Using Conjugates with Complex Numbers - Algebra 2

1. If $z_1 = 1 + 3i$ and $z_2 = 4 - 2i$, what is $\overline{z_1 + z_2} - \overline{z_1} - \overline{z_2}$?
   Answer: $0$ This question tests your understanding of complex number conjugates and whether conjugation distributes over addition. When you see conjugate operations combined with arithmetic, always check if there's a distributive property at work. Let's work through this step-by-step. First, find $z_1 + z_2 = (1 + 3i) + (4 - 2i) = 5 + i$. The conjugate of this sum is $\overline{z_1 + z_2} = 5 - i$. Next, find the individual conjugates: $\overline{z_1} = \overline{1 + 3i} = 1 - 3i$ and $\overline{z_2} = \overline{4 - 2i} = 4 + 2i$. Now substitute into the original expression: $\overline{z_1 + z_2} - \overline{z_1} - \overline{z_2} = (5 - i) - (1 - 3i) - (4 + 2i)$ $= 5 - i - 1 + 3i - 4 - 2i = 0$ The key insight is that $\overline{z_1 + z_2} = \overline{z_1} + \overline{z_2}$ for any complex numbers—conjugation distributes over addition. This means our expression always equals zero regardless of the specific values of $z_1$ and $z_2$. Choice A gives $-5 - i$, which you'd get if you made sign errors when distributing the subtraction. Choice B gives $5 + i$, which is just $z_1 + z_2$ itself—you might get this if you forgot to take conjugates entirely. Choice C gives $5 - i$, which is $\overline{z_1 + z_2}$—this results from ignoring the subtraction of the individual conjugates. Remember: conjugation distributes over addition and subtraction, so expressions like $\overline{z_1 + z_2} - \overline{z_1} - \overline{z_2}$ will always equal zero.
2. Use the key conjugate property $(a+bi)(a-bi)=a^2+b^2$ to compute $(4+3i)(4-3i)$. (This product is always real and equals the modulus squared.)
   Answer: $25$ 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 (4 + 3i)(4 - 3i), we apply the formula directly: a = 4 and b = 3, so the product equals 4² + 3² = 16 + 9 = 25. Choice C correctly computes 25 using the conjugate property formula a² + b². Choice B shows the intermediate step 16 - 9i² but fails to simplify: remember that i² = -1, so 16 - 9i² = 16 - 9(-1) = 16 + 9 = 25. Always complete the simplification when i² appears! The (a + bi)(a - bi) = a² + b² pattern is worth memorizing: it's how we eliminate i from denominators, and it equals the modulus squared! This real result is exactly why conjugates are perfect for division—we use them to make denominators real!
3. Use conjugates to simplify and express in $a+bi$ form: $\frac{7-3i}{2+3i}.$
   Answer: $\frac{5}{13} - \frac{27}{13}i$ 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): for 2 + 3i, conjugate is 2 - 3i. 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!). To divide (7 - 3i) by (2 + 3i) using conjugates: (1) Identify conjugate of denominator: conjugate of 2 + 3i is 2 - 3i. (2) Multiply numerator and denominator by this conjugate: [(7 - 3i)(2 - 3i)] / [(2 + 3i)(2 - 3i)]. (3) Multiply numerator using FOIL: (7 - 3i)(2 - 3i) = 14 - 21i - 6i + 9i² = 14 - 27i + 9(-1) = 14 - 27i - 9 = 5 - 27i. (4) Multiply denominator using conjugate property: (2 + 3i)(2 - 3i) = 4 + 9 = 13 (real!). (5) Divide: (5 - 27i)/13 = 5/13 - 27i/13 in standard a + bi form. Choice A correctly identifies 5/13 - 27i/13 as the simplified form. Choice B has the wrong sign on the imaginary part: the numerator multiplication gives -27i, not +27i—when you FOIL (7 - 3i)(2 - 3i), both -21i and -6i terms are negative, combining to -27i. Choice D shows the unsimplified expression—you must multiply out and simplify to get the final a + bi form. Division recipe using conjugates: multiply both numerator and denominator by the conjugate of the denominator, expand the numerator (will generally stay complex), use the conjugate property for the denominator (becomes real: a² + b²), then divide both parts of the numerator by this real denominator. Always check your arithmetic carefully, especially the signs when combining like terms!
4. Use conjugates to divide and write in standard form $a+bi$: $\frac{1+2i}{3-i}.$
   Answer: $\frac{1}{10}+\frac{7}{10}i$ 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. To divide $1 + 2i$ by $3 - i$: (1) Identify conjugate of denominator: conjugate of $3 - i$ is $3 + i$. (2) Multiply numerator and denominator by this conjugate: $[(1 + 2i)(3 + i)]$ divided by $[(3 - i)(3 + i)]$. (3) Multiply numerator using FOIL: $(1 + 2i)(3 + i) = 3 + i + 6i + 2i^2 = 3 + 7i + 2(-1) = 1 + 7i$. (4) Multiply denominator using conjugate property: $(3 - i)(3 + i) = 3^2 + 1^2 = 9 + 1 = 10$. (5) Divide: $(1 + 7i)/10 = 1/10 + 7i/10$. Choice A correctly identifies $1/10 + 7i/10$ by properly applying the conjugate multiplication process. Choice B has the wrong sign on the imaginary part, while choice C forgot to divide by the denominator. Always divide both real and imaginary parts by the real denominator!
5. What is the complex conjugate of $2-5i$?​
   Answer: $2+5i$ This question tests your understanding of what a complex conjugate is. The key concept is that the conjugate of a + bi is a - bi, flipping the sign of the imaginary part while keeping the real part the same. For example, conjugate of 3 + 4i is 3 - 4i; for -1 - 2i is -1 + 2i. Note that if there's no real part, conjugate of 5i is -5i. For 2 - 5i, flip the -5i to +5i, so 2 + 5i. Choice B correctly identifies the conjugate by flipping only the imaginary sign. Choice D might tempt if you forget to flip and just repeat the number, but remember to change the sign of the i term. To find the conjugate, imagine the complex plane and reflect over the real axis. The real part stays, imaginary sign flips—simple as that! Keep practicing, you're on the right track!
6. Let $z = 2 + 5i$ and $w = 1 - 3i$. Using the property $\overline{z+w}=\overline{z}+\overline{w}$, find $\overline{z+w}$.
   Answer: $3 - 2i$ This question tests your understanding of complex conjugates—pairs of complex numbers that differ only in the sign of their imaginary part—and how conjugate properties work with addition. The conjugate of a + bi is a - bi (flip only the sign of the imaginary part, keep real part the same): for 2 + 5i, conjugate is 2 - 5i; for 1 - 3i, conjugate is 1 + 3i. The property states that the conjugate of a sum equals the sum of the conjugates: conjugate of (z + w) = conjugate of z + conjugate of w. To solve: First find z + w = (2 + 5i) + (1 - 3i) = 3 + 2i. Then find the conjugate of this sum: conjugate of (3 + 2i) = 3 - 2i. Alternatively, use the property: conjugate of z = 2 - 5i, conjugate of w = 1 + 3i, so their sum is (2 - 5i) + (1 + 3i) = 3 - 2i. Choice C correctly identifies 3 - 2i as the conjugate of the sum. Choice A shows the sum z + w = 3 + 2i but forgets to take the conjugate—we need the conjugate of this sum, not the sum itself! Remember: the bar notation means 'take the conjugate,' which flips the sign of the imaginary part. Conjugate quick reference: to find conjugate of a + bi, keep a the same, flip the sign of bi to get a - bi. The conjugate property for sums is powerful: it lets you either add first then conjugate, or conjugate first then add—both give the same answer! This property extends to differences too: conjugate of (z - w) = conjugate of z - conjugate of w.
7. Let $z = 3 - 4i$. What is the complex conjugate $\overline{z}$? (Recall: the conjugate of $a+bi$ is $a-bi$.)
   Answer: $3 + 4i$ 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): for 3 + 4i, conjugate is 3 - 4i; for 2 - 5i, conjugate is 2 + 5i. For z = 3 - 4i, we identify a = 3 (real part) and b = -4 (coefficient of i), so the conjugate is 3 - (-4)i = 3 + 4i, keeping the real part 3 unchanged and flipping the sign of the imaginary part from -4i to +4i. Choice B correctly identifies the conjugate by flipping only the imaginary part's sign from negative to positive. Choice C incorrectly flips the sign of BOTH parts: conjugate of 3 - 4i is 3 + 4i, NOT -3 - 4i—only flip the imaginary part's sign, keep the real part unchanged! Conjugate quick reference: to find conjugate of a + bi, keep a the same, flip the sign of bi to get a - bi; to find conjugate of a - bi, keep a, flip to get a + bi. The pattern: positive imaginary becomes negative, negative imaginary becomes positive, real part never changes—visually on complex plane: conjugate is mirror image across horizontal (real) axis.
8. Simplify using conjugates and write your answer in $a+bi$ form: $\frac{3+i}{2-i}.$
   Answer: $1+i$ 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 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!). To divide (3 + i) by (2 - i) using conjugates: (1) Identify conjugate of denominator: conjugate of 2 - i is 2 + i. (2) Multiply numerator and denominator by this conjugate: [(3 + i)(2 + i)] / [(2 - i)(2 + i)]. (3) Multiply numerator using FOIL: (3 + i)(2 + i) = 6 + 3i + 2i + i² = 6 + 5i - 1 = 5 + 5i. (4) Multiply denominator using conjugate property: (2 - i)(2 + i) = 2² + 1² = 4 + 1 = 5. (5) Divide: (5 + 5i)/5 = 5/5 + 5i/5 = 1 + i. Choice A correctly shows 1 + i as the result of this division process. Choice B shows 1/5 + 1/5i, which makes the error of dividing by 5 twice or misunderstanding how to simplify the fraction—when the numerator is 5 + 5i and denominator is 5, we divide each part by 5 to get 1 + i, not 1/5 + 1/5i! Conjugate division tip: after multiplying by the conjugate, your denominator will always be a² + b² (a positive real number), making the final division straightforward. Always double-check by multiplying your answer by the original denominator—you should get back the original numerator!
9. For which value of $k$ will $\frac{3 + ki}{1 + 2i}$ result in a purely real number?
   Answer: $k = 6$ To find when the quotient is purely real, multiply by the conjugate: $\frac{3 + ki}{1 + 2i} \cdot \frac{1 - 2i}{1 - 2i} = \frac{(3 + ki)(1 - 2i)}{(1 + 2i)(1 - 2i)} = \frac{3 - 6i + ki - 2ki^2}{1 - 4i^2} = \frac{3 - 6i + ki + 2k}{1 + 4} = \frac{(3 + 2k) + (k - 6)i}{5}$. For this to be purely real, the imaginary part must be zero: $k - 6 = 0$, so $k = 6$. 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.
10. Use conjugates to rationalize the denominator and write in standard form $a+bi$: $\frac{2-i}{3+2i}.$
    Answer: $\frac{4}{13}-\frac{7}{13}i$ 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. To divide $2 - i$ by $3 + 2i$ using conjugates: (1) Identify conjugate of denominator: conjugate of $3 + 2i$ is $3 - 2i$. (2) Multiply numerator and denominator by this conjugate: $[(2 - i)(3 - 2i)]$ divided by $[(3 + 2i)(3 - 2i)]$. (3) Multiply numerator using FOIL: $(2 - i)(3 - 2i) = 6 - 4i - 3i + 2i^2 = 6 - 7i + 2(-1) = 4 - 7i$. (4) Multiply denominator using conjugate property: $(3 + 2i)(3 - 2i) = 3^2 + 2^2 = 9 + 4 = 13$. (5) Divide: $(4 - 7i)/13 = 4/13 - 7i/13$. Choice A correctly identifies $4/13 - 7i/13$ by properly multiplying and simplifying both numerator and denominator. Choice B has the wrong sign on the imaginary part, suggesting a sign error in the FOIL multiplication. Division recipe: always check your signs carefully when multiplying complex numbers!
11. Simplify using the conjugate multiplication property $(a+bi)(a-bi)=a^2+b^2$: $(6- i)(6+i).$
    Answer: $37$ 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!
12. Multiply a complex number by its conjugate to get a real number: Compute $(4 + 3i)(4 - 3i)$. (Hint: $(a+bi)(a-bi)=a^2+b^2$.)
    Answer: $25$ 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 multiply to a real number. The conjugate of a + bi is a - bi; multiplying them gives (a + bi)(a - bi) = a² + b², a real positive result since imaginary parts cancel. For (4 + 3i)(4 - 3i), it's 4² + 3² = 16 + 9 = 25, using the formula directly. You could also expand: 4*4 = 16, 4*(-3i) = -12i, 3i*4 = 12i, 3i*(-3i) = -9i² = 9 (since i² = -1), so 16 + 9 = 25. Choice C correctly applies the conjugate property to get the real number 25. Choice B forgets to simplify i² to -1, leaving it as 16 - 9i² without converting to 16 + 9 = 25—always remember i² = -1! The (a + bi)(a - bi) = a² + b² pattern is worth memorizing: it's the modulus squared and eliminates i. Practice with examples like (2 + i)(2 - i) = 4 + 1 = 5 to see the pattern—great job, keep practicing!
13. In an AC circuit, the impedance is $Z=4+3i$ ohms. A common calculation uses $Z\overline{Z}$, which equals the modulus squared. Compute $Z\overline{Z}$.
    Answer: $25$ 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!
14. The complex conjugate of $a+bi$ is $a-bi$ (flip the sign of the imaginary part only). What is the conjugate of $3-4i$?
    Answer: $3+4i$ 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): for 3 + 4i, conjugate is 3 - 4i; for 2 - 5i, conjugate is 2 + 5i. For the complex number 3 - 4i, we keep the real part 3 the same and flip the sign of the imaginary part from -4i to +4i, giving us 3 + 4i as the conjugate. Choice B correctly identifies 3 + 4i by flipping only the imaginary part's sign from negative to positive. Choice C incorrectly flips the sign of BOTH parts when finding the conjugate: conjugate of 3 - 4i is 3 + 4i, NOT -3 - 4i. Only flip the imaginary part's sign (the part with i), keep the real part unchanged! Conjugate quick reference: to find conjugate of a + bi, keep a the same, flip the sign of bi to get a - bi. To find conjugate of a - bi, keep a, flip to get a + bi.
15. Let $z = -3 + 8i$. Which value equals $z\overline{z}$? (Recall $z\overline{z}=a^2+b^2$, the modulus squared.)
    Answer: $73$ This question tests your understanding of complex conjugates—pairs of complex numbers that differ only in the sign of their imaginary part—and that z \overline{z} equals the modulus squared, a real number. The conjugate of a + bi is a - bi; z \overline{z} = (a + bi)(a - bi) = a² + b². For z = -3 + 8i, conjugate is -3 - 8i; product (-3)² + 8² =9 +64=73. Expanding: (-3)(-3)=9, (-3)(-8i)=24i, 8i*(-3)=-24i, 8i*(-8i)=-64i²=64; 9+64=73. Choice C correctly gives 73, the real positive value. Choice A is -9 +64=55, but that's not the product; perhaps partial calculation without multiplying properly. Always use the a² + b² formula for z \overline{z}, noting the real part's sign doesn't matter since it's squared. Try with 1 + i: (1 + i)(1 - i)=1+1=2—perfect practice!
16. Simplify and write in standard form $a+bi$: $\frac{2-i}{3+2i}.$
    Answer: $\frac{4}{13}-\frac{7}{13}i$ This question tests your understanding of using complex conjugates to divide complex numbers by rationalizing the denominator. The key concept is to multiply numerator and denominator by the conjugate of the denominator to make the denominator real. The conjugate of $c + di$ is $c - di$, and $(c + di)(c - di) = c^2 + d^2$. For example, to divide by $2 + i$, multiply by $2 - i$, den becomes $4 + 1 = 5$. For $(2 - i)/(3 + 2i)$, conjugate of denominator is $3 - 2i$, multiply: numerator $(2 - i)(3 - 2i) = 4 - 7i$, denominator $9 + 4 = 13$, so $\frac{4}{13} - \frac{7}{13} i$. Choice A correctly uses the conjugate and performs the algebra accurately, remembering $i^2 = -1$. Choice B fails by perhaps switching the signs in the imaginary part during multiplication, leading to a positive imaginary. When multiplying, use FOIL carefully and substitute $i^2 = -1$. Double-check signs and terms. You got this—practice a few more to master it!
17. Compute $(4 + 3i)(4 - 3i)$. (Use the key property $(a+bi)(a-bi)=a^2+b^2$.)
    Answer: $25$ This question tests your understanding of complex conjugates—pairs of complex numbers that differ only in the sign of their imaginary part—and how multiplying by a conjugate yields a real number, specifically the modulus squared. The key property is (a + bi)(a - bi) = a² + b², where the imaginary parts cancel out, and since i² = -1, it simplifies to a real positive result: for instance, (3 + 2i)(3 - 2i) = 9 + 4 = 13. For (4 + 3i)(4 - 3i), identify a = 4 and b = 3, so it's 4² + 3² = 16 + 9 = 25, using the conjugate property directly without expanding fully. Choice C correctly applies this property to get 25, recognizing it's the square of the modulus |4 + 3i|². A tempting distractor like Choice B might stop at 16 - 9i² without simplifying i² to -1, but remember to always convert i² terms: 16 - 9(-1) = 16 + 9 = 25. Memorize the (a + bi)(a - bi) = a² + b² pattern—it's a shortcut for rationalizing denominators in division too! Keep practicing these multiplications, and you'll master complex numbers in no time.
18. In AC circuits, an impedance can be written as $Z=4+3i$ ohms. Compute $Z \overline{Z}$.
    Answer: $25$ 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 $Z \overline{Z} = |Z|^2$, which is a real number representing the square of the modulus in contexts like AC circuits. For $Z = 4 + 3i$, the conjugate $\overline{Z} = 4 - 3i$, and $Z \overline{Z} = (4 + 3i)(4 - 3i) = 16 + 9 = 25$, using $(a + bi)(a - bi) = a^2 + b^2$. Choice C correctly computes this as 25, recognizing it's the modulus squared. A tempting distractor like Choice D might forget to simplify with $i^2 = -1$ 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: $|Z| = \sqrt{a^2 + b^2}$, so $|Z|^2 = a^2 + b^2$ directly. You're making fantastic progress—try it with $Z = 5 + 12i$ to get $25 + 144 = 169$!
19. Let $z=4+3i$. Which expression equals $z\overline{z}$ (the modulus squared $|z|^2$), and what is its value?
    Answer: $(4+3i)(4-3i)=25$ This question tests your understanding of complex conjugates—pairs of complex numbers that differ only in the sign of their imaginary part—and how multiplying a complex number by its conjugate gives the modulus squared. The conjugate of a + bi is a - bi; 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!), which equals |z|², the square of the modulus (distance from origin). For z = 4 + 3i, the conjugate z̄ = 4 - 3i, so z·z̄ = (4 + 3i)(4 - 3i) = 4² + 3² = 16 + 9 = 25; this represents |z|² = |4 + 3i|² = 25, meaning the distance from origin to the point (4, 3) is √25 = 5. Choice A correctly shows both the expression (4 + 3i)(4 - 3i) = 25 and its value. Choice C shows (4 + 3i)(4 + 3i), which is z·z (squaring the number), not z·z̄—this would give 16 + 24i + 9i² = 16 + 24i - 9 = 7 + 24i, a complex result rather than the real positive value we need! The product z·z̄ = |z|² is fundamental in complex analysis: it's always real and non-negative, equals zero only when z = 0, and geometrically represents the square of the distance from the origin. This property is crucial for defining division (multiply numerator and denominator by conjugate of denominator) and for proving many theorems about complex functions!
20. Let $z=1-2i$ and $w=3+i$. Compute $\overline{zw}$. (You may use $\overline{zw}=\overline{z}\,\overline{w}$.)
    Answer: $5+5i$ 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!

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