Understanding Complex Numbers - Algebra 2

1. A student claims that for any complex number $z = a + bi$, the expression $z + \overline{z}$ always equals $2a$. To verify this claim using $z = -3 + 7i$, what calculation should be performed?
   Answer: $(-3 + 7i) + (-3 - 7i) = -6$ The complex conjugate of $z = -3 + 7i$ is $\overline{z} = -3 - 7i$. So $z + \overline{z} = (-3 + 7i) + (-3 - 7i) = -6 + 0i = -6 = 2(-3) = 2a$, confirming the claim. Choice B incorrectly uses $3 + 7i$ as the conjugate. Choice C uses $z + z$ instead of $z + \overline{z}$. Choice D incorrectly uses $3 - 7i$ as the conjugate (wrong sign on real part).
2. If $i^{4k+2} = -1$ where $k$ is an integer, and $z = 3 + 4i^{4k+1}$, what is $z$ in standard form?
   Answer: $3 + 4i$ This question tests your understanding of powers of the imaginary unit $i$ and how they cycle in a predictable pattern. When working with complex numbers involving powers of $i$, remember that $i$ follows a repeating cycle every four powers: $i^1 = i,i^2 = -1,i^3 = -i$, and $i^4 = 1$. Since you're given that $i^{4k+2} = -1$, let's verify this makes sense. Using the cyclical pattern, $i^{4k+2} = i^{4k} \cdot i^2 = (i^4)^k \cdot i^2 = 1^k \cdot (-1) = -1$. This confirms the given information is consistent. Now to find $z = 3 + 4i^{4k+1}$, you need to determine what $i^{4k+1}$ equals. Using the same approach: $i^{4k+1} = i^{4k} \cdot i^1 = (i^4)^k \cdot i = 1^k \cdot i = i$. Therefore, $z = 3 + 4i^{4k+1} = 3 + 4i$, which is answer choice B. Looking at the wrong answers: A) $3 - 4i$ results from incorrectly thinking $i^{4k+1} = -i$. C) $-1 + 4i$ comes from mistakenly using the given value $i^{4k+2} = -1$ in place of the constant term. D) $3 + 4$ ignores the imaginary unit entirely, treating $i^{4k+1}$ as simply 1. Study tip: Memorize the four-step cycle of powers of $i$. When you see any power of $i$, divide the exponent by 4 and use the remainder to determine which value in the cycle applies.
3. If $i^n = -1$ where $n$ is a positive integer, which statement about $n$ must be true?
   Answer: $n$ is divisible by 4 with remainder 2 The powers of $i$ cycle every 4: $i^1 = i$, $i^2 = -1$, $i^3 = -i$, $i^4 = 1$. For $i^n = -1$, we need $n \equiv 2 \pmod{4}$, meaning $n$ leaves remainder 2 when divided by 4. Choice A correctly states this. Choice B is incorrect because not all even numbers work (e.g., $i^4 = 1$, not $-1$). Choice C says 'odd multiple of 2' which is contradictory since multiples of 2 are even. Choice D is wrong since many non-prime numbers work (e.g., $n = 6, 10, 14$).
4. A complex number $z$ satisfies $z^2 = -9$. If $z = a + bi$ where $a$ and $b$ are real, which statement about the possible values of $a$ and $b$ is correct?
   Answer: $a = 0$ and $b = \pm 3$ When you encounter a complex equation like $z^2 = -9$, you're looking for values that make a perfect square equal to a negative number — something impossible with real numbers alone. This is where complex numbers with imaginary components become essential. To solve $z^2 = -9$, let's substitute $z = a + bi$ and expand: $(a + bi)^2 = a^2 + 2abi + (bi)^2 = a^2 + 2abi + b^2i^2$. Since $i^2 = -1$, this becomes $a^2 + 2abi - b^2 = (a^2 - b^2) + 2abi$. Setting this equal to $-9$: $(a^2 - b^2) + 2abi = -9 + 0i$. For complex numbers to be equal, their real parts must equal and their imaginary parts must equal. This gives us: - Real parts: $a^2 - b^2 = -9$ - Imaginary parts: $2ab = 0$ From $2ab = 0$, either $a = 0$ or $b = 0$. If $a = 0$, then $-b^2 = -9$, so $b^2 = 9$ and $b = \pm 3$. If $b = 0$, then $a^2 = -9$, which has no real solutions since $a$ must be real. Choice A is wrong because $a^2 + b^2 = 9$ describes a circle of solutions, but we found only two specific points. Choice B incorrectly assumes $b = 0$, which leads to $a^2 = -9$ (impossible for real $a$). Choice C gives incorrect values that don't satisfy our constraint equations. Remember: when solving $z^n = \text{negative real}$, look for purely imaginary solutions first — they're often the key to problems involving negative perfect squares.
5. The equation $x^2 + 4x + 13 = 0$ has complex solutions. When written in the form $a \pm bi$, what are the values of $a$ and $b$?
   Answer: $a = -2, b = 3$ When you encounter a quadratic equation that doesn't factor nicely, you'll need the quadratic formula to find complex solutions. The key insight is recognizing when the discriminant (the part under the square root) will be negative, producing imaginary numbers. For $x^2 + 4x + 13 = 0$, apply the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 1,b = 4$, and $c = 13$. Substituting: $x = \frac{-4 \pm \sqrt{16 - 52}}{2} = \frac{-4 \pm \sqrt{-36}}{2}$ Since $\sqrt{-36} = \sqrt{36} \cdot \sqrt{-1} = 6i$, we get: $x = \frac{-4 \pm 6i}{2} = -2 \pm 3i$ This gives us $a = -2$ and $b = 3$, confirming answer C. Let's examine why the other options are incorrect. Choice A gives $a = -2$ correctly but has $b = \sqrt{13}$, which comes from mistakenly using $\sqrt{13}$ instead of properly simplifying $\sqrt{-36}$. Choice B has $a = 2$, which would result from forgetting the negative sign in the quadratic formula's numerator. Choice D shows $a = -4$, which happens when you forget to divide the entire numerator by 2. Remember this pattern: when the discriminant is negative, you'll always get complex conjugate pairs in the form $a \pm bi$. The real part $a$ comes from $\frac{-b}{2a}$, and the imaginary coefficient $b$ comes from simplifying the square root of the negative discriminant.

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