{19}=i^{4\times4 + 3}=(i^4)^4\times i^3\). Since \(i^4 = 1\) and \(i^3=-i\), we have \(i^{19}=1^4\times(-i)=-i\). Then \(-2i^{19}=-2\times(-i) = 2i\).

Step3: Simplify \(i^{116}\)

Divide 116 by 4: \(116\div4=29\) with a remainder of 0. So \(i^{116}=i^{4\times29}=(i^4)^{29}\). Since \(i^4 = 1\), we have \(i^{116}=1^{29}=1\). Then \(-10i^{116}=-10\times1=-10\).

Step4: Simplify \(i^{17}\)

Divide 17 by 4: \(17\div4 = 4\) with a remainder of 1. So \(i^{17}=i^{4\times4+1}=(i^4)^4\times i^1\). Since \(i^4 = 1\) and \(i^1 = i\), we have \(i^{17}=1^4\times i=i\). Then \(-10i^{17}=-10\times i=-10i\).

Step5: Simplify \(i^{85}\)

Divide 85 by 4: \(85\div4 = 21\) with a remainder of 1. So \(i^{85}=i^{4\times21 + 1}=(i^4)^{21}\times i^1\). Since \(i^4 = 1\) and \(i^1 = i\), we have \(i^{85}=1^{21}\times i=i\). Then \(i^{85}=i\).

Step6: Combine like terms

Now we substitute the simplified terms back into the original expression:

$$
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$$

Answer:

\(-10 - 7i\)