Question

drag the tiles to the correct boxes to complete the pairs. match each complex number with its equivalent expression. i -1 -i 1 i¹⁵⁷ i¹⁰² i³¹⁵ i⁷⁶

Explanation:

To solve for the powers of \(i\), we use the property that \(i\) has a cyclicity of 4, meaning \(i^n\) repeats every 4 exponents. The cycle is \(i^1 = i\), \(i^2 = -1\), \(i^3 = -i\), \(i^4 = 1\), and then it repeats. So we can find the remainder when the exponent is divided by 4 to determine the equivalent value.

Step 1: Simplify \(i^{157}\)

First, divide 157 by 4: \(157 \div 4 = 39\) with a remainder of 1 (since \(4 \times 39 = 156\), and \(157 - 156 = 1\)). So \(i^{157}=i^{4\times39 + 1}=(i^4)^{39}\times i^1\). Since \(i^4 = 1\), this simplifies to \(1^{39}\times i = i\).

Step 2: Simplify \(i^{102}\)

Divide 102 by 4: \(102 \div 4 = 25\) with a remainder of 2 (since \(4 \times 25 = 100\), and \(102 - 100 = 2\)). So \(i^{102}=i^{4\times25 + 2}=(i^4)^{25}\times i^2\). Since \(i^4 = 1\) and \(i^2 = -1\), this simplifies to \(1^{25}\times (-1)= -1\).

Step 3: Simplify \(i^{315}\)

Divide 315 by 4: \(315 \div 4 = 78\) with a remainder of 3 (since \(4 \times 78 = 312\), and \(315 - 312 = 3\)). So \(i^{315}=i^{4\times78 + 3}=(i^4)^{78}\times i^3\). Since \(i^4 = 1\) and \(i^3 = -i\), this simplifies to \(1^{78}\times (-i)= -i\).

Step 4: Simplify \(i^{76}\)

Divide 76 by 4: \(76 \div 4 = 19\) with a remainder of 0 (since \(4 \times 19 = 76\)). When the remainder is 0, it means the exponent is a multiple of 4, so \(i^{76}=i^{4\times19}=(i^4)^{19}\). Since \(i^4 = 1\), this simplifies to \(1^{19}=1\).

Answer:

• \(i^{157}\) matches with \(i\)
• \(i^{102}\) matches with \(-1\)
• \(i^{315}\) matches with \(-i\)
• \(i^{76}\) matches with \(1\)