Question

1. which of these is equivalent to $i^{75}$?

slide red circle in front of correct answer.
a $i$
b $-i$
c $1$
d $-1$

Explanation:

Step1: Recall the pattern of powers of \(i\)

The powers of \(i\) follow a cycle: \(i^1 = i\), \(i^2=- 1\), \(i^3 = i^2\times i=-i\), \(i^4=(i^2)^2 = 1\). The cycle repeats every \(4\) powers.

Step2: Divide the exponent by \(4\)

Divide \(75\) by \(4\): \(75\div4 = 18\) with a remainder. Using the formula \(a = bq + r\) (where \(a = 75\), \(b = 4\), \(q=18\)), we get \(75=4\times18 + 3\). So \(i^{75}=i^{4\times18 + 3}\).

Step3: Use the property \(i^{4k}=1\)

Since \(i^{4k}=1\) (for integer \(k\)), \(i^{4\times18 + 3}=(i^{4})^{18}\times i^{3}\). Substitute \(i^{4} = 1\), then \((i^{4})^{18}\times i^{3}=1^{18}\times i^{3}\).

Step4: Simplify the expression

Since \(1^{18}=1\) and \(i^{3}=-i\), \(i^{75}=-i\).

Answer:

B. \(-i\)