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Complex Numbers and Quadratic Equations - Power of i

Grade 11CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The imaginary unit ii is defined as the square root of −1-1, such that i=−1i = \sqrt{-1} and i2=−1i^2 = -1.

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The powers of ii are periodic and repeat their values every four steps: i1=ii^1 = i, i2=−1i^2 = -1, i3=−ii^3 = -i, and i4=1i^4 = 1.

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To find the value of ini^n for any positive integer nn, divide nn by 4 and find the remainder rr. Then in=iri^n = i^r.

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The sum of any four consecutive powers of ii is always zero: in+in+1+in+2+in+3=0i^n + i^{n+1} + i^{n+2} + i^{n+3} = 0.

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Negative powers of ii are calculated using the reciprocal property: i−n=1ini^{-n} = \frac{1}{i^n}.

📐Formulae

i1=ii^1 = i

i2=−1i^2 = -1

i3=−ii^3 = -i

i4=1i^4 = 1

i4n=1i^{4n} = 1

i4n+1=ii^{4n+1} = i

i4n+2=−1i^{4n+2} = -1

i4n+3=−ii^{4n+3} = -i

in+in+1+in+2+in+3=0i^n + i^{n+1} + i^{n+2} + i^{n+3} = 0

💡Examples

Problem 1:

Evaluate i243i^{243}.

Solution:

i243=i4×60+3=(i4)60⋅i3=(1)60⋅(−i)=−ii^{243} = i^{4 \times 60 + 3} = (i^4)^{60} \cdot i^3 = (1)^{60} \cdot (-i) = -i

Explanation:

Divide 243 by 4. The quotient is 60 and the remainder is 3. Since i4=1i^4 = 1, any power of i4i^4 is 1, leaving us with i3i^3, which equals −i-i.

Problem 2:

Find the value of i−35i^{-35}.

Solution:

i−35=1i35=1i4×8+3=1i3=1−ii^{-35} = \frac{1}{i^{35}} = \frac{1}{i^{4 \times 8 + 3}} = \frac{1}{i^3} = \frac{1}{-i} Multiply numerator and denominator by ii: 1⋅i−i⋅i=i−i2=i−(−1)=i\frac{1 \cdot i}{-i \cdot i} = \frac{i}{-i^2} = \frac{i}{-(-1)} = i

Explanation:

First, convert the negative exponent to a positive one in the denominator. Simplify i35i^{35} to i3i^3. To remove ii from the denominator, multiply by ii and use the fact that i2=−1i^2 = -1.

Problem 3:

Simplify the expression: i101+i102+i103+i104i^{101} + i^{102} + i^{103} + i^{104}.

Solution:

i101+i102+i103+i104=i101(1+i+i2+i3)=i101(1+i−1−i)=i101(0)=0i^{101} + i^{102} + i^{103} + i^{104} = i^{101}(1 + i + i^2 + i^3) = i^{101}(1 + i - 1 - i) = i^{101}(0) = 0

Explanation:

The sum of four consecutive powers of ii is zero. Factoring out the lowest power shows that the terms inside the parentheses cancel each other out.