Question

Match the LIST-I with LIST-II

LIST-I (Expressions)		LIST-II (Values)
A.	\( i^{49} \)	I.	1
B.	\( i^{38} \)	II.	\(-i\)
C.	\( i^{103} \)	III.	\(i\)
D.	\( i^{92} \)	IV.	\(-1\)

Choose the correct answer from the options given below:

• A. A - I, B - II, C - III, D - IV
• B. A - I, B - III, C - II, D - IV
• C. A - III, B - IV, C - II, D - I
• D. A - III, B - IV, C - I, D - II

Hint:

To quickly find the remainder when dividing a number by 4, you only need to look at its last two digits. For example, for \(i^{103}\), you just need the remainder of 03 divided by 4, which is 3.

Answer 1

The Correct Option is C

Step 1: Recall the cyclic property of powers of \(i\). \(i^1 = i\), \(i^2 = -1\), \(i^3 = -i\), \(i^4 = 1\). The pattern repeats every 4 powers. To find \(i^n\), we can evaluate \(i^k\) where \(k\) is the remainder of \(n\) divided by 4.
Step 2: Evaluate each expression. - A. \(i^{49}\): The remainder of 49 divided by 4 is 1. So, \(i^{49} = i^1 = i\). This matches III. - B. \(i^{38}\): The remainder of 38 divided by 4 is 2. So, \(i^{38} = i^2 = -1\). This matches IV. - C. \(i^{103}\): The remainder of 103 divided by 4 is 3. So, \(i^{103} = i^3 = -i\). This matches II. - D. \(i^{92}\): 92 is perfectly divisible by 4 (remainder is 0). We can treat this as \(i^4\). So, \(i^{92} = (i^4)^{23} = 1^{23} = 1\). This matches I.
Step 3: Formulate the correct sequence. The matches are: A\(\rightarrow\)III, B\(\rightarrow\)IV, C\(\rightarrow\)II, D\(\rightarrow\)I. This corresponds to option (3).