This question tests your understanding of complex conjugates—pairs of complex numbers that differ only in the sign of their imaginary part—and how to use them to divide complex numbers by rationalizing denominators. The conjugate of a + bi is a - bi (flip only the sign of the imaginary part, keep real part the same). An important property: taking the conjugate twice returns the original number, because flipping the sign twice brings you back where you started. For z = -6 + 2i, the conjugate is -6 - 2i (flip sign of +2i to -2i). Taking the conjugate again: conjugate of -6 - 2i is -6 + 2i (flip sign of -2i back to +2i). Choice C correctly shows -6 + 2i, which equals the original z. Choice A incorrectly changes the sign of the real part when taking conjugates. Remember: conjugates only flip the imaginary part's sign! This double-conjugate property confirms that conjugation is its own inverse operation.

This question tests your understanding of complex conjugates—pairs of complex numbers that differ only in the sign of their imaginary part—and their product being real and nonnegative. The conjugate of a + bi is a - bi; their product is a² + b², always real and positive (or zero). For (6 - 2i)(6 + 2i), it's $6^2 + 2^2 = 36 + 4 = 40$. You can expand: $6 \times 6 = 36$, $6 \times 2i = 12i$, $-2i \times 6 = -12i$, $-2i \times 2i = -4i^2 = 4$; $36 + 4 = 40$, imaginaries cancel. Choice C correctly computes 40 using the formula. Choice A is 32, perhaps from 36 -4 instead of +4, forgetting i²=-1 makes -(-4)=+4. The pattern (a + bi)(a - bi) = a² + b² is key for modulus and rationalizing. Memorize it and practice with $(5 + i)(5 - i) = 25 + 1 = 26$—you've got this!

This question tests your understanding of complex conjugates—pairs of complex numbers that differ only in the sign of their imaginary part—and how to use them to divide complex numbers by rationalizing denominators. The conjugate of a + bi is a - bi (flip only the sign of the imaginary part, keep real part the same); when dividing complex numbers, we multiply by the conjugate of the denominator to make it real. To divide (1 + 2i) by (3 - i) using conjugates: (1) Identify conjugate of denominator: conjugate of 3 - i is 3 + i. (2) Multiply numerator and denominator by this conjugate: [(1 + 2i)(3 + i)] / [(3 - i)(3 + i)]. (3) Multiply numerator using FOIL: (1 + 2i)(3 + i) = 3 + i + 6i + 2i² = 3 + 7i + 2(-1) = 3 + 7i - 2 = 1 + 7i. (4) Multiply denominator: (3 - i)(3 + i) = 3² + 1² = 9 + 1 = 10. (5) Divide: (1 + 7i)/10 = 1/10 + 7i/10. Choice A correctly shows 1/10 + 7i/10 as the result in standard a + bi form. Choice B shows 1/10 - 7i/10, which has the wrong sign for the imaginary part—when we computed the numerator as 1 + 7i, dividing by 10 preserves the positive sign: (1 + 7i)/10 = 1/10 + 7i/10, not 1/10 - 7i/10! Division tip: after finding your answer, verify by multiplying it by the original denominator—you should get back the original numerator. Here: (1/10 + 7i/10)(3 - i) = 3/10 - i/10 + 21i/10 - 7i²/10 = 3/10 + 20i/10 + 7/10 = 10/10 + 20i/10 = 1 + 2i ✓

This question tests your understanding of complex conjugates—pairs of complex numbers that differ only in the sign of their imaginary part—and the property that the conjugate of a quotient is the quotient of the conjugates: $\overline{(z/w)} = \overline{z} / \overline{w}$. To compute $\overline{((2 - i)/(3 + 2i))}$, first simplify the division or use the property directly: conjugate of numerator $2 - i$ is $2 + i$, conjugate of denominator $3 + 2i$ is $3 - 2i$, so $\overline{z/w} = (2 + i)/(3 - 2i)$. Choice B correctly applies this property, giving $(2 + i)/(3 - 2i)$. A tempting distractor like Choice D might flip the sign in the denominator incorrectly to $3 + 2i$ instead of $3 - 2i$, but remember to flip only the imaginary sign for each conjugate. The strategy is to handle conjugates of operations: for products, $\overline{zw} = \overline{z} \overline{w}$; for quotients, as above—practice by verifying with numbers, like computing the quotient first then conjugating to confirm. Keep up the excellent work; these properties save time in complex calculations!

This question tests your understanding of what a complex conjugate is. The key concept is that the conjugate of a + bi is a - bi, flipping the sign of the imaginary part while keeping the real part the same. For example, conjugate of 3 + 4i is 3 - 4i; for -1 - 2i is -1 + 2i. Note that if there's no real part, conjugate of 5i is -5i. For 2 - 5i, flip the -5i to +5i, so 2 + 5i. Choice B correctly identifies the conjugate by flipping only the imaginary sign. Choice D might tempt if you forget to flip and just repeat the number, but remember to change the sign of the i term. To find the conjugate, imagine the complex plane and reflect over the real axis. The real part stays, imaginary sign flips—simple as that! Keep practicing, you're on the right track!

This question tests your understanding of using complex conjugates to divide complex numbers by rationalizing the denominator. The key concept is to multiply numerator and denominator by the conjugate of the denominator to make the denominator real. The conjugate of $c + di$ is $c - di$, and $(c + di)(c - di) = c^2 + d^2$. For example, to divide by $2 + i$, multiply by $2 - i$, den becomes $4 + 1 = 5$. For $(2 - i)/(3 + 2i)$, conjugate of denominator is $3 - 2i$, multiply: numerator $(2 - i)(3 - 2i) = 4 - 7i$, denominator $9 + 4 = 13$, so $4/13 - 7/13 i$. Choice A correctly uses the conjugate and performs the algebra accurately, remembering $i^2 = -1$. Choice B fails by perhaps switching the signs in the imaginary part during multiplication, leading to a positive imaginary. When multiplying, use FOIL carefully and substitute $i^2 = -1$. Double-check signs and terms. You got this—practice a few more to master it!

This question tests your understanding of complex conjugates—pairs of complex numbers that differ only in the sign of their imaginary part—and how to use them to divide complex numbers by rationalizing denominators. The conjugate of a + bi is a - bi (flip only the sign of the imaginary part, keep real part the same): for 2 + 3i, conjugate is 2 - 3i. The magic property: when you multiply a complex number by its conjugate, you always get a real positive result: (a + bi)(a - bi) = a² + b² (the imaginary parts cancel!). To divide (7 - 3i) by (2 + 3i) using conjugates: (1) Identify conjugate of denominator: conjugate of 2 + 3i is 2 - 3i. (2) Multiply numerator and denominator by this conjugate: [(7 - 3i)(2 - 3i)] / [(2 + 3i)(2 - 3i)]. (3) Multiply numerator using FOIL: (7 - 3i)(2 - 3i) = 14 - 21i - 6i + 9i² = 14 - 27i + 9(-1) = 14 - 27i - 9 = 5 - 27i. (4) Multiply denominator using conjugate property: (2 + 3i)(2 - 3i) = 4 + 9 = 13 (real!). (5) Divide: (5 - 27i)/13 = 5/13 - 27i/13 in standard a + bi form. Choice A correctly identifies 5/13 - 27i/13 as the simplified form. Choice B has the wrong sign on the imaginary part: the numerator multiplication gives -27i, not +27i—when you FOIL (7 - 3i)(2 - 3i), both -21i and -6i terms are negative, combining to -27i. Choice D shows the unsimplified expression—you must multiply out and simplify to get the final a + bi form. Division recipe using conjugates: multiply both numerator and denominator by the conjugate of the denominator, expand the numerator (will generally stay complex), use the conjugate property for the denominator (becomes real: a² + b²), then divide both parts of the numerator by this real denominator. Always check your arithmetic carefully, especially the signs when combining like terms!

This question tests your understanding of complex conjugates—pairs of complex numbers that differ only in the sign of their imaginary part—and how to use them to divide complex numbers by rationalizing denominators. The conjugate of a + bi is a - bi (flip only the sign of the imaginary part, keep real part the same): for 3 + 4i, conjugate is 3 - 4i; for 2 - 5i, conjugate is 2 + 5i. For the complex number 3 - 4i, we keep the real part 3 the same and flip the sign of the imaginary part from -4i to +4i, giving us 3 + 4i as the conjugate. Choice B correctly identifies 3 + 4i by flipping only the imaginary part's sign from negative to positive. Choice C incorrectly flips the sign of BOTH parts when finding the conjugate: conjugate of 3 - 4i is 3 + 4i, NOT -3 - 4i. Only flip the imaginary part's sign (the part with i), keep the real part unchanged! Conjugate quick reference: to find conjugate of a + bi, keep a the same, flip the sign of bi to get a - bi. To find conjugate of a - bi, keep a, flip to get a + bi.

This question tests your understanding of complex conjugates—pairs of complex numbers that differ only in the sign of their imaginary part—and how to use them to divide complex numbers by rationalizing denominators. The conjugate of a + bi is a - bi (flip only the sign of the imaginary part, keep real part the same): for 3 + 2i, conjugate is 3 - 2i. The magic property: when you multiply a complex number by its conjugate, you always get a real positive result: (a + bi)(a - bi) = a² + b² (the imaginary parts cancel!). To divide (2 - i) by (3 + 2i) using conjugates: (1) Identify conjugate of denominator: conjugate of 3 + 2i is 3 - 2i. (2) Multiply numerator and denominator by this conjugate: [(2 - i)(3 - 2i)] / [(3 + 2i)(3 - 2i)]. (3) Multiply numerator using FOIL: (2 - i)(3 - 2i) = 6 - 4i - 3i + 2i² = 6 - 7i + 2(-1) = 4 - 7i. (4) Multiply denominator using conjugate property: (3 + 2i)(3 - 2i) = 9 + 4 = 13 (real!). (5) Divide: (4 - 7i)/13 = 4/13 - 7i/13 in standard a + bi form. Choice A correctly identifies 4/13 - 7i/13 as the simplified form. Choice B has the wrong sign on the imaginary part: 4/13 + 7i/13 instead of minus—check your FOIL multiplication of (2 - i)(3 - 2i) carefully! The -3i and -4i terms combine to give -7i, not +7i. Choice D shows the unsimplified fraction with conjugate in denominator—you must complete the multiplication to get a real denominator, then divide both parts by that real number. Division recipe using conjugates: multiply both top and bottom by conjugate of denominator, simplify numerator (stays complex), simplify denominator (becomes real using a² + b²), then divide both real and imaginary parts by the real denominator. Always express final answer as a + bi, not as a fraction with complex denominator!

This question tests your understanding of complex conjugates—pairs of complex numbers that differ only in the sign of their imaginary part—and the property that the conjugate of a sum equals the sum of the conjugates. The conjugate of a + bi is a - bi (flip only the sign of the imaginary part, keep real part the same); an important property: conjugate of (z + w) = conjugate of z + conjugate of w, written as z̄ + w̄ = z̄ + w̄. First, let's find z + w: z + w = (2 + 5i) + (1 - 3i) = 3 + 2i; then the conjugate of this sum is 3 - 2i. Alternatively, using the property: z̄ = 2 - 5i and w̄ = 1 + 3i, so z̄ + w̄ = (2 - 5i) + (1 + 3i) = 3 - 2i—same result! Choice C correctly identifies 3 - 2i as the conjugate of the sum. Choice A shows 3 + 2i, which is actually z + w itself, not its conjugate—remember to flip the sign of the imaginary part after finding the sum! The property z̄ + w̄ = z̄ + w̄ extends to any finite sum and is part of what makes the conjugate operation "linear"—it distributes over addition. This property, along with z̄ · w̄ = z̄ · w̄ for products, makes conjugates behave nicely in algebraic manipulations and is fundamental in many proofs about complex numbers!