Math 2 Quiz: Complex Number Operations
10 questions · exam conditions
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Complex Number OperationsQuestion 1 of 10

If z1=2+3iz_1 = 2 + 3i and z2=1−2iz_2 = 1 - 2i, what is the coefficient of ii in the expansion of (z1+z2)2(z_1 + z_2)^2?

22
44
66
88
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Math 2 Quiz

Math 2 Quiz: Complex Number Operations

Practice Complex Number Operations in Math 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 Complex Number Operations, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 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

If z1=2+3iz_1 = 2 + 3i and z2=1−2iz_2 = 1 - 2i, what is the coefficient of ii in the expansion of (z1+z2)2(z_1 + z_2)^2?

  1. 22
  2. 44
  3. 66 (correct answer)
  4. 88
Explanation: First calculate z1+z2=(2+3i)+(1−2i)=3+iz_1 + z_2 = (2+3i) + (1-2i) = 3 + i. Then (z1+z2)2=(3+i)2=9+6i+i2=9+6i−1=8+6i(z_1 + z_2)^2 = (3+i)^2 = 9 + 6i + i^2 = 9 + 6i - 1 = 8 + 6i. The coefficient of ii is 66.

Question 2

Let u=3+2iu = 3 + 2i and v=1−iv = 1 - i. If uv+uvˉ=a+biuv + u\bar{v} = a + bi where aa and bb are real, what is a+ba + b?

  1. 88
  2. 99
  3. 1010 (correct answer)
  4. 1111
Explanation: First calculate uv=(3+2i)(1−i)=3−3i+2i−2i2=3−i+2=5−iuv = (3+2i)(1-i) = 3 - 3i + 2i - 2i^2 = 3 - i + 2 = 5 - i. Next, vˉ=1−i‾=1+i\bar{v} = \overline{1-i} = 1+i, so uvˉ=(3+2i)(1+i)=3+3i+2i+2i2=3+5i−2=1+5iu\bar{v} = (3+2i)(1+i) = 3 + 3i + 2i + 2i^2 = 3 + 5i - 2 = 1 + 5i. Therefore, uv+uvˉ=(5−i)+(1+5i)=6+4iuv + u\bar{v} = (5-i) + (1+5i) = 6 + 4i. So a=6a = 6 and b=4b = 4, giving a+b=10a + b = 10.

Question 3

If z=2+i1−2iz = \frac{2+i}{1-2i}, what is the value of z+zˉz + \bar{z}?

  1. −45-\frac{4}{5}
  2. 00 (correct answer)
  3. 45\frac{4}{5}
  4. 85\frac{8}{5}
Explanation: To find zz, multiply numerator and denominator by the conjugate of the denominator: z=2+i1−2i⋅1+2i1+2i=(2+i)(1+2i)(1−2i)(1+2i)=2+4i+i+2i21+4=2+5i−25=5i5=iz = \frac{2+i}{1-2i} \cdot \frac{1+2i}{1+2i} = \frac{(2+i)(1+2i)}{(1-2i)(1+2i)} = \frac{2+4i+i+2i^2}{1+4} = \frac{2+5i-2}{5} = \frac{5i}{5} = i. Therefore, zˉ=i‾=−i\bar{z} = \overline{i} = -i, and z+zˉ=i+(−i)=0z + \bar{z} = i + (-i) = 0.

Question 4

If w1=4−3iw_1 = 4 - 3i and w2=−1+2iw_2 = -1 + 2i, which expression represents the conjugate of w1w2w_1w_2?

  1. w1‾⋅w2‾\overline{w_1} \cdot \overline{w_2} (correct answer)
  2. w1⋅w2w_1 \cdot w_2
  3. w1‾+w2‾\overline{w_1} + \overline{w_2}
  4. −w1‾⋅w2‾-\overline{w_1} \cdot \overline{w_2}
Explanation: This tests the property that the conjugate of a product equals the product of the conjugates: w1w2‾=w1‾⋅w2‾\overline{w_1w_2} = \overline{w_1} \cdot \overline{w_2}. To verify: w1w2=(4−3i)(−1+2i)=−4+8i+3i−6i2=−4+11i+6=2+11iw_1w_2 = (4-3i)(-1+2i) = -4 + 8i + 3i - 6i^2 = -4 + 11i + 6 = 2 + 11i, so w1w2‾=2−11i\overline{w_1w_2} = 2 - 11i. Also, w1‾=4+3i\overline{w_1} = 4 + 3i and w2‾=−1−2i\overline{w_2} = -1 - 2i, giving w1‾⋅w2‾=(4+3i)(−1−2i)=−4−8i−3i−6i2=−4−11i+6=2−11i\overline{w_1} \cdot \overline{w_2} = (4+3i)(-1-2i) = -4 - 8i - 3i - 6i^2 = -4 - 11i + 6 = 2 - 11i. These match, confirming the property.

Question 5

Let z=a+biz = a + bi where aa and bb are real numbers. If z2+z+1=0z^2 + z + 1 = 0, what is the value of ∣z∣2|z|^2?

  1. 12\frac{1}{2}
  2. 11 (correct answer)
  3. 32\frac{3}{2}
  4. 22
Explanation: From z2+z+1=0z^2 + z + 1 = 0, we get z2=−z−1z^2 = -z - 1. Multiplying both sides by zˉ\bar{z} (the complex conjugate): z2zˉ=(−z−1)zˉ=−zzˉ−zˉ=−∣z∣2−zˉz^2\bar{z} = (-z-1)\bar{z} = -z\bar{z} - \bar{z} = -|z|^2 - \bar{z}. Since z2zˉ=z⋅zzˉ=z∣z∣2z^2\bar{z} = z \cdot z\bar{z} = z|z|^2, we have z∣z∣2=−∣z∣2−zˉz|z|^2 = -|z|^2 - \bar{z}. This gives z∣z∣2+∣z∣2=−zˉz|z|^2 + |z|^2 = -\bar{z}, or ∣z∣2(z+1)=−zˉ|z|^2(z + 1) = -\bar{z}. Taking the conjugate of the original equation: z2+z+1‾=zˉ2+zˉ+1=0\overline{z^2 + z + 1} = \bar{z}^2 + \bar{z} + 1 = 0. From z2+z+1=0z^2 + z + 1 = 0, we can use the quadratic formula: z=−1±1−42=−1±i32z = \frac{-1 \pm \sqrt{1-4}}{2} = \frac{-1 \pm i\sqrt{3}}{2}. For either root, ∣z∣2=(−12)2+(±32)2=14+34=1|z|^2 = \left(\frac{-1}{2}\right)^2 + \left(\pm\frac{\sqrt{3}}{2}\right)^2 = \frac{1}{4} + \frac{3}{4} = 1.

Question 6

Let α=1+i1−i\alpha = \frac{1+i}{1-i} and β=1−i1+i\beta = \frac{1-i}{1+i}. What is the value of α2+β2\alpha^2 + \beta^2?

  1. −4-4
  2. 22
  3. 00
  4. −2-2 (correct answer)
Explanation: This problem tests your ability to work with complex numbers and recognize important relationships between conjugate expressions. When you see complex fractions like these, start by simplifying each term. To simplify α=1+i1−i\alpha = \frac{1+i}{1-i}, multiply both numerator and denominator by the conjugate of the denominator: 1+i1−i⋅1+i1+i=(1+i)21−i2=1+2i+i21−(−1)=1+2i−12=2i2=i\frac{1+i}{1-i} \cdot \frac{1+i}{1+i} = \frac{(1+i)^2}{1-i^2} = \frac{1+2i+i^2}{1-(-1)} = \frac{1+2i-1}{2} = \frac{2i}{2} = i. Similarly, for β=1−i1+i\beta = \frac{1-i}{1+i}, multiply by 1−i1−i\frac{1-i}{1-i}: (1−i)21+i2=1−2i+i22=1−2i−12=−2i2=−i\frac{(1-i)^2}{1+i^2} = \frac{1-2i+i^2}{2} = \frac{1-2i-1}{2} = \frac{-2i}{2} = -i. Therefore, α2+β2=i2+(−i)2=−1+(−1)=−2\alpha^2 + \beta^2 = i^2 + (-i)^2 = -1 + (-1) = -2. Choice A (-4) might result from incorrectly calculating (2i)2+(−2i)2(2i)^2 + (-2i)^2 without the division by 2. Choice B (2) could come from forgetting that i2=−1i^2 = -1 and treating it as positive 1. Choice C (0) might occur if you mistakenly think i2i^2 and (−i)2(-i)^2 cancel each other out rather than both equaling -1. When working with complex number ratios, always rationalize by multiplying by the conjugate of the denominator. Also remember that i2=−1i^2 = -1, and (−i)2=i2=−1(-i)^2 = i^2 = -1, not −i2-i^2.

Question 7

For complex numbers s=1+2is = 1 + 2i and t=3−it = 3 - i, which of the following equals ∣st∣2|st|^2?

  1. ∣s∣2+∣t∣2|s|^2 + |t|^2
  2. (∣s∣−∣t∣)2(|s| - |t|)^2
  3. (∣s∣+∣t∣)2(|s| + |t|)^2
  4. ∣s∣2⋅∣t∣2|s|^2 \cdot |t|^2 (correct answer)
Explanation: When you encounter problems involving the modulus (absolute value) of products of complex numbers, remember that there's a fundamental property that makes these calculations much simpler than computing the product first. The key insight is that ∣st∣=∣s∣⋅∣t∣|st| = |s| \cdot |t| for any complex numbers ss and tt. Therefore, ∣st∣2=(∣s∣⋅∣t∣)2=∣s∣2⋅∣t∣2|st|^2 = (|s| \cdot |t|)^2 = |s|^2 \cdot |t|^2. Let's verify this with our given values. First, find the moduli: ∣s∣=∣1+2i∣=12+22=5|s| = |1 + 2i| = \sqrt{1^2 + 2^2} = \sqrt{5} and ∣t∣=∣3−i∣=32+(−1)2=10|t| = |3 - i| = \sqrt{3^2 + (-1)^2} = \sqrt{10}. So ∣s∣2⋅∣t∣2=5⋅10=50|s|^2 \cdot |t|^2 = 5 \cdot 10 = 50. Now let's see why the other options fail. Choice A gives ∣s∣2+∣t∣2=5+10=15|s|^2 + |t|^2 = 5 + 10 = 15, which confuses addition with multiplication. Choice B yields (∣s∣−∣t∣)2=(5−10)2(|s| - |t|)^2 = (\sqrt{5} - \sqrt{10})^2, which is negative since 5<10\sqrt{5} < \sqrt{10}, and squares of real numbers can't be negative. Choice C gives (∣s∣+∣t∣)2=(5+10)2=5+10+250=15+102(|s| + |t|)^2 = (\sqrt{5} + \sqrt{10})^2 = 5 + 10 + 2\sqrt{50} = 15 + 10\sqrt{2}, which incorrectly treats the modulus of a product like the modulus of a sum. The correct answer is D: ∣s∣2⋅∣t∣2|s|^2 \cdot |t|^2. Study tip: Memorize that ∣zw∣=∣z∣⋅∣w∣|zw| = |z| \cdot |w| for complex numbers. This property often eliminates the need to multiply complex numbers explicitly, saving time and reducing errors.

Question 8

If z=a+biz = a + bi where aa and bb are real, and z⋅z‾+z+z‾=7z \cdot \overline{z} + z + \overline{z} = 7, what is the value of a2+b2+2aa^2 + b^2 + 2a?

  1. 55
  2. 77 (correct answer)
  3. 99
  4. 1111
Explanation: We have z=a+biz = a + bi and z‾=a−bi\overline{z} = a - bi. First, z⋅z‾=(a+bi)(a−bi)=a2+b2z \cdot \overline{z} = (a+bi)(a-bi) = a^2 + b^2. Also, z+z‾=(a+bi)+(a−bi)=2az + \overline{z} = (a+bi) + (a-bi) = 2a. So the equation becomes a2+b2+2a=7a^2 + b^2 + 2a = 7. Therefore, a2+b2+2a=7a^2 + b^2 + 2a = 7. The answer is directly given by the constraint equation. Choice A might come from students who forget the factor of 2 in z+z‾z + \overline{z}. Choice C might result from adding an extra 2 somewhere. Choice D might come from computational errors in expanding the products.

Question 9

Let w=2+3iw = 2 + 3i and v=1−2iv = 1 - 2i. If p=w⋅v‾+w‾⋅vp = w \cdot \overline{v} + \overline{w} \cdot v, what type of number is pp?

  1. A purely imaginary number with positive imaginary part
  2. A purely imaginary number with negative imaginary part
  3. A real number (correct answer)
  4. A complex number that is neither real nor purely imaginary
Explanation: We have w=2+3iw = 2 + 3i, so w‾=2−3i\overline{w} = 2 - 3i. Also v=1−2iv = 1 - 2i, so v‾=1+2i\overline{v} = 1 + 2i. Now compute w⋅v‾=(2+3i)(1+2i)=2+4i+3i+6i2=2+7i−6=−4+7iw \cdot \overline{v} = (2+3i)(1+2i) = 2 + 4i + 3i + 6i^2 = 2 + 7i - 6 = -4 + 7i. And w‾⋅v=(2−3i)(1−2i)=2−4i−3i+6i2=2−7i−6=−4−7i\overline{w} \cdot v = (2-3i)(1-2i) = 2 - 4i - 3i + 6i^2 = 2 - 7i - 6 = -4 - 7i. Therefore p=(−4+7i)+(−4−7i)=−8p = (-4+7i) + (-4-7i) = -8. Since p=−8p = -8 is a real number, the answer is C. Notice that p=wv‾+w‾v=wv‾+wv‾‾p = w\overline{v} + \overline{w}v = w\overline{v} + \overline{w\overline{v}}, which is always real because it has the form z+z‾z + \overline{z} for z=wv‾z = w\overline{v}. Choice A and B would appeal to students who make computational errors. Choice D would appeal to students who don't recognize the general pattern.

Question 10

Let aa and bb be real numbers such that (a+bi)2=−5+12i(a + bi)^2 = -5 + 12i. What is the value of a2+b2a^2 + b^2?

  1. 13\sqrt{13}
  2. 1313 (correct answer)
  3. 55
  4. 169\sqrt{169}
Explanation: Expanding (a+bi)2=a2+2abi+(bi)2=a2−b2+2abi=−5+12i(a+bi)^2 = a^2 + 2abi + (bi)^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 need to find a2+b2a^2 + b^2. From ab=6ab = 6, we get b=6ab = \frac{6}{a}. Substituting into the first equation: a2−36a2=−5a^2 - \frac{36}{a^2} = -5. Multiplying by a2a^2: a4+5a2−36=0a^4 + 5a^2 - 36 = 0. Let u=a2u = a^2: u2+5u−36=0u^2 + 5u - 36 = 0. Factoring: (u+9)(u−4)=0(u+9)(u-4) = 0. Since u=a2≥0u = a^2 \geq 0, we have u=4u = 4, so a2=4a^2 = 4. Then b2=36a2=364=9b^2 = \frac{36}{a^2} = \frac{36}{4} = 9. Therefore a2+b2=4+9=13a^2 + b^2 = 4 + 9 = 13. Choice A gives the modulus of the original complex number, choice C is just the real part's absolute value, and choice D is 169=13\sqrt{169} = 13 but in the wrong form.