Imaginary Unit i - Math 2

If $w = i^{2023} + i^{2024} + i^{2025} + i^{2026}$ , what is the value of $w$ ?
Answer: $0$ The powers of i repeat every 4: $i^1 = i$ , $i^2 = -1$ , $i^3 = -i$ , $i^4 = 1$ . To find the remainder when dividing the exponent by 4: $2023 = 4(505) + 3$ , so $i^{2023} = i^3 = -i$ . $2024 = 4(506) + 0$ , so $i^{2024} = i^0 = 1$ . $2025 = 4(506) + 1$ , so $i^{2025} = i^1 = i$ . $2026 = 4(506) + 2$ , so $i^{2026} = i^2 = -1$ . Therefore, $w = -i + 1 + i + (-1) = 0$ .
For how many integer values of $n$ with $1 \leq n \leq 20$ is $i^n + i^{n+1} + i^{n+2} + i^{n+3}$ equal to zero?
Answer: $20$ The expression $i^n + i^{n+1} + i^{n+2} + i^{n+3} = i^n(1 + i + i^2 + i^3) = i^n(1 + i - 1 - i) = i^n \cdot 0 = 0$ . This equals zero regardless of the value of $n$ . Therefore, for all integer values $n$ with $1 \leq n \leq 20$ , 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.
If $z = i^{100} + i^{101} + i^{102} + i^{103}$ , what is $z^2$ ?
Answer: $0$ When you encounter powers of the imaginary unit $i$ , remember that $i$ follows a cyclical pattern every 4 powers: $i^1 = i$ , $i^2 = -1$ , $i^3 = -i$ , and $i^4 = 1$ . This cycle then repeats. To find each term in $z$ , divide the exponent by 4 and use the remainder. For $i^{100}$ : $100 ÷ 4 = 25$ remainder $0$ , so $i^{100} = i^0 = 1$ . For $i^{101}$ : $101 ÷ 4 = 25$ remainder $1$ , so $i^{101} = i^1 = i$ . For $i^{102}$ : $102 ÷ 4 = 25$ remainder $2$ , so $i^{102} = i^2 = -1$ . For $i^{103}$ : $103 ÷ 4 = 25$ remainder $3$ , so $i^{103} = i^3 = -i$ . Therefore: $z = 1 + i + (-1) + (-i) = 1 + i - 1 - i = 0$ Since $z = 0$ , we have $z^2 = 0^2 = 0$ , making (C) correct. (A) $-1$ might tempt you if you incorrectly calculated $z$ or confused it with $i^2$ . (B) $1$ could result from miscounting the cycle or thinking $z^2$ always equals $1$ for complex numbers. (D) $4$ might occur if you added the four terms without considering the signs or thought each term equaled $1$ . Study tip: For any four consecutive powers of $i$ , their sum is always zero because the cycle $i, -1, -i, 1$ adds to zero. Recognizing this pattern saves time on similar problems.
What is the value of $(1+i)^8$ ?
Answer: $16$ First find $(1+i)^2 = 1 + 2i + i^2 = 1 + 2i - 1 = 2i$ . Then $(1+i)^4 = ((1+i)^2)^2 = (2i)^2 = 4i^2 = -4$ . Finally, $(1+i)^8 = ((1+i)^4)^2 = (-4)^2 = 16$ . Choice B incorrectly computes $(1+i)^4 = 4i$ , choice C gets $(1+i)^4 = 4$ instead of $-4$ , and choice D computes $(1+i)^8$ as if it were $2^8$ .
Which of the following equals $i^{-15}$ ?
Answer: $i$ Since $i^{-15} = \frac{1}{i^{15}}$ , we first find $i^{15}$ . Since $15 = 4(3) + 3$ , we have $i^{15} = i^3 = -i$ . Therefore $i^{-15} = \frac{1}{-i} = \frac{1}{-i} \cdot \frac{i}{i} = \frac{i}{-i^2} = \frac{i}{-(-1)} = \frac{i}{1} = i$ . Choice B results from forgetting the negative sign when computing $\frac{1}{-i}$ , choice C comes from incorrectly thinking $i^{15} = i^3 = 1$ , and choice D comes from computing $i^{15}$ incorrectly as $1$ then taking the reciprocal incorrectly.
If $i^n = -i$ where $n$ is a positive integer, which of the following could be the remainder when $n^2$ is divided by 8?
Answer: $1$ or $5$ Since $i^n = -i$ and the powers of i cycle every 4 with $i^3 = -i$ , we need $n \equiv 3 \pmod{4}$ . So $n = 4k + 3$ for some non-negative integer $k$ . Then $n^2 = (4k + 3)^2 = 16k^2 + 24k + 9 = 8(2k^2 + 3k + 1) + 1$ . For $k$ even, say $k = 2m$ : $n^2 = 8(8m^2 + 6m + 1) + 1 \equiv 1 \pmod{8}$ . For $k$ odd, say $k = 2m + 1$ : $n^2 = 8(8m^2 + 10m + 3) + 5 \equiv 5 \pmod{8}$ . Therefore the remainder is either 1 or 5. Choice B incorrectly assumes $n \equiv 1$ or $n \equiv 3 \pmod{4}$ . Choice C assumes $n \equiv 3 \pmod{4}$ but miscalculates $n^2 \pmod{8}$ . Choice D assumes different values of $n \pmod{4}$ .
If $w = i^{17} \cdot i^{23} \cdot i^{31}$ and $w^k = 1$ for some positive integer $k$ , what is the smallest possible value of $k$ ?
Answer: $4$ First, simplify $w = i^{17} \cdot i^{23} \cdot i^{31} = i^{17+23+31} = i^{71}$ . Since $71 = 4 \times 17 + 3$ , we have $i^{71} = i^3 = -i$ . So $w = -i$ . We need the smallest positive integer $k$ such that $w^k = (-i)^k = 1$ . Since $(-i)^1 = -i$ , $(-i)^2 = (-1)^2 \cdot i^2 = 1 \cdot (-1) = -1$ , $(-i)^3 = (-i)^2 \cdot (-i) = (-1)(-i) = i$ , and $(-i)^4 = ((-i)^2)^2 = (-1)^2 = 1$ , the smallest value is $k = 4$ . Choice A assumes $(-i)^2 = 1$ by ignoring the negative sign. Choice B assumes $(-i)^3 = 1$ by miscalculating. Choice D assumes the period is 8 instead of 4.
If $\sum_{k=0}^{n} i^k = 0$ , what is the smallest positive integer $n$ for which this is true?
Answer: $n = 3$ We compute: $\sum_{k=0}^{n} i^k = 1 + i + i^2 + i^3 + \cdots + i^n$ . For $n=0$ : sum is $1$ . For $n=1$ : sum is $1+i$ . For $n=2$ : sum is $1+i+(-1) = i$ . For $n=3$ : sum is $1+i+(-1)+(-i) = 0$ . Therefore the smallest positive integer $n$ is $3$ . Choice A gives sum $= i \neq 0$ , choice C would be the next value where the sum is $0$ (since the cycle repeats every 4), and choice D gives sum $= 0$ but is not the smallest such $n$ .
If $p$ and $q$ are positive integers such that $i^p \cdot i^q = -1$ , what is the smallest possible value of $p + q$ ?
Answer: $2$ When you encounter problems involving powers of the imaginary unit $i$ , remember that $i$ follows a cyclical pattern: $i^1 = i$ , $i^2 = -1$ , $i^3 = -i$ , and $i^4 = 1$ . This pattern repeats every 4 powers. Using the property of exponents, $i^p \cdot i^q = i^{p+q}$ . So you need $i^{p+q} = -1$ . From the cyclical pattern, $i^n = -1$ when $n$ leaves a remainder of 2 when divided by 4. In other words, $p + q \equiv 2 \pmod{4}$ . The smallest positive value where $p + q \equiv 2 \pmod{4}$ is $p + q = 2$ . This works when $p = 1$ and $q = 1$ : $i^1 \cdot i^1 = i^2 = -1$ . ✓ Now let's check why the other answers don't work: A) If $p + q = 4$ , then $i^4 = 1$ , not $-1$ . B) If $p + q = 3$ , then $i^3 = -i$ , not $-1$ . D) If $p + q = 5$ , then $i^5 = i^1 = i$ , not $-1$ . The correct answer is C. Study tip: Master the cycle of powers of $i$ by memorizing that $i^{4k+1} = i$ , $i^{4k+2} = -1$ , $i^{4k+3} = -i$ , and $i^{4k} = 1$ . For any power of $i$ , just find the remainder when the exponent is divided by 4, then use the appropriate value from the basic cycle.
What is the value of $\frac{i^{2022} + i^{2023}}{i^{2020} + i^{2021}}$ ?
Answer: $-1$ Since $2020 = 4(505)$ , we have $i^{2020} = 1$ , $i^{2021} = i$ , $i^{2022} = -1$ , and $i^{2023} = -i$ . The expression becomes $\frac{-1 + (-i)}{1 + i} = \frac{-1-i}{1+i}$ . Multiplying numerator and denominator by the conjugate: $\frac{(-1-i)(1-i)}{(1+i)(1-i)} = \frac{-1+i-i+i^2}{1-i^2} = \frac{-1-1}{1-(-1)} = \frac{-2}{2} = -1$ . 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.
If $z = \frac{1 + i}{1 - i}$ , what is the value of $z^{2024}$ ?
Answer: $1$ First, simplify $z = \frac{1 + i}{1 - i}$ by multiplying numerator and denominator by the conjugate of the denominator: $z = \frac{(1 + i)(1 + i)}{(1 - i)(1 + i)} = \frac{1 + 2i + i^2}{1 - i^2} = \frac{1 + 2i - 1}{1 - (-1)} = \frac{2i}{2} = i$ . Therefore $z^{2024} = i^{2024}$ . Since $2024 = 4 \times 506$ , we have $i^{2024} = (i^4)^{506} = 1^{506} = 1$ . Choice B assumes $z = -i$ from a sign error. Choice C assumes the student thinks $z^{2024} = z$ . Choice D assumes both errors.

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