Math 2 Quiz: Imaginary Unit I
11 questions · exam conditions
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Imaginary Unit IQuestion 1 of 11

Which of the following equals i−15i^{-15}?

ii
−i-i
11
−1-1
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Math 2 Quiz

Math 2 Quiz: Imaginary Unit I

Practice Imaginary Unit I in Math 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Imaginary Unit I, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 2.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

Which of the following equals i−15i^{-15}?

  1. ii (correct answer)
  2. −i-i
  3. 11
  4. −1-1
Explanation: Since i−15=1i15i^{-15} = \frac{1}{i^{15}}, we first find i15i^{15}. Since 15=4(3)+315 = 4(3) + 3, we have i15=i3=−ii^{15} = i^3 = -i. Therefore i−15=1−i=1−i⋅ii=i−i2=i−(−1)=i1=ii^{-15} = \frac{1}{-i} = \frac{1}{-i} \cdot \frac{i}{i} = \frac{i}{-i^2} = \frac{i}{-(-1)} = \frac{i}{1} = i. Choice B results from forgetting the negative sign when computing 1−i\frac{1}{-i}, choice C comes from incorrectly thinking i15=i3=1i^{15} = i^3 = 1, and choice D comes from computing i15i^{15} incorrectly as 11 then taking the reciprocal incorrectly.

Question 2

If w=i17⋅i23⋅i31w = i^{17} \cdot i^{23} \cdot i^{31} and wk=1w^k = 1 for some positive integer kk, what is the smallest possible value of kk?

  1. 22
  2. 33
  3. 44 (correct answer)
  4. 88
Explanation: First, simplify w=i17⋅i23⋅i31=i17+23+31=i71w = i^{17} \cdot i^{23} \cdot i^{31} = i^{17+23+31} = i^{71}. Since 71=4×17+371 = 4 \times 17 + 3, we have i71=i3=−ii^{71} = i^3 = -i. So w=−iw = -i. We need the smallest positive integer kk such that wk=(−i)k=1w^k = (-i)^k = 1. Since (−i)1=−i(-i)^1 = -i, (−i)2=(−1)2⋅i2=1⋅(−1)=−1(-i)^2 = (-1)^2 \cdot i^2 = 1 \cdot (-1) = -1, (−i)3=(−i)2⋅(−i)=(−1)(−i)=i(-i)^3 = (-i)^2 \cdot (-i) = (-1)(-i) = i, and (−i)4=((−i)2)2=(−1)2=1(-i)^4 = ((-i)^2)^2 = (-1)^2 = 1, the smallest value is k=4k = 4. Choice A assumes (−i)2=1(-i)^2 = 1 by ignoring the negative sign. Choice B assumes (−i)3=1(-i)^3 = 1 by miscalculating. Choice D assumes the period is 8 instead of 4.

Question 3

If w=i2023+i2024+i2025+i2026w = i^{2023} + i^{2024} + i^{2025} + i^{2026}, what is the value of ww?

  1. 00 (correct answer)
  2. 11
  3. ii
  4. −1-1
Explanation: The powers of i repeat every 4: i1=ii^1 = i, i2=−1i^2 = -1, i3=−ii^3 = -i, i4=1i^4 = 1. To find the remainder when dividing the exponent by 4: 2023=4(505)+32023 = 4(505) + 3, so i2023=i3=−ii^{2023} = i^3 = -i. 2024=4(506)+02024 = 4(506) + 0, so i2024=i0=1i^{2024} = i^0 = 1. 2025=4(506)+12025 = 4(506) + 1, so i2025=i1=ii^{2025} = i^1 = i. 2026=4(506)+22026 = 4(506) + 2, so i2026=i2=−1i^{2026} = i^2 = -1. Therefore, w=−i+1+i+(−1)=0w = -i + 1 + i + (-1) = 0.

Question 4

If in=−ii^n = -i where nn is a positive integer, which of the following could be the remainder when n2n^2 is divided by 8?

  1. 11 or 55 (correct answer)
  2. 11 or 77
  3. 33 or 77
  4. 33 or 55
Explanation: Since in=−ii^n = -i and the powers of i cycle every 4 with i3=−ii^3 = -i, we need n≡3(mod4)n \equiv 3 \pmod{4}. So n=4k+3n = 4k + 3 for some non-negative integer kk. Then n2=(4k+3)2=16k2+24k+9=8(2k2+3k+1)+1n^2 = (4k + 3)^2 = 16k^2 + 24k + 9 = 8(2k^2 + 3k + 1) + 1. For kk even, say k=2mk = 2m: n2=8(8m2+6m+1)+1≡1(mod8)n^2 = 8(8m^2 + 6m + 1) + 1 \equiv 1 \pmod{8}. For kk odd, say k=2m+1k = 2m + 1: n2=8(8m2+10m+3)+5≡5(mod8)n^2 = 8(8m^2 + 10m + 3) + 5 \equiv 5 \pmod{8}. Therefore the remainder is either 1 or 5. Choice B incorrectly assumes n≡1n \equiv 1 or n≡3(mod4)n \equiv 3 \pmod{4}. Choice C assumes n≡3(mod4)n \equiv 3 \pmod{4} but miscalculates n2(mod8)n^2 \pmod{8}. Choice D assumes different values of n(mod4)n \pmod{4}.

Question 5

If ∑k=0nik=0\sum_{k=0}^{n} i^k = 0, what is the smallest positive integer nn for which this is true?

  1. n=2n = 2
  2. n=3n = 3 (correct answer)
  3. n=4n = 4
  4. n=7n = 7
Explanation: We compute: ∑k=0nik=1+i+i2+i3+⋯+in\sum_{k=0}^{n} i^k = 1 + i + i^2 + i^3 + \cdots + i^n. For n=0n=0: sum is 11. For n=1n=1: sum is 1+i1+i. For n=2n=2: sum is 1+i+(−1)=i1+i+(-1) = i. For n=3n=3: sum is 1+i+(−1)+(−i)=01+i+(-1)+(-i) = 0. Therefore the smallest positive integer nn is 33. Choice A gives sum =i≠0= i \neq 0, choice C would be the next value where the sum is 00 (since the cycle repeats every 4), and choice D gives sum =0= 0 but is not the smallest such nn.

Question 6

What is the value of i2022+i2023i2020+i2021\frac{i^{2022} + i^{2023}}{i^{2020} + i^{2021}}?

  1. 11
  2. −1-1 (correct answer)
  3. ii
  4. −i-i
Explanation: Since 2020=4(505)2020 = 4(505), we have i2020=1i^{2020} = 1, i2021=ii^{2021} = i, i2022=−1i^{2022} = -1, and i2023=−ii^{2023} = -i. The expression becomes −1+(−i)1+i=−1−i1+i\frac{-1 + (-i)}{1 + i} = \frac{-1-i}{1+i}. Multiplying numerator and denominator by the conjugate: (−1−i)(1−i)(1+i)(1−i)=−1+i−i+i21−i2=−1−11−(−1)=−22=−1\frac{(-1-i)(1-i)}{(1+i)(1-i)} = \frac{-1+i-i+i^2}{1-i^2} = \frac{-1-1}{1-(-1)} = \frac{-2}{2} = -1. Choice A comes from an error in signs, choice C comes from not rationalizing the denominator correctly, and choice D comes from a computational error in the conjugate multiplication.

Question 7

What is the value of (1+i)8(1+i)^8?

  1. 1616 (correct answer)
  2. 16i16i
  3. −16-16
  4. 256256
Explanation: First find (1+i)2=1+2i+i2=1+2i−1=2i(1+i)^2 = 1 + 2i + i^2 = 1 + 2i - 1 = 2i. Then (1+i)4=((1+i)2)2=(2i)2=4i2=−4(1+i)^4 = ((1+i)^2)^2 = (2i)^2 = 4i^2 = -4. Finally, (1+i)8=((1+i)4)2=(−4)2=16(1+i)^8 = ((1+i)^4)^2 = (-4)^2 = 16. Choice B incorrectly computes (1+i)4=4i(1+i)^4 = 4i, choice C gets (1+i)4=4(1+i)^4 = 4 instead of −4-4, and choice D computes (1+i)8(1+i)^8 as if it were 282^8.

Question 8

If z=i100+i101+i102+i103z = i^{100} + i^{101} + i^{102} + i^{103}, what is z2z^2?

  1. −1-1
  2. 11
  3. 00 (correct answer)
  4. 44
Explanation: When you encounter powers of the imaginary unit ii, remember that ii follows a cyclical pattern every 4 powers: i1=ii^1 = i, i2=−1i^2 = -1, i3=−ii^3 = -i, and i4=1i^4 = 1. This cycle then repeats. To find each term in zz, divide the exponent by 4 and use the remainder. For i100i^{100}: 100÷4=25100 ÷ 4 = 25 remainder 00, so i100=i0=1i^{100} = i^0 = 1. For i101i^{101}: 101÷4=25101 ÷ 4 = 25 remainder 11, so i101=i1=ii^{101} = i^1 = i. For i102i^{102}: 102÷4=25102 ÷ 4 = 25 remainder 22, so i102=i2=−1i^{102} = i^2 = -1. For i103i^{103}: 103÷4=25103 ÷ 4 = 25 remainder 33, so i103=i3=−ii^{103} = i^3 = -i. Therefore: z=1+i+(−1)+(−i)=1+i−1−i=0z = 1 + i + (-1) + (-i) = 1 + i - 1 - i = 0 Since z=0z = 0, we have z2=02=0z^2 = 0^2 = 0, making (C) correct. (A) −1-1 might tempt you if you incorrectly calculated zz or confused it with i2i^2. (B) 11 could result from miscounting the cycle or thinking z2z^2 always equals 11 for complex numbers. (D) 44 might occur if you added the four terms without considering the signs or thought each term equaled 11. Study tip: For any four consecutive powers of ii, their sum is always zero because the cycle i,−1,−i,1i, -1, -i, 1 adds to zero. Recognizing this pattern saves time on similar problems.

Question 9

If pp and qq are positive integers such that ip⋅iq=−1i^p \cdot i^q = -1, what is the smallest possible value of p+qp + q?

  1. 44
  2. 33
  3. 22 (correct answer)
  4. 55
Explanation: When you encounter problems involving powers of the imaginary unit ii, remember that ii follows a cyclical pattern: i1=ii^1 = i, i2=−1i^2 = -1, i3=−ii^3 = -i, and i4=1i^4 = 1. This pattern repeats every 4 powers. Using the property of exponents, ip⋅iq=ip+qi^p \cdot i^q = i^{p+q}. So you need ip+q=−1i^{p+q} = -1. From the cyclical pattern, in=−1i^n = -1 when nn leaves a remainder of 2 when divided by 4. In other words, p+q≡2(mod4)p + q \equiv 2 \pmod{4}. The smallest positive value where p+q≡2(mod4)p + q \equiv 2 \pmod{4} is p+q=2p + q = 2. This works when p=1p = 1 and q=1q = 1: i1⋅i1=i2=−1i^1 \cdot i^1 = i^2 = -1. ✓ Now let's check why the other answers don't work: A) If p+q=4p + q = 4, then i4=1i^4 = 1, not −1-1. B) If p+q=3p + q = 3, then i3=−ii^3 = -i, not −1-1. D) If p+q=5p + q = 5, then i5=i1=ii^5 = i^1 = i, not −1-1. The correct answer is C. Study tip: Master the cycle of powers of ii by memorizing that i4k+1=ii^{4k+1} = i, i4k+2=−1i^{4k+2} = -1, i4k+3=−ii^{4k+3} = -i, and i4k=1i^{4k} = 1. For any power of ii, just find the remainder when the exponent is divided by 4, then use the appropriate value from the basic cycle.

Question 10

For how many integer values of nn with 1≤n≤201 \leq n \leq 20 is in+in+1+in+2+in+3i^n + i^{n+1} + i^{n+2} + i^{n+3} equal to zero?

  1. 44
  2. 55
  3. 1616
  4. 2020 (correct answer)
Explanation: The expression in+in+1+in+2+in+3=in(1+i+i2+i3)=in(1+i−1−i)=in⋅0=0i^n + i^{n+1} + i^{n+2} + i^{n+3} = i^n(1 + i + i^2 + i^3) = i^n(1 + i - 1 - i) = i^n \cdot 0 = 0. This equals zero regardless of the value of nn. Therefore, for all integer values nn with 1≤n≤201 \leq n \leq 20, the expression equals zero. That's all 20 values. Choice A might come from thinking only multiples of 4 work, choice B from an off-by-one error, and choice C from misunderstanding which values make the expression zero.

Question 11

If z=1+i1−iz = \frac{1 + i}{1 - i}, what is the value of z2024z^{2024}?

  1. 11 (correct answer)
  2. −1-1
  3. ii
  4. −i-i
Explanation: First, simplify z=1+i1−iz = \frac{1 + i}{1 - i} by multiplying numerator and denominator by the conjugate of the denominator: z=(1+i)(1+i)(1−i)(1+i)=1+2i+i21−i2=1+2i−11−(−1)=2i2=iz = \frac{(1 + i)(1 + i)}{(1 - i)(1 + i)} = \frac{1 + 2i + i^2}{1 - i^2} = \frac{1 + 2i - 1}{1 - (-1)} = \frac{2i}{2} = i. Therefore z2024=i2024z^{2024} = i^{2024}. Since 2024=4×5062024 = 4 \times 506, we have i2024=(i4)506=1506=1i^{2024} = (i^4)^{506} = 1^{506} = 1. Choice B assumes z=−iz = -i from a sign error. Choice C assumes the student thinks z2024=zz^{2024} = z. Choice D assumes both errors.