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 \(i\)

The powers of \(i\) repeat every 4: \(i^1 = i\), \(i^2 = -1\), \(i^3 = -i\), \(i^4 = 1\), and then the cycle repeats. So we can find the remainder when the exponent is divided by 4 to simplify each term.

Step2: Simplify \(-2i^{19}\)

Divide 19 by 4: \(19 \div 4 = 4\) with a remainder of 3. So \(i^{19} = i^{4\times4 + 3} = (i^4)^4 \times i^3 = 1^4 \times (-i) = -i\). Then \(-2i^{19} = -2\times(-i) = 2i\).

Step3: Simplify \(-10i^{116}\)

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

Step4: Simplify \(-10i^{17}\)

Divide 17 by 4: \(17 \div 4 = 4\) with a remainder of 1. So \(i^{17} = i^{4\times4 + 1} = (i^4)^4 \times i^1 = 1^4 \times i = i\). Then \(-10i^{17} = -10\times i = -10i\).

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

Divide 85 by 4: \(85 \div 4 = 21\) with a remainder of 1. So \(i^{85} = i^{4\times21 + 1} = (i^4)^{21} \times i^1 = 1^{21} \times i = i\).

Step6: Combine all terms

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

$$\begin{align*} &-2i^{19} - 10i^{116} - 10i^{17} + i^{85}\\ =& 2i - 10 - 10i + i\\ =& -10 + (2i - 10i + i)\\ =& -10 - 7i \end{align*}$$

\]

Answer:

\(-10 - 7i\)