Question

find the power of i.
(i^{-17})
(i^{-17} = square)
(simplify your answer. type your answer in the form (a + bi).)

Explanation:

Step1: Recall the property of negative exponents

For any non - zero complex number \(z\) and integer \(n\), \(z^{-n}=\frac{1}{z^{n}}\). So, \(i^{-17}=\frac{1}{i^{17}}\).

Step2: Recall the cyclicity of powers of \(i\)

The powers of \(i\) have a cyclic pattern: \(i^{1}=i\), \(i^{2}=- 1\), \(i^{3}=i^{2}\cdot i=-i\), \(i^{4}=(i^{2})^{2}=(-1)^{2} = 1\), and then the cycle repeats every 4 powers. That is, \(i^{n + 4k}=i^{n}\) for any integer \(k\) and \(n = 1,2,3,4\).

To find the equivalent of \(i^{17}\) in the cycle, we divide the exponent by 4 and find the remainder. We use the division algorithm: \(17=4\times4 + 1\). So, \(i^{17}=i^{4\times4+1}=(i^{4})^{4}\cdot i^{1}\). Since \(i^{4} = 1\), then \((i^{4})^{4}=1^{4}=1\), and \(i^{17}=1\times i=i\).

Step3: Simplify \(\frac{1}{i}\)

We know that to simplify \(\frac{1}{i}\), we multiply the numerator and denominator by the conjugate of \(i\), which is \(-i\) (because the conjugate of a complex number \(a + bi\) is \(a - bi\), and for \(i=0 + 1i\), the conjugate is \(0-1i=-i\)).

So, \(\frac{1}{i}=\frac{1\times(-i)}{i\times(-i)}=\frac{-i}{-i^{2}}\). Since \(i^{2}=-1\), then \(\frac{-i}{-(-1)}=\frac{-i}{1}=-i\).

Answer:

\(0 - i\) (or simply \(-i\))