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ACT Math

ACT Math Help: Complex Numbers

Review real example questions for Complex Numbers in ACT Math.

Question 1

For the complex number iii, where i2=−1i^2 = -1i2=−1, what is the simplified form of 3+i1−i\dfrac{3 + i}{1 - i}1−i3+i​?

  1. 1+2i1 + 2i1+2i
  2. 2+i2 + i2+i
  3. 1−2i1 - 2i1−2i
  4. 2−i2 - i2−i
Explanation: The correct answer is A (1 + 2i). Multiply by the conjugate of the denominator: (3 + i)/(1 − i) × (1 + i)/(1 + i). Numerator: (3 + i)(1 + i) = 3 + 3i + i + i² = 3 + 4i − 1 = 2 + 4i. Denominator: (1 − i)(1 + i) = 1 − i² = 1 + 1 = 2. Result: (2 + 4i)/2 = 1 + 2i. B (2 + i) likely comes from incomplete multiplication or forgetting to divide. C (1 − 2i) comes from a sign error in the numerator expansion. D (2 − i) comes from multiplying by the wrong conjugate (1 − i) instead of (1 + i). The key technique: always multiply by the conjugate to eliminate i from the denominator, remembering that i² = −1.

Question 2

A complex impedance is given by −6+7i-6+7i−6+7i. What is the complex conjugate of −6+7i-6+7i−6+7i (flip the sign of the imaginary part only)?

  1. 6+7i6+7i6+7i
  2. −6−7i-6-7i−6−7i
  3. 6−7i6-7i6−7i
  4. −6+7i-6+7i−6+7i
Explanation: This problem asks for the complex conjugate of −6+7i-6 + 7i−6+7i, which is found by changing the sign of the imaginary part only. The real part is −6-6−6, and the imaginary part 7i7i7i becomes −7i-7i−7i. Thus, the conjugate is −6−7i-6 - 7i−6−7i. Choice D might result from incorrectly flipping the sign of the real part instead of the imaginary part.

Question 3

To combine two complex measurements, subtract one from the other. What is (8+5i)−(3−9i)(8+5i)-(3-9i)(8+5i)−(3−9i) written in standard form a+bia+bia+bi?

  1. 5−4i5-4i5−4i
  2. 11+14i11+14i11+14i
  3. 5+14i5+14i5+14i
  4. 11−4i11-4i11−4i
Explanation: Subtracting complex numbers requires distributing the negative sign and combining like terms. (8+5i)−(3−9i)=8+5i−3+9i(8 + 5i) - (3 - 9i) = 8 + 5i - 3 + 9i(8+5i)−(3−9i)=8+5i−3+9i. Combining real parts: 8−3=58 - 3 = 58−3=5, and combining imaginary parts: 5i+9i=14i5i + 9i = 14i5i+9i=14i. The result in standard form is 5+14i5 + 14i5+14i.

Question 4

A complex impedance is modeled as z=(2+3i)+(7−10i)z=(2+3i)+(7-10i)z=(2+3i)+(7−10i). What is zzz in standard form a+bia+bia+bi after combining real and imaginary parts?

  1. −5−7i-5-7i−5−7i
  2. 5−13i5-13i5−13i
  3. 9+13i9+13i9+13i
  4. 9−7i9-7i9−7i
Explanation: Adding complex numbers involves combining real and imaginary parts separately. For (2+3i)+(7−10i)(2 + 3i) + (7 - 10i)(2+3i)+(7−10i), we add the real parts: 2+7=92 + 7 = 92+7=9, and add the imaginary parts: 3i+(−10i)=−7i3i + (-10i) = -7i3i+(−10i)=−7i. The result in standard form is 9−7i9 - 7i9−7i. This represents combining two complex impedances in an AC circuit.

Question 5

Let xxx be a real number. In the product (x+2i)(3−i)(x + 2i)(3 - i)(x+2i)(3−i), what is the real part after multiplying using FOIL and applying i2=−1i^2 = -1i2=−1?

  1. 3x3x3x
  2. 3x−23x-23x−2
  3. 3x+23x+23x+2
  4. −3x+2-3x+2−3x+2
Explanation: Using FOIL to multiply: (x+2i)(3−i)=x⋅3+x⋅(−i)+(2i)⋅3+(2i)⋅(−i)=3x−xi+6i−2i2(x + 2i)(3 - i) = x \cdot 3 + x \cdot (-i) + (2i) \cdot 3 + (2i) \cdot (-i) = 3x - xi + 6i - 2i^2(x+2i)(3−i)=x⋅3+x⋅(−i)+(2i)⋅3+(2i)⋅(−i)=3x−xi+6i−2i2. Since i2=−1i^2 = -1i2=−1, this becomes 3x−xi+6i−2(−1)=3x−xi+6i+23x - xi + 6i - 2(-1) = 3x - xi + 6i + 23x−xi+6i−2(−1)=3x−xi+6i+2. The real part consists of terms without iii: 3x+23x + 23x+2.

Question 6

Given i=−1i = \sqrt{-1}i=−1​, what is the simplified form of (4+5i)−(1−2i)(4 + 5i) - (1 - 2i)(4+5i)−(1−2i)?

  1. 3+3i3 + 3i3+3i
  2. 3+7i3 + 7i3+7i
  3. 5+3i5 + 3i5+3i
  4. 5+7i5 + 7i5+7i
Explanation: This is a complex numbers question testing subtraction with distribution. Choice B (3 + 7i) is correct — distribute the negative: (4 + 5i) − (1 − 2i) = 4 + 5i − 1 + 2i. Key step: −(−2i) = +2i. Combine real parts: 4 − 1 = 3. Combine imaginary parts: 5i + 2i = 7i. Result: 3 + 7i. Choice A (3 + 3i) correctly subtracts the real parts but fails to distribute the negative on the imaginary term: 5i − 2i = 3i instead of 5i + 2i = 7i. Choice C (5 + 3i) adds the real parts instead of subtracting: 4 + 1 = 5, and also gets the imaginary term wrong. Choice D (5 + 7i) adds real parts (correctly gets +7i from the imaginary) — two separate errors that partially cancel. Pro tip: When subtracting a complex number, rewrite it as addition of the negative first: (4 + 5i) + (−1 + 2i). This prevents sign errors by making every operation an addition. The most common mistake is treating −(−2i) as −2i instead of +2i.

Question 7

For the complex number iii, where i2=−1i^2 = -1i2=−1, what is the value of (3+2i)−(5−4i)(3 + 2i) - (5 - 4i)(3+2i)−(5−4i)?

  1. −2−2i-2 - 2i−2−2i
  2. −2+6i-2 + 6i−2+6i
  3. 8−2i8 - 2i8−2i
  4. −2−6i-2 - 6i−2−6i
Explanation: This is a complex numbers question testing subtraction with distribution. Choice B (−2 + 6i) is correct — distribute the negative sign: (3 + 2i) − (5 − 4i) = 3 + 2i − 5 + 4i. The critical step: −(−4i) = +4i. Combine real parts: 3 − 5 = −2. Combine imaginary parts: 2i + 4i = 6i. Result: −2 + 6i. Choice A (−2 − 2i) gets the real part right but subtracts the imaginary parts without distributing the negative: treating it as 2i − 4i = −2i instead of 2i + 4i = 6i. Choice C (8 − 2i) adds the real parts instead of subtracting: 3 + 5 = 8, and also handles the imaginary term incorrectly. Choice D (−2 − 6i) gets the real part right but applies the sign error in the opposite direction — treating −(−4i) as −6i. Pro tip: When subtracting a complex number, rewrite the subtraction as adding the negative first: (3 + 2i) − (5 − 4i) becomes (3 + 2i) + (−5 + 4i). Then combine real and imaginary parts separately.

Question 8

A point in the complex plane is represented by 3−4i3-4i3−4i. What is the absolute value of (3−4i)(3-4i)(3−4i)? Use ∣a+bi∣=a2+b2|a+bi|=\sqrt{a^2+b^2}∣a+bi∣=a2+b2​ and simplify.

  1. 7\sqrt{7}7​
  2. 111
  3. 555
  4. 777
Explanation: The magnitude formula is ∣a+bi∣=a2+b2|a + bi| = \sqrt{a^2 + b^2}∣a+bi∣=a2+b2​. For (3−4i)(3 - 4i)(3−4i), we have a=3a = 3a=3 and b=−4b = -4b=−4, so the magnitude is 32+(−4)2=9+16=25=5\sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 532+(−4)2​=9+16​=25​=5. This represents the distance from the origin to the point (3,−4)(3, -4)(3,−4) in the complex plane, forming a 3-4-5 right triangle.

Question 9

Which expression is equivalent to (1+6i)(4−i)(1 + 6i)(4 - i)(1+6i)(4−i)?

  1. 4 - 19i
  2. 10 - 23i
  3. 4 + 19i
  4. 10 + 23i
Explanation: This problem requires multiplying two complex numbers using the FOIL method. (1 + 6i)(4 - i) = 1(4) + 1(-i) + 6i(4) + 6i(-i) = 4 - i + 24i - 6i². Since i² equals negative one, this becomes 4 + 23i - 6(-1) = 4 + 23i + 6 = 10 + 23i. Choice C likely forgot that i² equals negative one.

Question 10

What is the real part of (3+5i)(2−3i)(3 + 5i)(2 - 3i)(3+5i)(2−3i)?

  1. 14
  2. 9
  3. 10
  4. 21
Explanation: This is multiplication of complex numbers using FOIL. (3+5i)(2−3i)=3⋅2+3⋅(−3i)+5i⋅2+5i⋅(−3i)=6−9i+10i−15i2(3 + 5i)(2 - 3i) = 3 \cdot 2 + 3 \cdot (-3i) + 5i \cdot 2 + 5i \cdot (-3i) = 6 - 9i + 10i - 15i^2(3+5i)(2−3i)=3⋅2+3⋅(−3i)+5i⋅2+5i⋅(−3i)=6−9i+10i−15i2. Since i2=−1i^2 = -1i2=−1, this becomes 6+i−15(−1)=6+i+15=21+i6 + i - 15(-1) = 6 + i + 15 = 21 + i6+i−15(−1)=6+i+15=21+i. The real part is 212121.