Complex Numbers - ACT Math

1. To remove imaginary terms from a denominator later, you plan to multiply by a conjugate. What is the complex conjugate of $8 - 11i$?
   Answer: $8 + 11i$ The operation is finding the complex conjugate of 8 - 11i. The conjugate changes the sign of the imaginary part, so for a + bi it becomes a - bi, but here it's 8 + (-11)i, so conjugate is 8 + 11i. This is useful for rationalizing denominators. Choice D is the original number, a common distractor if forgetting to change the sign.
2. In simplifying a signal calculation, you need to multiply two complex numbers. What is $(3+4i)(2-5i)$? Use FOIL and apply that $i^2=-1$ to write the result in standard form $a+bi$.
   Answer: $26-7i$ This requires multiplying two complex numbers using FOIL method. $(3+4i)(2-5i) = 3 \cdot 2 + 3 \cdot (-5i) + 4i \cdot 2 + 4i \cdot (-5i) = 6 - 15i + 8i - 20i^2$. Since $i^2 = -1$, this becomes $6 - 15i + 8i - 20(-1) = 6 - 15i + 8i + 20 = 26 - 7i$. The result in standard form is $26 - 7i$.
3. A complex impedance is modeled as $z=(2+3i)+(7-10i)$. What is $z$ in standard form $a+bi$ after combining real and imaginary parts?
   Answer: $9-7i$ Adding complex numbers involves combining real and imaginary parts separately. For $(2 + 3i) + (7 - 10i)$, we add the real parts: $2 + 7 = 9$, and add the imaginary parts: $3i + (-10i) = -7i$. The result in standard form is $9 - 7i$. This represents combining two complex impedances in an AC circuit.
4. A student evaluates a product involving a variable real part. What is the real part of $(x+3i)(2-i)$? (Assume $x$ is a real number and use $i^2=-1$.)
   Answer: $2x+3$ To find the real part of $(x+3i)(2-i)$, we use FOIL: $2x - xi + 6i - 3i^2$. Since $i^2 = -1$, we have $-3i^2 = -3(-1) = 3$. The expression becomes $2x + 3 + (-x + 6)i$. The real part is $2x + 3$.
5. For a complex number $-9+12i$, compute its magnitude to determine its distance from the origin. What is the absolute value of $(-9+12i)$? Use $|a+bi|=\sqrt{a^2+b^2}$ and simplify.
   Answer: $15$ The magnitude formula gives $|a + bi| = \sqrt{a^2 + b^2}$. For $(-9 + 12i)$, we have $a = -9$ and $b = 12$, so the magnitude is $\sqrt{(-9)^2 + 12^2} = \sqrt{81 + 144} = \sqrt{225} = 15$. The absolute value represents the distance from the origin to the point $(-9, 12)$ in the complex plane.
6. What is the absolute value of $4 + 3i$?
   Answer: 5 The absolute value (magnitude) of a complex number $a + bi$ equals $\sqrt{a^2 + b^2}$. For $4 + 3i$, we calculate $\sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$. The absolute value is $5$.
7. What is the complex conjugate of $6 - 8i$?
   Answer: 6 + 8i The complex conjugate of a complex number changes the sign of the imaginary part. For a complex number a + bi, the conjugate is a - bi. The complex conjugate of 6 - 8i is 6 - (-8i) = 6 + 8i. We change the sign of the imaginary term from negative to positive.
8. What is $(4 + i)(3 - 2i)$?
   Answer: 14 - 5i This problem requires multiplying two complex numbers using the FOIL method. (4 + i)(3 - 2i) = 4(3) + 4(-2i) + i(3) + i(-2i) = 12 - 8i + 3i - 2i². Since i² equals negative one, this becomes 12 - 5i - 2(-1) = 12 - 5i + 2 = 14 - 5i. The result is in standard form a + bi.
9. What is the real part of $ (5 + 2i)(4 - 3i) $?
   Answer: 26 This is multiplication of complex numbers using FOIL. $(5 + 2i)(4 - 3i) = 5 \cdot 4 + 5 \cdot (-3i) + 2i \cdot 4 + 2i \cdot (-3i) = 20 - 15i + 8i - 6i^2$. Since $ $i^2$ = -1 $, this becomes $ 20 - 7i - 6(-1) = 20 - 7i + 6 = 26 - 7i $. The real part is $ 26 $.
10. Which expression is equivalent to $(1 + 3i)(2 + 5i)$?
    Answer: -13 + 11i This is multiplication of complex numbers using FOIL. $(1 + 3i)(2 + 5i) = 1 \cdot 2 + 1 \cdot 5i + 3i \cdot 2 + 3i \cdot 5i = 2 + 5i + 6i + 15i^2$. Since $i^2 = -1$, this becomes $2 + 11i + 15(-1) = 2 + 11i - 15 = -13 + 11i$.
11. What is $(2 - 3i)(3 + 2i)$?
    Answer: $12 - 5i$ This is multiplication of complex numbers using FOIL. $(2 - 3i)(3 + 2i) = 2 \cdot 3 + 2 \cdot 2i + (-3i) \cdot 3 + (-3i) \cdot 2i = 6 + 4i - 9i - 6i^2$. Since $i^2 = -1$, this becomes $6 - 5i - 6(-1) = 6 - 5i + 6 = 12 - 5i$.
12. In an AC circuit calculation, a technician multiplies two complex impedances. What is $(3 + 4i)(2 - 5i)$? Use FOIL and apply $i^2 = -1$ to write the result in standard form $a + bi$.
    Answer: $26 - 7i$ The operation is multiplication of two complex numbers, (3 + 4i) and (2 - 5i). Using the FOIL method, expand as follows: first terms 3*2 = 6, outer terms 3*(-5i) = -15i, inner terms 4i*2 = 8i, and last terms 4i*(-5i) = -20i². Since i squared equals negative one, -20i² = -20*(-1) = 20. Combining the real parts 6 + 20 = 26 and the imaginary parts -15i + 8i = -7i gives 26 - 7i. Choice C is a common error from adding instead of subtracting the imaginary parts, and choice D forgets to apply i squared equals negative one.
13. In simplifying a polynomial with complex coefficients, you multiply two binomials. What is $(1 - 6i)(-3 - 2i)$? Use FOIL and apply $i^2=-1$ to write the result as $a + bi$.
    Answer: $-15 + 16i$ The operation is multiplication of (1 - 6i) and (-3 - 2i). Using FOIL: first 1*(-3) = -3, outer 1*(-2i) = -2i, inner (-6i)*(-3) = 18i, last (-6i)*(-2i) = 12i². Since i squared equals negative one, 12i² = 12*(-1) = -12. Combine real parts -3 - 12 = -15 and imaginary parts -2i + 18i = 16i to get -15 + 16i. Choice B might result from a sign error in the imaginary terms.
14. What is the complex conjugate of $6 + i$?
    Answer: 6 - i Finding the complex conjugate means changing the sign of the imaginary part while keeping the real part the same. For the complex number $6 + i$, we keep the real part as $6$ and change the sign of the imaginary part from $+i$ to $-i$. Therefore, the complex conjugate is $6 - i$.
15. What is $(4 + 2i) - (1 - 3i)$?
    Answer: 3 + 5i This is subtraction of complex numbers, where we subtract corresponding parts. $(4 + 2i) - (1 - 3i) = (4 - 1) + (2i - (-3i)) = 3 + (2i + 3i) = 3 + 5i$. We subtract the real parts and subtract the imaginary parts separately.
16. What is $(-2 + 5i) + (3 - 4i)$?
    Answer: $1 + i$ This is addition of complex numbers, where we combine like terms. To add $(-2 + 5i) + (3 - 4i)$, we add the real parts: $-2 + 3 = 1$ and add the imaginary parts: $5i + (-4i) = i$. Therefore, the result is $1 + i$.
17. What is the absolute value of $6 - 8i$?
    Answer: 10 The absolute value (magnitude) of a complex number $a + bi$ equals $\sqrt{a^2 + b^2}$. For $6 - 8i$, we calculate $\sqrt{6^2 + (-8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10$. The absolute value is $10$.
18. To simplify an expression, evaluate $(2 + 3i) - (7 - 5i)$ and write the result in standard form $a + bi$.
    Answer: $-5 + 8i$ The operation is subtraction of two complex numbers, (2 + 3i) - (7 - 5i). Distribute the negative: 2 + 3i - 7 + 5i. Combine real parts: 2 - 7 = -5. Combine imaginary parts: 3i + 5i = 8i. The result is -5 + 8i. Choice A might result from subtracting the imaginary parts incorrectly.
19. A signal processing step multiplies two complex gains. Using FOIL and applying $i^2=-1$, what is $(4-3i)(2+5i)$ in standard form $a+bi$?​​
    Answer: $23+14i$ This problem requires multiplying two complex numbers using the FOIL method: (4-3i)(2+5i). First, 4*2 = 8; outer, 4*5i = 20i; inner, -3i*2 = -6i; last, -3i*5i = -15i². Since i² equals -1, -15i² = -15*(-1) = 15. Combining terms: 8 + 15 + (20i - 6i) = 23 + 14i. Choice B might come from forgetting that i² equals -1 and treating -15i² as -15, leading to -7 + 14i.
20. When dividing by a complex number, you often use its conjugate. What is the complex conjugate of $-6+13i$?
    Answer: $-6-13i$ The complex conjugate of a complex number $a+bi$ is $a-bi$, where we change the sign of the imaginary part only. For $-6+13i$, the conjugate is $-6-13i$. The real part stays the same at $-6$, while the imaginary part changes from $+13i$ to $-13i$.

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