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

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

6−2i6-2i
−6−2i-6-2i
−6+2i-6+2i
6+2i6+2i
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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.

How to use this quiz

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.

All questions

Question 1

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 2

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!

Question 3

Let z=3−4iz = 3 - 4i. The complex conjugate of zz, written z‾\overline{z}, is found by flipping the sign of the imaginary part only. What is z‾\overline{z}?

  1. 3+4i3 + 4i (correct answer)
  2. −3−4i-3 - 4i
  3. 3−4i3 - 4i
  4. −3+4i-3 + 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 find the conjugate of a given complex number. 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 z = 3 - 4i, we identify the real part as 3 and the imaginary part as -4i; to find the conjugate, we keep the real part 3 unchanged and flip the sign of the imaginary part from -4i to +4i, giving us 3 + 4i. Choice B correctly identifies the conjugate by keeping the real part 3 the same and changing only the sign of the imaginary part from negative to positive. Choice C incorrectly flips the sign of BOTH parts: the conjugate of 3 - 4i is 3 + 4i, NOT -3 - 4i—remember to only flip the imaginary part's sign! Conjugate quick reference: to find conjugate of a + bi, keep a the same, flip the sign of bi to get a - bi; visually on the complex plane, the conjugate is the mirror image across the horizontal (real) axis. The conjugate operation is its own inverse: if you take the conjugate twice, you get back to the original number, which makes sense since flipping a sign twice returns it to its original state!

Question 4

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 5

For which value of kk will 3+ki1+2i\frac{3 + ki}{1 + 2i} result in a purely real number?

  1. k=6k = 6 (correct answer)
  2. k=−6k = -6
  3. k=32k = \frac{3}{2}
  4. k=−32k = -\frac{3}{2}
Explanation: To find when the quotient is purely real, multiply by the conjugate: 3+ki1+2i⋅1−2i1−2i=(3+ki)(1−2i)(1+2i)(1−2i)=3−6i+ki−2ki21−4i2=3−6i+ki+2k1+4=(3+2k)+(k−6)i5\frac{3 + ki}{1 + 2i} \cdot \frac{1 - 2i}{1 - 2i} = \frac{(3 + ki)(1 - 2i)}{(1 + 2i)(1 - 2i)} = \frac{3 - 6i + ki - 2ki^2}{1 - 4i^2} = \frac{3 - 6i + ki + 2k}{1 + 4} = \frac{(3 + 2k) + (k - 6)i}{5}. For this to be purely real, the imaginary part must be zero: k−6=0k - 6 = 0, so k=6k = 6. Choice B gives the negative of the correct answer. Choice C results from setting the real part equal to zero instead. Choice D combines both errors.

Question 6

If z=a+biz = a + bi where aa and bb are real, and z⋅z‾=25z \cdot \overline{z} = 25, and z+z‾=6z + \overline{z} = 6, what are the possible values of zz?

  1. 3+4i3 + 4i only
  2. 3−4i3 - 4i only
  3. 3+4i3 + 4i or 3−4i3 - 4i (correct answer)
  4. 4+3i4 + 3i or 4−3i4 - 3i
Explanation: From z⋅z‾=25z \cdot \overline{z} = 25, we get ∣z∣2=a2+b2=25|z|^2 = a^2 + b^2 = 25. From z+z‾=6z + \overline{z} = 6, we get (a+bi)+(a−bi)=2a=6(a + bi) + (a - bi) = 2a = 6, so a=3a = 3. Substituting into the first equation: 32+b2=253^2 + b^2 = 25, so 9+b2=259 + b^2 = 25, giving b2=16b^2 = 16 and b=±4b = \pm 4. Therefore both z=3+4iz = 3 + 4i and z=3−4iz = 3 - 4i satisfy the conditions. Choice D has the values of aa and bb swapped.

Question 7

If z=−2+5iz = -2 + 5i, what is ∣z∣2|z|^2 calculated using the conjugate method?

  1. 2929 (correct answer)
  2. 29\sqrt{29}
  3. 2121
  4. 33
Explanation: The modulus squared of a complex number can be found by multiplying the number by its conjugate: ∣z∣2=z⋅z‾|z|^2 = z \cdot \overline{z}. The conjugate of z=−2+5iz = -2 + 5i is z‾=−2−5i\overline{z} = -2 - 5i. Therefore: ∣z∣2=(−2+5i)(−2−5i)=(−2)2−(5i)2=4−25i2=4−25(−1)=4+25=29|z|^2 = (-2 + 5i)(-2 - 5i) = (-2)^2 - (5i)^2 = 4 - 25i^2 = 4 - 25(-1) = 4 + 25 = 29. Choice B gives the modulus itself, not its square. Choice C results from incorrectly computing (−2)2+52−2(−2)(5)=4+25−(−20)=49(-2)^2 + 5^2 - 2(-2)(5) = 4 + 25 - (-20) = 49, then subtracting something. Choice D results from 5−2=35 - 2 = 3, a completely incorrect approach.

Question 8

In an AC circuit, the impedance is Z=4+3iZ=4+3i ohms. A common calculation uses ZZ‾Z\overline{Z}, which equals the modulus squared. Compute ZZ‾Z\overline{Z}.

  1. 2525 (correct answer)
  2. 77
  3. 11
  4. 16+9i16+9i
Explanation: This question tests your understanding of complex conjugates—pairs of complex numbers that differ only in the sign of their imaginary part—and applying z \overline{z} as the modulus squared in contexts like impedance. The conjugate of a + bi is a - bi; z \overline{z} = a² + b², a real number representing |z|². For Z=4+3i, conjugate 4-3i; product 4² + 3²=16+9=25. In AC circuits, this relates to power calculations, but the math is the same. Choice C correctly computes 25. Choice D leaves it as 16+9i, perhaps forgetting to multiply or apply the formula—remember, the product is real! Use the quick formula a² + b² for any z \overline{z}. Try with 3+4i: 9+16=25, which is |z|²=5²— you're mastering this!

Question 9

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. 413+713i\frac{4}{13}+\frac{7}{13}i
  3. 713−413i\frac{7}{13}-\frac{4}{13}i
  4. 4−7i3+2i\frac{4-7i}{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 the denominator in division, writing the result in standard a+bia + bi form. For (2−i)/(3+2i)(2 - i)/(3 + 2i), conjugate of denominator is 3−2i3 - 2i; multiply: numerator (2−i)(3−2i)=6−4i−3i+2i2=6−7i+2(−1)=6−7i−2=4−7i(2 - i)(3 - 2i) = 6 - 4i - 3i + 2i^2 = 6 - 7i + 2(-1) = 6 - 7i - 2 = 4 - 7i; denominator (3+2i)(3−2i)=9+4=13(3 + 2i)(3 - 2i) = 9 + 4 = 13; so (4−7i)/13=4/13−7/13i(4 - 7i)/13 = 4/13 - 7/13 i. Choice A correctly performs this calculation, arriving at 4/13−7/13i4/13 - 7/13 i. A tempting distractor like Choice B might get the imaginary sign wrong, perhaps by not flipping correctly in multiplication, but double-check FOIL and signs: real parts 2∗3+(−i)(−2i)=6+2i2=6−2=42*3 + (-i)(-2i) = 6 + 2i^2 = 6 - 2 = 4, imaginary −4i−3i=−7i-4i -3i = -7i. Always divide both real and imaginary parts by the real denominator after rationalizing. Great effort—practice with (1+i)/(1−i)(1 + i)/(1 - i) to get (1+i)2/(1+1)=(1+2i−1)/2=2i/2=i(1 + i)^2 / (1 + 1) = (1 + 2i - 1)/2 = 2i/2 = i, building confidence!

Question 10

Compute the product of a complex number and its conjugate, and note it equals the modulus squared. Find (5−4i)(5+4i)(5-4i)(5+4i).

  1. 4141 (correct answer)
  2. 25−1625-16
  3. 99
  4. 41i41i
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!). For (5 - 4i)(5 + 4i), notice these are conjugates of each other, so we apply the formula: a = 5 and b = 4 (taking the absolute value), giving us 5² + 4² = 25 + 16 = 41. Choice C correctly computes 41 using the conjugate property formula. Choice B shows the intermediate step 25 - 16 but incorrectly subtracts instead of adding: the formula is a² + b², not a² - b². The (a + bi)(a - bi) = a² + b² pattern always adds the squares! This real result equals the modulus squared of the complex number, which is why conjugates are perfect for eliminating imaginary parts from denominators.

Question 11

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 12

Let z=1−2iz=1-2i and w=3+iw=3+i. Compute zw‾\overline{zw}. (You may use zw‾=z‾ w‾\overline{zw}=\overline{z}\,\overline{w}.)

  1. 5−5i5-5i
  2. 5+5i5+5i (correct answer)
  3. −5+5i-5+5i
  4. −5−5i-5-5i
Explanation: This question tests your understanding of complex conjugates—pairs of complex numbers that differ only in the sign of their imaginary part—and how to use them to divide complex numbers by rationalizing denominators. The property states that the conjugate of a product equals the product of the conjugates: conjugate of (zw) = (conjugate of z)(conjugate of w). First, let's find zw: (1 - 2i)(3 + i) = 3 + i - 6i - 2i² = 3 - 5i + 2 = 5 - 5i (using i² = -1). The conjugate of (5 - 5i) is 5 + 5i, flipping the sign of the imaginary part from -5i to +5i. Choice B correctly identifies this as 5 + 5i. Choice A incorrectly keeps the imaginary part negative: the conjugate of 5 - 5i is 5 + 5i, not 5 - 5i—you must flip the sign! We can verify using the property: conjugate of z = 1 + 2i, conjugate of w = 3 - i, so their product is (1 + 2i)(3 - i) = 3 - i + 6i - 2i² = 3 + 5i + 2 = 5 + 5i, confirming our answer. This property is very useful when dealing with products of complex numbers and extends to any finite product!

Question 13

Compute the product of a complex number and its conjugate: (4+3i)(4−3i)(4 + 3i)(4 - 3i). (Recall (a+bi)(a−bi)=a2+b2(a+bi)(a-bi)=a^2+b^2, which is real and equals ∣a+bi∣2|a+bi|^2.)

  1. 77
  2. 16−9i216 - 9i^2
  3. 2525 (correct answer)
  4. 16−916 - 9
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 when multiplying them together. 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. 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 (4 + 3i)(4 - 3i), we can use the formula directly: a = 4 and b = 3, so the product equals 4² + 3² = 16 + 9 = 25. Choice C correctly gives 25 as the result of this conjugate multiplication. Choice B shows 16 - 9i² but fails to simplify: since i² = -1, this becomes 16 - 9(-1) = 16 + 9 = 25, so it's an incomplete answer that hasn't been fully evaluated. The (a + bi)(a - bi) = a² + b² pattern is worth memorizing: it always gives a real positive number equal to the modulus squared |a + bi|²!

Question 14

In AC circuits, an impedance can be written as Z=4+3iZ=4+3i ohms. Compute ZZ‾Z \overline{Z}.

  1. 7i7i
  2. 16+9i16+9i
  3. 2525 (correct answer)
  4. 16−916-9
Explanation: This question tests your understanding of complex conjugates—pairs of complex numbers that differ only in the sign of their imaginary part—and applying the property ZZ‾=∣Z∣2Z \overline{Z} = |Z|^2, which is a real number representing the square of the modulus in contexts like AC circuits. For Z=4+3iZ = 4 + 3i, the conjugate Z‾=4−3i\overline{Z} = 4 - 3i, and ZZ‾=(4+3i)(4−3i)=16+9=25Z \overline{Z} = (4 + 3i)(4 - 3i) = 16 + 9 = 25, using (a+bi)(a−bi)=a2+b2(a + bi)(a - bi) = a^2 + b^2. Choice C correctly computes this as 25, recognizing it's the modulus squared. A tempting distractor like Choice D might forget to simplify with i2=−1i^2 = -1 and stop at 16 - 9, but always remember the conjugate product eliminates imaginaries to give a real result. Use this pattern in physics or engineering: ∣Z∣=a2+b2|Z| = \sqrt{a^2 + b^2}, so ∣Z∣2=a2+b2|Z|^2 = a^2 + b^2 directly. You're making fantastic progress—try it with Z=5+12iZ = 5 + 12i to get 25+144=16925 + 144 = 169!

Question 15

Which of the following expressions is equivalent to (2+i1−i)‾\overline{\left(\frac{2 + i}{1 - i}\right)}?

  1. 2+i‾1−i‾\frac{\overline{2 + i}}{\overline{1 - i}}
  2. 2−i1+i\frac{2 - i}{1 + i} (correct answer)
  3. 1+i2−i\frac{1 + i}{2 - i}
  4. 1−i‾2+i‾\frac{\overline{1 - i}}{\overline{2 + i}}
Explanation: Using the property (zw)‾=z‾w‾\overline{\left(\frac{z}{w}\right)} = \frac{\overline{z}}{\overline{w}}, we have (2+i1−i)‾=2+i‾1−i‾=2−i1+i\overline{\left(\frac{2 + i}{1 - i}\right)} = \frac{\overline{2 + i}}{\overline{1 - i}} = \frac{2 - i}{1 + i}. Choice A shows the property but isn't simplified to have actual conjugates computed. Choice C inverts the fraction incorrectly. Choice D has the conjugates in the wrong positions (numerator and denominator swapped).

Question 16

Let z=3+4iz = 3 + 4i. Using the fact that (a+bi)(a−bi)=a2+b2(a+bi)(a-bi)=a^2+b^2 (which equals ∣z∣2|z|^2), what is zz‾z\overline{z}?​

  1. 2525 (correct answer)
  2. 3−4i3-4i
  3. 9+16i9+16i
  4. 77
Explanation: This question tests your understanding of complex conjugates and how their product with the complex number gives the modulus squared. The conjugate of a + bi is a - bi, and the magic property is that (a + bi)(a - bi) = a² + b², a real positive number, as the imaginary parts cancel. For example, (3 + 4i)(3 - 4i) = 9 + 16 = 25. This is exactly |z|². To compute z\overline{z} for z = 3 + 4i, identify \overline{z} = 3 - 4i, then multiply to get 25. Choice B correctly applies this property to get 25. Choice A might come from a² - b² = 9 - 16 = -7, but remember it's plus b² since i² = -1 turns the negative to positive. The pattern (a + bi)(a - bi) = a² + b² is worth memorizing! It's the sum of squares. Always check if the result is real; if not, you probably didn't use the conjugate correctly. Keep up the good work!

Question 17

If w=3−2i2+iw = \frac{3 - 2i}{2 + i}, what is the value of w+w‾w + \overline{w}?

  1. 22
  2. 65\frac{6}{5}
  3. 45\frac{4}{5}
  4. 85\frac{8}{5} (correct answer)
Explanation: When working with complex numbers in fraction form, you'll often need to find the conjugate and perform operations. The key insight here is that w+w‾w + \overline{w} always gives you twice the real part of ww, so you need to find the real part of this complex fraction. To find ww, multiply both numerator and denominator by the conjugate of the denominator. The conjugate of 2+i2 + i is 2−i2 - i: w=3−2i2+i⋅2−i2−i=(3−2i)(2−i)(2+i)(2−i)w = \frac{3 - 2i}{2 + i} \cdot \frac{2 - i}{2 - i} = \frac{(3 - 2i)(2 - i)}{(2 + i)(2 - i)} The denominator becomes: (2+i)(2−i)=4−i2=4−(−1)=5(2 + i)(2 - i) = 4 - i^2 = 4 - (-1) = 5 The numerator becomes: (3−2i)(2−i)=6−3i−4i+2i2=6−7i−2=4−7i(3 - 2i)(2 - i) = 6 - 3i - 4i + 2i^2 = 6 - 7i - 2 = 4 - 7i Therefore: w=4−7i5=45−75iw = \frac{4 - 7i}{5} = \frac{4}{5} - \frac{7}{5}i The conjugate is: w‾=45+75i\overline{w} = \frac{4}{5} + \frac{7}{5}i So: w+w‾=45−75i+45+75i=85w + \overline{w} = \frac{4}{5} - \frac{7}{5}i + \frac{4}{5} + \frac{7}{5}i = \frac{8}{5} Choice A (22) might come from incorrectly simplifying the original fraction. Choice B (65\frac{6}{5}) could result from using only the real part of the numerator before rationalization. Choice C (45\frac{4}{5}) is just the real part of ww, forgetting to double it when adding the conjugate. Remember: z+z‾=2⋅Re(z)z + \overline{z} = 2 \cdot \text{Re}(z). This shortcut can save time once you find the real part of any complex number.

Question 18

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 19

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 20

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!