Question

expression and write the answer in complex standard form.
\\(\frac{i^{1066} + 4i^{1067}}{4i^{1068} + i^{1069}}\\)

Explanation:

Step1: Recall the cycle of \(i\)

The imaginary unit \(i\) has a cycle of 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 \(i^n\).

For \(i^{1066}\): Divide 1066 by 4. \(1066 \div 4 = 266\) with a remainder of 2. So \(i^{1066} = i^{4\times266 + 2} = (i^4)^{266} \times i^2 = 1^{266} \times (-1) = -1\).

For \(i^{1067}\): Divide 1067 by 4. \(1067 \div 4 = 266\) with a remainder of 3. So \(i^{1067} = i^{4\times266 + 3} = (i^4)^{266} \times i^3 = 1^{266} \times (-i) = -i\).

For \(i^{1068}\): Divide 1068 by 4. \(1068 \div 4 = 267\) with a remainder of 0. So \(i^{1068} = i^{4\times267} = (i^4)^{267} = 1^{267} = 1\).

For \(i^{1069}\): Divide 1069 by 4. \(1069 \div 4 = 267\) with a remainder of 1. So \(i^{1069} = i^{4\times267 + 1} = (i^4)^{267} \times i^1 = 1^{267} \times i = i\).

Step2: Substitute the simplified \(i^n\) into the expression

Substitute the values we found into the numerator and denominator:

Numerator: \(i^{1066} + 4i^{1067} = -1 + 4(-i) = -1 - 4i\)

Denominator: \(4i^{1068} + i^{1069} = 4(1) + i = 4 + i\)

So the expression becomes \(\frac{-1 - 4i}{4 + i}\).

Step3: Rationalize the denominator

To rationalize the denominator, multiply the numerator and denominator by the conjugate of the denominator, which is \(4 - i\).

\[

$$\begin{align*} \frac{-1 - 4i}{4 + i} \times \frac{4 - i}{4 - i} &= \frac{(-1 - 4i)(4 - i)}{(4 + i)(4 - i)} \\ &= \frac{-1\times4 + (-1)\times(-i) - 4i\times4 + (-4i)\times(-i)}{4^2 - i^2} \\ &= \frac{-4 + i - 16i + 4i^2}{16 - (-1)} \\ &= \frac{-4 - 15i + 4(-1)}{17} \quad (\text{since } i^2 = -1) \\ &= \frac{-4 - 15i - 4}{17} \\ &= \frac{-8 - 15i}{17} \\ &= -\frac{8}{17} - \frac{15}{17}i \end{align*}$$

\]

Answer:

\(-\dfrac{8}{17} - \dfrac{15}{17}i\)