Algebra 2 Quiz: Using Conjugates With Complex Numbers
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Using Conjugates With Complex NumbersQuestion 1 of 20

What is the complex conjugate of 2−5i2-5i?

−2+5i-2+5i
2−5i2-5i
2+5i2+5i
−2−5i-2-5i
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Algebra 2 Quiz

Algebra 2 Quiz: Using Conjugates With Complex Numbers

Practice Using Conjugates With Complex Numbers in Algebra 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 Using Conjugates With Complex Numbers, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra 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

What is the complex conjugate of 2−5i2-5i?

  1. −2+5i-2+5i
  2. 2−5i2-5i
  3. 2+5i2+5i (correct answer)
  4. −2−5i-2-5i
Explanation: 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!

Question 2

Use conjugates to rationalize the denominator and write in standard form a+bia+bi:

2−i3+2i.\frac{2-i}{3+2i}.

  1. 413−713i\frac{4}{13}-\frac{7}{13}i (correct answer)
  2. 813−113i\frac{8}{13}-\frac{1}{13}i
  3. 2−i3+2i\frac{2-i}{3+2i}
  4. 413+713i\frac{4}{13}+\frac{7}{13}i
Explanation: 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. To divide 2−i2 - i by 3+2i3 + 2i using conjugates: (1) Identify conjugate of denominator: conjugate of 3+2i3 + 2i is 3−2i3 - 2i. (2) Multiply numerator and denominator by this conjugate: [(2−i)(3−2i)][(2 - i)(3 - 2i)] divided by [(3+2i)(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(2 - i)(3 - 2i) = 6 - 4i - 3i + 2i^2 = 6 - 7i + 2(-1) = 4 - 7i. (4) Multiply denominator using conjugate property: (3+2i)(3−2i)=32+22=9+4=13(3 + 2i)(3 - 2i) = 3^2 + 2^2 = 9 + 4 = 13. (5) Divide: (4−7i)/13=4/13−7i/13(4 - 7i)/13 = 4/13 - 7i/13. Choice A correctly identifies 4/13−7i/134/13 - 7i/13 by properly multiplying and simplifying both numerator and denominator. Choice B has the wrong sign on the imaginary part, suggesting a sign error in the FOIL multiplication. Division recipe: always check your signs carefully when multiplying complex numbers!

Question 3

Let z=2+5iz=2+5i and w=1−3iw=1-3i. Using the property z+w‾=z‾+w‾\overline{z+w}=\overline{z}+\overline{w}, what is z+w‾\overline{z+w}?

  1. 3+2i3+2i
  2. −3+2i-3+2i
  3. 3−2i3-2i (correct answer)
  4. 1+8i1+8i
Explanation: 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!

Question 4

Use conjugates to divide and write in standard form a+bia+bi:

1+2i3−i.\frac{1+2i}{3-i}.

  1. 1+2i3−i\frac{1+2i}{3-i}
  2. 110−710i\frac{1}{10}-\frac{7}{10}i
  3. 15+75i\frac{1}{5}+\frac{7}{5}i
  4. 110+710i\frac{1}{10}+\frac{7}{10}i (correct answer)
Explanation: 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. To divide 1+2i1 + 2i by 3−i3 - i: (1) Identify conjugate of denominator: conjugate of 3−i3 - i is 3+i3 + i. (2) Multiply numerator and denominator by this conjugate: [(1+2i)(3+i)][(1 + 2i)(3 + i)] divided by [(3−i)(3+i)][(3 - i)(3 + i)]. (3) Multiply numerator using FOIL: (1+2i)(3+i)=3+i+6i+2i2=3+7i+2(−1)=1+7i(1 + 2i)(3 + i) = 3 + i + 6i + 2i^2 = 3 + 7i + 2(-1) = 1 + 7i. (4) Multiply denominator using conjugate property: (3−i)(3+i)=32+12=9+1=10(3 - i)(3 + i) = 3^2 + 1^2 = 9 + 1 = 10. (5) Divide: (1+7i)/10=1/10+7i/10(1 + 7i)/10 = 1/10 + 7i/10. Choice A correctly identifies 1/10+7i/101/10 + 7i/10 by properly applying the conjugate multiplication process. Choice B has the wrong sign on the imaginary part, while choice C forgot to divide by the denominator. Always divide both real and imaginary parts by the real denominator!

Question 5

Which expression is always real and nonnegative for real numbers aa and bb (not both zero), and equals the modulus squared ∣a+bi∣2|a+bi|^2?

Choose the correct expression.

  1. (a+bi)(a+bi)(a+bi)(a+bi)
  2. (a+bi)(a−bi)(a+bi)(a-bi) (correct answer)
  3. (a−bi)(a−bi)(a-bi)(a-bi)
  4. (a+bi)+(a−bi)(a+bi)+(a-bi)
Explanation: 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 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!). Let's verify each option: (A) (a + bi)(a + bi) = a² + 2abi - b² is complex (has 2abi term), not always real. (B) (a + bi)(a - bi) = a² - abi + abi - b²i² = a² + b² is always real and equals |a + bi|². (C) (a - bi)(a - bi) = a² - 2abi - b² is complex (has -2abi term), not always real. (D) (a + bi) + (a - bi) = 2a is real but doesn't equal the modulus squared. Choice B correctly identifies (a + bi)(a - bi) as the expression that's always real, nonnegative (since it's a sum of squares), and equals the modulus squared. Choice A multiplies the complex number by itself instead of its conjugate, which generally gives a complex result with imaginary part 2ab—not what we want! The (a + bi)(a - bi) = a² + b² pattern is worth memorizing: it's how we eliminate i from denominators, and it equals the modulus squared! This connection between conjugates and modulus is fundamental in complex analysis.

Question 6

Use the conjugate method to simplify and write in a+bia+bi form: 2−i3+2i.\frac{2-i}{3+2i}.

  1. 413−713i\frac{4}{13} - \frac{7}{13}i (correct answer)
  2. 413+713i\frac{4}{13} + \frac{7}{13}i
  3. 813−113i\frac{8}{13} - \frac{1}{13}i
  4. 2−i3−2i\frac{2-i}{3-2i}
Explanation: This question tests your understanding of complex conjugates—pairs of complex numbers that differ only in the sign of their imaginary part—and using them to rationalize denominators. The conjugate of a + bi is a - bi; multiply numerator and denominator by it to eliminate i below. For (2 - i)/(3 + 2i), conjugate of denominator 3 - 2i; numerator (2 - i)(3 - 2i)=23=6, 2(-2i)=-4i, -i3=-3i, -i(-2i)=2i²=-2; real 6-2=4, imaginary -4i-3i=-7i, so 4 - 7i. Denominator (3 + 2i)(3 - 2i)=9 +4=13. Thus (4 - 7i)/13 = 4/13 - 7/13 i. Choice A correctly rationalizes to 4/13 - 7/13 i. Choice B has +7/13 i, perhaps from sign error in imaginary part during multiplication—check FOIL signs carefully! Follow the division steps: identify conjugate, multiply top and bottom, expand, simplify. Example: (1 - i)/(1 + i) * (1 - i)/(1 - i) = (1 - 2i + i²)/(1 +1) = (1 - 2i -1)/2 = -2i/2 = -i—keep practicing, you're improving!

Question 7

Let z=7−4iz = 7 - 4i. What is the complex conjugate z‾\overline{z}? (Recall: the conjugate of a+bia+bi is a−bia-bi.)

  1. −7+4i-7 + 4i
  2. 7+4i7 + 4i (correct answer)
  3. −7−4i-7 - 4i
  4. 7−4i7 - 4i
Explanation: 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 identify them directly from the definition. The conjugate of a + bi is a - bi, where you flip only the sign of the imaginary part while keeping the real part the same: for example, the conjugate of 3 + 4i is 3 - 4i, and for 2 - 5i, it's 2 + 5i. For z = 7 - 4i, the conjugate is 7 + 4i because we change -4i to +4i, leaving the real part 7 unchanged—this is like reflecting the point across the real axis in the complex plane. Choice B correctly identifies the conjugate by flipping only the imaginary part's sign, resulting in 7 + 4i. A tempting distractor like Choice A or C might flip the signs of both parts, but remember, only the imaginary part's sign changes—the real part stays the same, so -7 + 4i or -7 - 4i would be incorrect. To find the conjugate quickly, always keep the real number as is and switch the sign before the i; practicing with a few examples like 1 + i (conjugate 1 - i) or -3 - 2i (conjugate -3 + 2i) will make this second nature. You're doing great—keep visualizing that mirror flip across the real axis!

Question 8

Let z=2+5iz=2+5i and w=1−3iw=1-3i. Find (z+w)‾\overline{(z+w)}.

(Use the property z+w‾=z‾+w‾\overline{z+w}=\overline{z}+\overline{w}.)

  1. 3−2i3-2i (correct answer)
  2. −3+2i-3+2i
  3. 3+2i3+2i
  4. −3−2i-3-2i
Explanation: 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), and there's a useful property: the conjugate of a sum equals the sum of the conjugates, or symbolically, conjugate of (z + w) = conjugate of z + conjugate of w. First, let's find z + w: (2 + 5i) + (1 - 3i) = 3 + 2i (add real parts: 2 + 1 = 3, add imaginary parts: 5i - 3i = 2i). Now find the conjugate of this sum: conjugate of (3 + 2i) = 3 - 2i, flipping only the sign of the imaginary part from +2i to -2i. Choice A correctly identifies this as 3 - 2i. Choice C incorrectly keeps the imaginary part positive: the conjugate of 3 + 2i is 3 - 2i, not 3 + 2i—you must flip the sign of the imaginary part! We could also verify using the property: conjugate of z = 2 - 5i, conjugate of w = 1 + 3i, so their sum is (2 - 5i) + (1 + 3i) = 3 - 2i, confirming our answer. This property extends to any finite sum of complex numbers and is very useful for simplifying conjugate calculations!

Question 9

Let z=−6+2iz=-6+2i. Using the fact that taking the conjugate twice returns the original number, compute z‾‾\overline{\overline{z}}.

  1. 6−2i6-2i
  2. −6−2i-6-2i
  3. −6+2i-6+2i (correct answer)
  4. 6+2i6+2i
Explanation: 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.

Question 10

Use conjugates to simplify and express in a+bia+bi form: 7−3i2+3i.\frac{7-3i}{2+3i}.

  1. 513−2713i\frac{5}{13} - \frac{27}{13}i (correct answer)
  2. 513+2713i\frac{5}{13} + \frac{27}{13}i
  3. 2713−513i\frac{27}{13} - \frac{5}{13}i
  4. 5−27i2+3i\frac{5-27i}{2+3i}
Explanation: 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!

Question 11

Compute the product and note it is always real and nonnegative: (6−2i)(6+2i).(6-2i)(6+2i).

  1. 3232
  2. 36−436 - 4
  3. 4040 (correct answer)
  4. 36+4i36 + 4i
Explanation: 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 62+22=36+4=406^2 + 2^2 = 36 + 4 = 40. You can expand: 6×6=366 \times 6 = 36, 6×2i=12i6 \times 2i = 12i, −2i×6=−12i-2i \times 6 = -12i, −2i×2i=−4i2=4-2i \times 2i = -4i^2 = 4; 36+4=4036 + 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(5 + i)(5 - i) = 25 + 1 = 26—you've got this!

Question 12

For z=−6+8iz=-6+8i, compute zz‾z\overline{z}. (Recall zz‾=a2+b2=∣z∣2z\overline{z}=a^2+b^2=|z|^2, a real positive number.)

  1. 2828
  2. 100100 (correct answer)
  3. −100-100
  4. 36−6436-64
Explanation: This question tests your understanding of complex conjugates—pairs of complex numbers that differ only in the sign of their imaginary part—and the fundamental property that z times its conjugate equals the modulus squared. The conjugate of a + bi is a - bi (flip only the sign of the imaginary part, keep real part the same): for -6 + 8i, conjugate is -6 - 8i. 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!). For z = -6 + 8i, we compute z·conjugate(z) = (-6 + 8i)(-6 - 8i). Using the formula directly: a² + b² = (-6)² + 8² = 36 + 64 = 100. Alternatively, multiply out: (-6 + 8i)(-6 - 8i) = 36 + 48i - 48i - 64i² = 36 - 64(-1) = 36 + 64 = 100. This equals |z|², the square of the modulus! Choice B correctly identifies 100 as the product. Choice A gives 28, which might come from incorrectly computing 6² - 8² = 36 - 64 = -28 and taking absolute value—but the formula is a² + b², always adding! Choice C shows -100, but z·conjugate(z) is always positive (it's a sum of squares). Choice D shows 36 - 64 = -28, again subtracting instead of adding—remember the conjugate product formula always adds: a² + b². The z·conjugate(z) = |z|² property is fundamental: it's always real, always non-negative, and equals zero only when z = 0. This property makes conjugates essential for division (creating real denominators) and for finding distances in the complex plane. Practice: for any z = a + bi, quickly compute |z|² = a² + b² without expanding the full product!

Question 13

Compute using conjugates and express in a+bia+bi form: 1+2i3−i.\frac{1+2i}{3-i}.

  1. 110+710i\frac{1}{10}+\frac{7}{10}i (correct answer)
  2. 110−710i\frac{1}{10}-\frac{7}{10}i
  3. 1+2i3+i\frac{1+2i}{3+i}
  4. 14+12i\frac{1}{4}+\frac{1}{2}i
Explanation: 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 ✓

Question 14

Simplify and write in standard form a+bia+bi: 2−i3+2i.\frac{2-i}{3+2i}.

  1. 2−i3−2i\frac{2-i}{3-2i}
  2. 713−413i\frac{7}{13}-\frac{4}{13}i
  3. 413−713i\frac{4}{13}-\frac{7}{13}i (correct answer)
  4. 413+713i\frac{4}{13}+\frac{7}{13}i
Explanation: 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+dic + di is c−dic - di, and (c+di)(c−di)=c2+d2(c + di)(c - di) = c^2 + d^2. For example, to divide by 2+i2 + i, multiply by 2−i2 - i, den becomes 4+1=54 + 1 = 5. For (2−i)/(3+2i)(2 - i)/(3 + 2i), conjugate of denominator is 3−2i3 - 2i, multiply: numerator (2−i)(3−2i)=4−7i(2 - i)(3 - 2i) = 4 - 7i, denominator 9+4=139 + 4 = 13, so 413−713i\frac{4}{13} - \frac{7}{13} i. Choice A correctly uses the conjugate and performs the algebra accurately, remembering i2=−1i^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 i2=−1i^2 = -1. Double-check signs and terms. You got this—practice a few more to master it!

Question 15

Compute the product of a complex number and its conjugate: (5+2i)(5−2i).(5+2i)(5-2i). (Recall (a+bi)(a−bi)=a2+b2(a+bi)(a-bi)=a^2+b^2, which is always real and positive.)

  1. 2121
  2. 2929 (correct answer)
  3. 25−4i25-4i
  4. 25+425+4
Explanation: This question tests your understanding of complex conjugates—pairs of complex numbers that differ only in the sign of their imaginary part—and the special property that occurs when multiplying a complex number by its conjugate. The conjugate of a + bi is a - bi (flip only the sign of the imaginary part, keep real part the same); 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 compute (5 + 2i)(5 - 2i), we can use the formula directly: a = 5 and b = 2, so the product equals 5² + 2² = 25 + 4 = 29; alternatively, we can multiply it out: (5 + 2i)(5 - 2i) = 25 - 10i + 10i - 4i² = 25 - 4(-1) = 25 + 4 = 29. Choice B correctly applies the conjugate multiplication formula to get the real positive result 29. Choice C shows 25 - 4i, which incorrectly treats this as regular multiplication without recognizing that the imaginary parts cancel completely when multiplying conjugates—the result must be a real number with no imaginary part! The (a + bi)(a - bi) = a² + b² pattern is worth memorizing: it's how we eliminate i from denominators in division problems, and it equals the modulus squared |z|² of the complex number. This property has a beautiful geometric interpretation: if z represents a point in the complex plane, then |z|² = z·z̄ represents the square of its distance from the origin!

Question 16

Compute (2−i3+2i)‾\overline{\left(\dfrac{2-i}{3+2i}\right)}

  1. 2−i3−2i\dfrac{2-i}{3-2i}
  2. 2+i3−2i\dfrac{2+i}{3-2i} (correct answer)
  3. −2+i−3−2i\dfrac{-2+i}{-3-2i}
  4. 2+i3+2i\dfrac{2+i}{3+2i}
Explanation: 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: (z/w)‾=z‾/w‾\overline{(z/w)} = \overline{z} / \overline{w}. To compute ((2−i)/(3+2i))‾\overline{((2 - i)/(3 + 2i))}, first simplify the division or use the property directly: conjugate of numerator 2−i2 - i is 2+i2 + i, conjugate of denominator 3+2i3 + 2i is 3−2i3 - 2i, so z/w‾=(2+i)/(3−2i)\overline{z/w} = (2 + i)/(3 - 2i). Choice B correctly applies this property, giving (2+i)/(3−2i)(2 + i)/(3 - 2i). A tempting distractor like Choice D might flip the sign in the denominator incorrectly to 3+2i3 + 2i instead of 3−2i3 - 2i, but remember to flip only the imaginary sign for each conjugate. The strategy is to handle conjugates of operations: for products, zw‾=z‾w‾\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!

Question 17

A complex number is z=6−iz=6- i. Which expression equals ∣z∣2|z|^2 (the modulus squared) using conjugates?​

  1. z‾\overline{z}
  2. z+z‾z+\overline{z}
  3. zz‾z\overline{z} (correct answer)
  4. z−z‾z-\overline{z}
Explanation: This question tests knowing that |z|² = z \overline{z}. Key concept: modulus squared is a² + b², which is z times conjugate. For example, for i, |i|²=1, i * (-i) = -i² =1. Here for 6 - i, it's (6 - i)(6 + i)=36 + 1=37. Choice C correctly selects z \overline{z}. Choice B is z + \overline{z} = 2 Re(z)=12, not squared. Remember |z|² is always positive real. Use this in polar forms too. Super!

Question 18

Use conjugates to rationalize the denominator and write in a+bia+bi form: 2−i3+2i\frac{2-i}{3+2i}

  1. 413−713i\frac{4}{13}-\frac{7}{13}i (correct answer)
  2. 413+713i\frac{4}{13}+\frac{7}{13}i
  3. 713−413i\frac{7}{13}-\frac{4}{13}i
  4. 2−i3−2i\frac{2-i}{3-2i}
Explanation: 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. To divide (2 - i) by (3 + 2i) using conjugates: (1) Multiply numerator and denominator by conjugate of denominator 3 - 2i: [(2 - i)(3 - 2i)] / [(3 + 2i)(3 - 2i)]. (2) Multiply numerator using FOIL: (2 - i)(3 - 2i) = 6 - 4i - 3i + 2i² = 6 - 7i - 2 = 4 - 7i. (3) Multiply denominator using conjugate property: (3 + 2i)(3 - 2i) = 3² + 2² = 9 + 4 = 13. (4) Divide: (4 - 7i)/13 = 4/13 - 7i/13. Choice A correctly computes this as 4/13 - 7i/13 in standard a + bi form. Choice B has the wrong sign on the imaginary part: when we computed (2 - i)(3 - 2i), the middle terms give us -4i - 3i = -7i, not +7i—be careful with signs when multiplying! Division recipe using conjugates: multiply both numerator and denominator by the conjugate of the denominator, simplify the numerator (stays complex), simplify the denominator (becomes real), then divide both parts.

Question 19

Let z=2−5iz=2-5i. What is z‾‾\overline{\overline{z}}? (Recall that taking the conjugate twice returns the original complex number.)

  1. 2+5i2+5i
  2. −2−5i-2-5i
  3. 2−5i2-5i (correct answer)
  4. −2+5i-2+5i
Explanation: This question tests your understanding of complex conjugates—pairs of complex numbers that differ only in the sign of their imaginary part—and the important property that taking the conjugate twice returns the original number. The conjugate of a + bi is a - bi (flip only the sign of the imaginary part, keep real part the same); if you apply the conjugate operation twice, you flip the sign and then flip it back, returning to where you started! For z = 2 - 5i, first find z̄: the conjugate of 2 - 5i is 2 + 5i (keep 2, change -5i to +5i). Then find the conjugate of z̄: the conjugate of 2 + 5i is 2 - 5i (keep 2, change +5i back to -5i). So z̄̄ = 2 - 5i = z, confirming the conjugate operation is its own inverse! Choice C correctly shows 2 - 5i, which equals the original z. Choice A shows 2 + 5i, which is z̄ (the conjugate applied once), not z̄̄ (the conjugate applied twice)—remember we need to apply the conjugate operation twice, not just once! The property z̄̄ = z (conjugate of conjugate equals original) is fundamental: it shows conjugation is an "involution" (self-inverse operation). Geometrically, this makes sense: if conjugate reflects across the real axis, doing it twice brings you back to the starting point—like flipping a pancake twice returns it to its original orientation!

Question 20

Simplify using the conjugate multiplication property (a+bi)(a−bi)=a2+b2(a+bi)(a-bi)=a^2+b^2: (6−i)(6+i).(6- i)(6+i).

  1. 3737 (correct answer)
  2. 3535
  3. 36+136+1
  4. 36−1i36-1i
Explanation: This question tests your understanding of complex conjugates—pairs of complex numbers that differ only in the sign of their imaginary part—and the special property that multiplying conjugates gives a real result. The conjugate of a + bi is a - bi (flip only the sign of the imaginary part, keep real part the same); 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!). For (6 - i)(6 + i), we recognize these as conjugates of each other, so we can apply the formula directly: a = 6 and b = 1 (since 6 - i = 6 - 1i), giving us 6² + 1² = 36 + 1 = 37. Alternatively, multiply it out: (6 - i)(6 + i) = 36 + 6i - 6i - i² = 36 - (-1) = 36 + 1 = 37. Choice B correctly identifies 37 as the product of these conjugates. Choice A shows 35, which might come from incorrectly computing 6² - 1² = 36 - 1 = 35—remember the formula is a² + b², not a² - b²! The sum of squares, not difference, because when you expand (a + bi)(a - bi), the cross terms cancel and i² = -1 makes the last term positive. The (a + bi)(a - bi) = a² + b² pattern appears everywhere in complex number theory: it's the basis for rationalizing denominators, computing moduli, and even in the proof that every polynomial with real coefficients can be factored into linear and quadratic factors with real coefficients!