Algebra 2 Quiz: Operations With Complex Numbers
20 questions · exam conditions
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Operations With Complex NumbersQuestion 1 of 20

Let z1=6−2iz_1 = 6 - 2i and z2=1+5iz_2 = 1 + 5i. What is z1−z2z_1 - z_2 expressed in standard form a+bia + bi?

7−5i7 - 5i
5−7i5 - 7i
5+3i5 + 3i
−5−7i-5 - 7i
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Algebra 2 Quiz

Algebra 2 Quiz: Operations With Complex Numbers

Practice Operations 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 Operations 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 z1=6−2iz_1 = 6 - 2i and z2=1+5iz_2 = 1 + 5i. What is z1−z2z_1 - z_2 expressed in standard form a+bia + bi?

  1. 7−5i7 - 5i
  2. 5−7i5 - 7i (correct answer)
  3. 5+3i5 + 3i
  4. −5−7i-5 - 7i
Explanation: This question tests your understanding of complex numbers—numbers in form a + bi where i is the imaginary unit with i squared = -1—and how to perform operations (add, subtract, multiply, divide) while keeping results in standard a + bi form. Complex numbers extend the real number system to include square roots of negative numbers using i = square root of -1, so i squared = -1. The standard form a + bi has a as the real part and b as the coefficient of the imaginary part (bi). Operations work like algebra with the crucial rule: whenever you see i squared, replace it with -1! For addition and subtraction, combine real parts together and imaginary parts together: (3 + 2i) - (1 - 5i) = (3 - 1) + (2 - (-5))i = 2 + 7i. To subtract z₂ = 1 + 5i from z₁ = 6 - 2i: subtract real parts: 6 - 1 = 5, subtract imaginary parts: -2i - 5i = -7i, so z₁ - z₂ = 5 - 7i. Choice A correctly performs the complex number subtraction and expresses the result in standard form a + bi with proper handling of subtracting both parts. Choice C makes a sign error when subtracting the imaginary parts: -2i - 5i = -7i, not -5i—when subtracting positive 5i from negative 2i, you get more negative! Complex number operations summary: ADDITION/SUBTRACTION—combine real parts, combine imaginary parts: (a + bi) plus or minus (c + di) = (a plus or minus c) + (b plus or minus d)i. These patterns are consistent! Remember that subtraction distributes the negative sign to both parts of the second complex number: (6 - 2i) - (1 + 5i) = 6 - 2i - 1 - 5i.

Question 2

Using the fact that (a+bi)(a−bi)=a2+b2(a+bi)(a-bi)=a^2+b^2, compute (2+5i)(2−5i)(2+5i)(2-5i).

  1. −21-21
  2. 2929 (correct answer)
  3. 4+25i4+25i
  4. 4−25i4-25i
Explanation: This question tests your understanding of complex numbers—numbers in form a+bia + bi where ii is the imaginary unit with i2=−1i^2 = -1—and how to compute the product with a conjugate using the difference of squares. Complex numbers extend the real number system to include square roots of negative numbers using i=−1i = \sqrt{-1}, so i2=−1i^2 = -1. The formula (a+bi)(a−bi)=a2+b2(a + bi)(a - bi) = a^2 + b^2 eliminates the imaginary part; for example, (3+2i)(3−2i)=9+4=13(3 + 2i)(3 - 2i) = 9 + 4 = 13. For (2+5i)(2−5i)(2 + 5i)(2 - 5i), it's 4+25=294 + 25 = 29, since (5i)2=25i2=25×(−1)=−25(5i)^2 = 25i^2 = 25 \times (-1) = -25, so 4−(−25)=294 - (-25) = 29. Choice B correctly computes the product using the conjugate formula and simplifies with i2=−1i^2 = -1. Choice A forgets to replace i2i^2 with -1: treating (5i)2(5i)^2 as +25 instead of -25 leads to 4+(−25)4 + (-25) wait no, but actually misapplying as 4−25=−214 - 25 = -21—the key is the difference of squares gives positive! This property is key for moduli and division: ∣z∣2=z×conjugate(z)|z|^2 = z \times \text{conjugate}(z)—you're grasping these concepts beautifully, keep it up!

Question 3

Subtract and express in standard form a+bia+bi: (6−4i)−(1+9i)(6-4i)-(1+9i).

  1. 5−13i5-13i (correct answer)
  2. 5+5i5+5i
  3. 7−13i7-13i
  4. −5−13i-5-13i
Explanation: This question tests your understanding of complex numbers—numbers in the form a+bia + bi where ii is the imaginary unit with i2=−1i^2 = -1—and how to perform subtraction while expressing results in standard a+bia + bi form. Complex numbers extend the real number system to include square roots of negative numbers using i=−1i = \sqrt{-1}, so i2=−1i^2 = -1. The standard form a+bia + bi has aa as the real part and bb as the coefficient of the imaginary part (bibi). For subtraction, distribute the negative: (6−4i)−(1+9i)=6−4i−1−9i(6 - 4i) - (1 + 9i) = 6 - 4i - 1 - 9i, then combine reals 6−1=56 - 1 = 5 and imaginaries −4−9=−13-4 - 9 = -13, so 5−13i5 - 13i—excellent work! A tempting distractor like choice D might forget to distribute the negative to both parts, but always apply the minus to real and imaginary alike. Remember this strategy: subtraction is just addition of the opposite—change signs and add! With practice, these operations will feel natural, and you're doing great.

Question 4

If z=a+biz = a + bi where aa and bb are real numbers, and z2=−5+12iz^2 = -5 + 12i, what is the value of a2+b2a^2 + b^2?

  1. 1313 (correct answer)
  2. 1717
  3. 2525
  4. 169169
Explanation: We have z2=(a+bi)2=a2+2abi+b2i2=a2−b2+2abi=−5+12iz^2 = (a+bi)^2 = a^2 + 2abi + b^2i^2 = a^2 - b^2 + 2abi = -5 + 12i. Equating real and imaginary parts: a2−b2=−5a^2 - b^2 = -5 and 2ab=122ab = 12, so ab=6ab = 6. We want to find a2+b2a^2 + b^2. Note that ∣z∣2=a2+b2|z|^2 = a^2 + b^2 and ∣z2∣=∣z∣2|z^2| = |z|^2. Since z2=−5+12iz^2 = -5 + 12i, we have ∣z2∣=(−5)2+122=25+144=169=13|z^2| = \sqrt{(-5)^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13. But this gives us ∣z∣2=13|z|^2 = \sqrt{13}, which seems wrong. Actually, ∣z2∣=∣z∣2|z^2| = |z|^2 is incorrect; we should have ∣z2∣=∣z∣2|z^2| = |z|^2. Let me reconsider: if z2=−5+12iz^2 = -5 + 12i, then ∣z2∣=25+144=13|z^2| = \sqrt{25 + 144} = 13. Since ∣z2∣=∣z∣2|z^2| = |z|^2, we have ∣z∣2=13|z|^2 = \sqrt{13}. No wait, ∣z2∣=(∣z∣)2|z^2| = (|z|)^2, so (∣z∣)2=13(|z|)^2 = 13, which means ∣z∣=13|z| = \sqrt{13}. Therefore, a2+b2=∣z∣2=13a^2 + b^2 = |z|^2 = 13. Choice B (17) might come from (−5)2−12=25−12=13(-5)^2 - 12 = 25 - 12 = 13, no that gives 13. Choice C (25) is (−5)2(-5)^2. Choice D (169) is (13)2(13)^2.

Question 5

Find the modulus (absolute value) of the complex number 3+4i3 + 4i. (Recall ∣a+bi∣=a2+b2|a+bi|=\sqrt{a^2+b^2}.)

  1. 5\sqrt{5}
  2. 55 (correct answer)
  3. 77
  4. 7\sqrt{7}
Explanation: This question tests your understanding of complex numbers—numbers in the form a+bia + bi where i is the imaginary unit with i2=−1i^2 = -1—and finding the modulus. Complex numbers extend the real number system to include square roots of negative numbers using i=−1i = \sqrt{-1}, so i2=−1i^2 = -1. The modulus ∣a+bi∣=a2+b2|a + bi| = \sqrt{a^2 + b^2}, like distance from origin. For 3+4i3 + 4i, ∣3+4i∣=9+16=25=5|3 + 4i| = \sqrt{9 + 16} = \sqrt{25} = 5—recognize the 3-4-5 triangle! Choice B correctly computes this. A tempting distractor like Choice C might add without squaring, getting 7. Modulus is always non-negative—keep going, you're doing great!

Question 6

Solve the quadratic equation x2+2x+10=0x^2 + 2x + 10 = 0 and express the solutions as complex numbers (standard form a+bia + bi).

  1. x=−1±3x = -1 \pm 3
  2. x=−1±3ix = -1 \pm 3i (correct answer)
  3. x=1±3ix = 1 \pm 3i
  4. x=−1±6 ix = -1 \pm \sqrt{6}\,i
Explanation: This question tests your understanding of complex numbers—numbers in the form a + bi where i is the imaginary unit with i squared = -1—and solving quadratics with complex roots. Complex numbers extend the real number system to include square roots of negative numbers using i = square root of -1, so i squared = -1. For x² + 2x + 10 = 0, discriminant = 4 - 40 = -36, sqrt(-36) = 6i; solutions x = [-2 ± 6i]/2 = -1 ± 3i. Choice A correctly expresses the roots in a + bi form. A tempting distractor like Choice C might use sqrt(6) instead of simplifying sqrt(36)=6. Complex roots come in conjugate pairs, so this makes sense—fantastic effort solving!

Question 7

Let z1=5+3iz_1 = 5 + 3i and z2=2−7iz_2 = 2 - 7i. What is z1+z2z_1 + z_2 in standard form a+bia+bi?

  1. 3−4i3-4i
  2. 7+10i7+10i
  3. 7−4i7-4i (correct answer)
  4. 3+10i3+10i
Explanation: This question tests your understanding of complex numbers—numbers in form a + bi where i is the imaginary unit with i squared = -1—and how to perform addition while keeping results in standard a + bi form. Complex numbers extend the real number system to include square roots of negative numbers using i = square root of -1, so i squared = -1. The standard form a + bi has a as the real part and b as the coefficient of the imaginary part (bi). For addition, combine real parts together and imaginary parts together: (5 + 3i) + (2 - 7i) = (5 + 2) + (3 - 7)i = 7 - 4i. Choice A correctly performs the complex number addition and expresses the result in standard form a + bi with proper combination of like terms. Choice C makes a sign error when combining imaginary parts: adding 3 + 7 instead of 3 - 7, but remember the sign of the imaginary coefficient in z2 is negative! Complex number addition summary: (a + bi) + (c + di) = (a + c) + (b + d)i—it's just like combining like terms in algebra, and you're doing great mastering this foundational skill!

Question 8

Multiply and simplify using i2=−1i^2=-1: (3+2i)(1−4i).(3+2i)(1-4i).

  1. 11−10i11-10i (correct answer)
  2. −5−10i-5-10i
  3. 11+10i11+10i
  4. 3−10i3-10i
Explanation: This question tests your understanding of complex numbers—numbers in the form a + bi where i is the imaginary unit with i2=−1i^2 = -1—and how to perform multiplication while simplifying using i2=−1i^2 = -1. Complex numbers extend the real number system to include square roots of negative numbers using i=−1i = \sqrt{-1}, so i2=−1i^2 = -1. Operations work like algebra with the crucial rule: whenever you see i2i^2, replace it with −1-1! For multiplication, use FOIL: for (3+2i)(1−4i)(3 + 2i)(1 - 4i), First: 3×1=33 \times 1 = 3, Outer: 3×(−4i)=−12i3 \times (-4i) = -12i, Inner: 2i×1=2i2i \times 1 = 2i, Last: 2i×(−4i)=−8i2=82i \times (-4i) = -8i^2 = 8 (since −8×−1=8-8 \times -1 = 8), combining to 3+8+(−12i+2i)=11−10i3 + 8 + (-12i + 2i) = 11 - 10i—way to go! Choice B might result from forgetting to replace i2i^2 with −1-1, leaving it as −8i2-8i^2 and not simplifying, but always make that substitution! Multiplication summary: distribute fully, then simplify i powers—it's consistent and powerful. You're progressing wonderfully toward fluency in complex numbers!

Question 9

Find the modulus (absolute value) of the complex number 3+4i3+4i: ∣3+4i∣.|3+4i|.

  1. 77
  2. 55 (correct answer)
  3. 7\sqrt{7}
  4. 5\sqrt{5}
Explanation: This question tests your understanding of complex numbers—numbers in the form a + bi where i is the imaginary unit with i2=−1i^2 = -1—and finding the modulus, which is like the distance from origin in the complex plane. Complex numbers extend the real number system to include square roots of negative numbers using i=−1i = \sqrt{-1}, so i2=−1i^2 = -1. The modulus ∣a+bi∣=a2+b2|a + bi| = \sqrt{a^2 + b^2}: for 3+4i3 + 4i, 9+16=25=5\sqrt{9 + 16} = \sqrt{25} = 5—perfect! Choice B correctly computes this Pythagorean-style value. Choice A might add without squaring, but always square and add inside the root. This concept ties into multiplication properties too—great job! You're strengthening your skills step by step.

Question 10

Multiply and simplify using i2=−1i^2 = -1: (3+2i)(1−4i)(3 + 2i)(1 - 4i). Express your answer in standard form a+bia + bi.

  1. 11+10i11 + 10i
  2. 11−10i11 - 10i (correct answer)
  3. 3−10i3 - 10i
  4. −5−10i-5 - 10i
Explanation: This question tests your understanding of complex numbers—numbers in form a + bi where i is the imaginary unit with i squared = -1—and how to perform operations (add, subtract, multiply, divide) while keeping results in standard a + bi form. Complex numbers extend the real number system to include square roots of negative numbers using i = square root of -1, so i squared = -1. The standard form a + bi has a as the real part and b as the coefficient of the imaginary part (bi). Operations work like algebra with the crucial rule: whenever you see i squared, replace it with -1! For multiplication, use FOIL or distribution, then simplify any i squared terms to -1. To multiply (3 + 2i)(1 - 4i): use FOIL: First: 3 × 1 = 3. Outer: 3 × (-4i) = -12i. Inner: 2i × 1 = 2i. Last: 2i × (-4i) = -8i². Combine: 3 - 12i + 2i - 8i² = 3 - 10i - 8i². Now replace i² with -1: 3 - 10i - 8(-1) = 3 - 10i + 8 = 11 - 10i. The key is replacing i squared with -1 at the end! Choice A correctly performs the complex number multiplication and expresses the result in standard form a + bi with proper handling of i squared = -1. Choice B forgets to replace i squared with -1 during multiplication: when you get terms like 2i × (-4i) = -8i squared, you must simplify to -8(-1) = +8, not leave as -8i squared or treat as -8. The i squared simplification is mandatory—it's not optional algebra! Complex number operations summary: MULTIPLICATION—use FOIL or distribution, then simplify i squared to -1: watch for i squared terms and replace immediately! These patterns are consistent! The i squared = -1 rule is your constant companion in complex multiplication.

Question 11

Let z1=5+3iz_1 = 5 + 3i and z2=2−7iz_2 = 2 - 7i, where i2=−1i^2 = -1. What is z1+z2z_1 + z_2 in standard form a+bia + bi?

  1. 7−4i7 - 4i (correct answer)
  2. 3−10i3 - 10i
  3. 7+10i7 + 10i
  4. 10−5i10 - 5i
Explanation: This question tests your understanding of complex numbers—numbers in form a + bi where i is the imaginary unit with i squared = -1—and how to perform addition while keeping results in standard a + bi form. Complex numbers extend the real number system to include square roots of negative numbers using i = square root of -1, so i squared = -1. The standard form a + bi has a as the real part and b as the coefficient of the imaginary part (bi). For addition, combine real parts together and imaginary parts together: (5 + 3i) + (2 - 7i) = (5 + 2) + (3 + (-7))i = 7 + (-4)i = 7 - 4i. Choice A correctly performs the complex number addition: real parts 5 + 2 = 7, and imaginary parts 3i + (-7i) = -4i, giving 7 - 4i in standard form. Choice B incorrectly subtracts the real parts (5 - 2 = 3) instead of adding them—remember, we're adding z₁ + z₂, not subtracting! When adding complex numbers, always add real to real and imaginary to imaginary: think of it like combining like terms in algebra where 'i' is just another variable that happens to have the special property i² = -1.

Question 12

If w1=1+2iw_1 = 1 + 2i and w2=3+iw_2 = 3 + i, what is the real part of w1w2‾−w1‾w2w_1 \overline{w_2} - \overline{w_1} w_2?

  1. 00 (correct answer)
  2. −6-6
  3. 66
  4. 1010
Explanation: First, find the conjugates: w1‾=1−2i\overline{w_1} = 1 - 2i and w2‾=3−i\overline{w_2} = 3 - i. Next, calculate w1w2‾=(1+2i)(3−i)=3−i+6i−2i2=3+5i+2=5+5iw_1 \overline{w_2} = (1+2i)(3-i) = 3 - i + 6i - 2i^2 = 3 + 5i + 2 = 5 + 5i. Then calculate w1‾w2=(1−2i)(3+i)=3+i−6i−2i2=3−5i+2=5−5i\overline{w_1} w_2 = (1-2i)(3+i) = 3 + i - 6i - 2i^2 = 3 - 5i + 2 = 5 - 5i. Therefore, w1w2‾−w1‾w2=(5+5i)−(5−5i)=5+5i−5+5i=10iw_1 \overline{w_2} - \overline{w_1} w_2 = (5+5i) - (5-5i) = 5 + 5i - 5 + 5i = 10i. The real part is 00. This result makes sense because w1w2‾−w1‾w2w_1 \overline{w_2} - \overline{w_1} w_2 is always purely imaginary (it equals 2i⋅Im(w1w2‾)2i \cdot \text{Im}(w_1 \overline{w_2})). Choice B (-6) might come from incorrectly calculating the imaginary parts. Choice C (6) could arise from a sign error. Choice D (10) is the coefficient of the imaginary part.

Question 13

When (2+3i)2(2 + 3i)^2 is written in standard form a+bia + bi, what is the value of a−ba - b?

  1. −10-10
  2. −8-8
  3. −17-17 (correct answer)
  4. −7-7
Explanation: Expanding (2+3i)2=(2+3i)(2+3i)=4+6i+6i+9i2=4+12i+9(−1)=4+12i−9=−5+12i(2 + 3i)^2 = (2 + 3i)(2 + 3i) = 4 + 6i + 6i + 9i^2 = 4 + 12i + 9(-1) = 4 + 12i - 9 = -5 + 12i. Therefore, a=−5a = -5 and b=12b = 12, so a−b=−5−12=−17a - b = -5 - 12 = -17. Choice A results from calculating a+b=−5+12=7a + b = -5 + 12 = 7 and then taking the negative. Choice B comes from the error 4−12=−84 - 12 = -8. Choice D is b−a=12−(−5)=17b - a = 12 - (-5) = 17 with a sign error.

Question 14

Divide and express in standard form a+bia + bi: 3+2i1−i\frac{3 + 2i}{1 - i} (Use the conjugate of the denominator.)

  1. 12+52i\frac{1}{2} + \frac{5}{2}i (correct answer)
  2. 52+12i\frac{5}{2} + \frac{1}{2}i
  3. 12−52i\frac{1}{2} - \frac{5}{2}i
  4. 1+5i1 + 5i
Explanation: This question tests your understanding of complex numbers—numbers in form a + bi where i is the imaginary unit with i squared = -1—and how to perform operations (add, subtract, multiply, divide) while keeping results in standard a + bi form. Complex numbers extend the real number system to include square roots of negative numbers using i = square root of -1, so i squared = -1. The standard form a + bi has a as the real part and b as the coefficient of the imaginary part (bi). Operations work like algebra with the crucial rule: whenever you see i squared, replace it with -1! For division like (3 + 2i) divided by (1 - i), multiply numerator and denominator by the conjugate of denominator (1 + i): numerator becomes (3 + 2i)(1 + i) = 3 + 3i + 2i + 2i² = 3 + 5i + 2(-1) = 3 + 5i - 2 = 1 + 5i. Denominator: (1 - i)(1 + i) = 1 - i² = 1 - (-1) = 1 + 1 = 2. Result: (1 + 5i) ÷ 2 = 1/2 + 5i/2 = 1/2 + 5/2i. Conjugate multiplication makes denominator real! Choice A correctly performs the complex number division and expresses the result in standard form a + bi with proper handling of the conjugate method. Choice C makes a sign error in the imaginary part: the result should be 1/2 + 5/2i, not 1/2 - 5/2i—check your arithmetic when multiplying the numerator! Complex number operations summary: DIVISION—multiply top and bottom by conjugate of denominator (flip imaginary sign), simplify using (a + bi)(a - bi) = a squared + b squared (always real), then express result as a + bi by dividing both parts by denominator. These patterns are consistent! The conjugate method always produces a real denominator, making division possible.

Question 15

Divide and express in standard form a+bia+bi by multiplying by the conjugate of the denominator: 3+2i1−i.\frac{3+2i}{1-i}.

  1. 32+22i\frac{3}{2}+\frac{2}{2}i
  2. 52+12i\frac{5}{2}+\frac{1}{2}i
  3. 12−52i\frac{1}{2}-\frac{5}{2}i
  4. 12+52i\frac{1}{2}+\frac{5}{2}i (correct answer)
Explanation: This question tests your understanding of complex numbers—numbers in the form a + bi where i is the imaginary unit with i squared = -1—and dividing by multiplying by the conjugate of the denominator. Complex numbers extend the real number system to include square roots of negative numbers using i = square root of -1, so i squared = -1. The conjugate of a - bi is a + bi, and multiplying by it makes the denominator real: for (3 + 2i)/(1 - i), use (1 + i), numerator 3 + 3i + 2i + 2i² = 3 + 5i - 2 = 1 + 5i, denominator 1 + 1 = 2, so 1/2 + (5/2)i—terrific! Choice A correctly handles the conjugate and simplifies to standard form. A distractor like choice C might flip signs incorrectly in the numerator—always compute carefully! This method works for all complex divisions—practice makes perfect. Keep going; you're gaining confidence!

Question 16

Let z=2+5iz = 2 + 5i. What is the product of zz and its conjugate z‾\overline{z}? (Recall (a+bi)(a−bi)=a2+b2(a+bi)(a-bi)=a^2+b^2.)

  1. 4+25i4 + 25i
  2. −21-21
  3. 2−5i2 - 5i
  4. 2929 (correct answer)
Explanation: This question tests your understanding of complex numbers—numbers in form a + bi where i is the imaginary unit with i squared = -1—and the special property that a complex number times its conjugate always gives a real number. Complex numbers extend the real number system to include square roots of negative numbers using i = square root of -1, so i squared = -1. The conjugate of z = a + bi is z̄ = a - bi, and their product is (a + bi)(a - bi) = a² - abi + abi - b²i² = a² - b²(-1) = a² + b². For z = 2 + 5i, the conjugate is z̄ = 2 - 5i, and their product is (2 + 5i)(2 - 5i) = 2² + 5² = 4 + 25 = 29. Choice B correctly calculates z × z̄ = 2² + 5² = 4 + 25 = 29, which is always a positive real number. Choice A incorrectly calculates 2² - 5² = 4 - 25 = -21, forgetting that the formula is a² + b², not a² - b², because the i² term contributes a positive value when simplified. The product of conjugates eliminates all imaginary parts, leaving only the sum of squares!

Question 17

What is the conjugate of the complex number −3+8i-3+8i?

  1. 3+8i3+8i
  2. 3−8i3-8i
  3. −3−8i-3-8i (correct answer)
  4. −8+3i-8+3i
Explanation: This question tests your understanding of complex numbers—numbers in form a+bia + bi where i is the imaginary unit with i2=−1i^2 = -1—and how to find the conjugate. Complex numbers extend the real number system to include square roots of negative numbers using i=−1i = \sqrt{-1}, so i2=−1i^2 = -1. The conjugate of a+bia + bi is a−bia - bi, flipping the sign of the imaginary part only; for example, conjugate of 3+4i3 + 4i is 3−4i3 - 4i. For −3+8i-3 + 8i, the conjugate is −3−8i-3 - 8i, keeping the real part the same and flipping the imaginary sign. Choice B correctly identifies the conjugate of the complex number with proper sign flip of only the imaginary part. Choice A flips the sign of the real part incorrectly: from −3+8i-3 + 8i to +3−8i+3 - 8i, but the conjugate changes only the imaginary sign, not the real! Remember, conjugates are useful in division and moduli: the product of a number and its conjugate is always real and positive—keep exploring these properties, you're making great progress!

Question 18

Let z1=6−2iz_1 = 6 - 2i and z2=1+5iz_2 = 1 + 5i. What is z1−z2z_1 - z_2 in standard form a+bia+bi?

  1. 5+3i5+3i
  2. 5−7i5-7i (correct answer)
  3. 7+3i7+3i
  4. 7−7i7-7i
Explanation: This question tests your understanding of complex numbers—numbers in form a + bi where i is the imaginary unit with i squared = -1—and how to perform subtraction while keeping results in standard a + bi form. Complex numbers extend the real number system to include square roots of negative numbers using i = square root of -1, so i squared = -1. The standard form a + bi has a as the real part and b as the coefficient of the imaginary part (bi). For subtraction, combine real parts and imaginary parts after distributing the negative: (6 - 2i) - (1 + 5i) = (6 - 1) + (-2 - 5)i = 5 - 7i. Choice B correctly performs the complex number subtraction and expresses the result in standard form a + bi with accurate sign handling. Choice A forgets to distribute the negative to the imaginary part: subtracting 5i means -5i, not +3i from incorrect combination—always distribute the minus sign! Complex number subtraction summary: (a + bi) - (c + di) = (a - c) + (b - d)i—these operations build your confidence in working with imaginary numbers, keep up the excellent work!

Question 19

Use i2=−1i^2 = -1 to simplify the product (3+2i)(1−4i)(3 + 2i)(1 - 4i) and write the result in the form a+bia + bi.

  1. 11−10i11 - 10i (correct answer)
  2. −5+10i-5 + 10i
  3. −5−10i-5 - 10i
  4. 11+10i11 + 10i
Explanation: This question tests your understanding of complex numbers—numbers in the form a + bi where i is the imaginary unit with i squared = -1—and how to perform multiplication while keeping results in standard a + bi form. Complex numbers extend the real number system to include square roots of negative numbers using i = square root of -1, so i squared = -1. Operations work like algebra with the crucial rule: whenever you see i squared, replace it with -1! For multiplication, use FOIL: (3 + 2i)(1 - 4i) = 31 + 3(-4i) + 2i1 + 2i(-4i) = 3 - 12i + 2i - 8i² = 3 - 10i + 8 (since -8*(-1)=8) = 11 - 10i. Choice A correctly performs the multiplication and simplifies i squared to -1, resulting in 11 - 10i. A tempting distractor like Choice C might forget to replace i squared with -1, leaving it as 3 - 10i - 8i² or mishandling signs. Always simplify i squared right away, and multiplication will be a breeze—keep up the great work!

Question 20

Multiply and simplify using i2=−1i^2 = -1: (3+2i)(1−4i)(3 + 2i)(1 - 4i). Write your answer in a+bia + bi form.

  1. 11−10i11 - 10i (correct answer)
  2. −5−10i-5 - 10i
  3. 11+10i11 + 10i
  4. −5+10i-5 + 10i
Explanation: This question tests your understanding of complex numbers—numbers in form a + bi where i is the imaginary unit with i squared = -1—and how to perform multiplication using FOIL while keeping results in standard a + bi form. Complex numbers extend the real number system to include square roots of negative numbers using i = square root of -1, so i squared = -1. For multiplication, use FOIL or distribution, then simplify any i squared terms to -1. To multiply (3 + 2i)(1 - 4i): First: 3 × 1 = 3. Outer: 3 × (-4i) = -12i. Inner: 2i × 1 = 2i. Last: 2i × (-4i) = -8i². Combine: 3 - 12i + 2i - 8i² = 3 - 10i - 8i². Now replace i² with -1: 3 - 10i - 8(-1) = 3 - 10i + 8 = 11 - 10i. Choice A correctly performs the multiplication and simplifies i² = -1 to get 11 - 10i in standard form. Choice B forgets to replace i² with -1: they get 3 - 10i - 8i² but leave it as -5 - 10i instead of recognizing -8i² = -8(-1) = +8, which adds to 3 to give 11. The i² simplification is mandatory—whenever you see i², immediately replace it with -1!