Operations with Complex Numbers - Algebra 2

Let $z_1 = 5 + 3i$ and $z_2 = 2 - 7i$ . What is $z_1 + z_2$ expressed in standard form $a + bi$ ?
Answer: $7 - 4i$ 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 = 4 - 3i. To add z₁ = 5 + 3i and z₂ = 2 - 7i: combine real parts: 5 + 2 = 7, combine imaginary parts: 3i + (-7i) = -4i, so z₁ + z₂ = 7 - 4i. Choice A correctly performs the complex number addition and expresses the result in standard form a + bi with proper handling of combining like terms. Choice B incorrectly subtracts the real parts (5 - 2 = 3) instead of adding them, while correctly adding the imaginary parts—remember addition means add both parts! 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! The key is keeping real and imaginary parts separate during operations, then combining at the end.
Find the modulus (absolute value) of the complex number $3 + 4i$ . (Recall $|a+bi|=\sqrt{a^2+b^2}$ .)
Answer: $5$ This question tests your understanding of complex numbers—numbers in the form $a + bi$ where i is the imaginary unit with $i^2 = -1$ —and finding the modulus. Complex numbers extend the real number system to include square roots of negative numbers using $i = \sqrt{-1}$ , so $i^2 = -1$ . The modulus $|a + bi| = \sqrt{a^2 + b^2}$ , like distance from origin. For $3 + 4i$ , $|3 + 4i| = \sqrt{9 + 16} = \sqrt{25} = 5$ —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!
Using the fact that $(a+bi)(a-bi)=a^2+b^2$ , compute $(2+5i)(2-5i)$ .
Answer: $29$ This question tests your understanding of complex numbers—numbers in form $a + bi$ where $i$ is the imaginary unit with $i^2 = -1$ —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 $i = \sqrt{-1}$ , so $i^2 = -1$ . The formula $(a + bi)(a - bi) = a^2 + b^2$ eliminates the imaginary part; for example, $(3 + 2i)(3 - 2i) = 9 + 4 = 13$ . For $(2 + 5i)(2 - 5i)$ , it's $4 + 25 = 29$ , since $(5i)^2 = 25i^2 = 25 \times (-1) = -25$ , so $4 - (-25) = 29$ . Choice B correctly computes the product using the conjugate formula and simplifies with $i^2 = -1$ . Choice A forgets to replace $i^2$ with -1: treating $(5i)^2$ as +25 instead of -25 leads to $4 + (-25)$ wait no, but actually misapplying as $4 - 25 = -21$ —the key is the difference of squares gives positive! This property is key for moduli and division: $|z|^2 = z \times \text{conjugate}(z)$ —you're grasping these concepts beautifully, keep it up!
Multiply and simplify using $i^2 = -1$ : $(3 + 2i)(1 - 4i)$ . Express your answer in standard form $a + bi$ .
Answer: $11 - 10i$ 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.
Divide and express in standard form $a + bi$ : $\frac{3 + 2i}{1 - i}$ (Use the conjugate of the denominator.)
Answer: $\frac{1}{2} + \frac{5}{2}i$ 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.
Solve the quadratic and express all solutions as complex numbers: $x^2+2x+10=0.$
Answer: $x=-1\pm 3i$ This question tests your understanding of complex numbers—numbers in the form $a + bi$ where $i$ is the imaginary unit with $i^2 = -1$ —and solving quadratics with complex roots using the quadratic formula. Complex numbers extend the real number system to include square roots of negative numbers using $i = \sqrt{-1}$ , so $i^2 = -1$ . For $x^2 + 2x + 10 = 0$ , discriminant $4 - 40 = -36$ , $\sqrt{-36} = 6i$ , so $x = \frac{-2 \pm 6i}{2} = -1 \pm 3i$ —excellent! Choice A correctly expresses the solutions in standard form. Choice C might mishandle the square root of -36 as $\sqrt{36} i$ but forget to adjust, but it's $\sqrt{36} i = 6i$ . Remember: negative discriminant means $\pm i \sqrt{|d|}/(2a)$ after real part— you've got the pattern! This makes all quadratics solvable; you're doing wonderfully.
Solve the quadratic equation $x^2 + 2x + 10 = 0$ and express the solutions as complex numbers (standard form $a + bi$ ).
Answer: $x = -1 \pm 3i$ 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!
Let $z = 2 + 5i$ . What is the product of $z$ and its conjugate $\overline{z}$ ? (Recall $(a+bi)(a-bi)=a^2+b^2$ .)
Answer: $29$ 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!
Find the modulus (absolute value) of the complex number $z = 3 + 4i$ . (Recall $|a+bi| = \sqrt{a^2+b^2}$ .)
Answer: $5$ 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 modulus (absolute value) of a complex number z = a + bi is |z| = √(a² + b²), which represents the distance from the origin to the point (a, b) in the complex plane. For z = 3 + 4i: |z| = √(3² + 4²) = √(9 + 16) = √25 = 5. The modulus is always a non-negative real number! Choice B correctly calculates the modulus using the formula |a + bi| = √(a² + b²) and simplifies √25 = 5. Choice A incorrectly calculates √(3² + 4²) as √7 instead of √25—remember to add the squares first: 9 + 16 = 25, then take the square root! Don't take square roots separately. Complex number operations summary: The modulus |a + bi| = √(a² + b²) measures the distance from origin and satisfies |z₁z₂| = |z₁||z₂| and |z/w| = |z|/|w|. These patterns are consistent! The modulus connects to the conjugate: |z|² = z × z̄ where z̄ is the conjugate, so for 3 + 4i: |z|² = (3 + 4i)(3 - 4i) = 9 + 16 = 25, confirming |z| = 5.
If $z = a + bi$ where $a$ and $b$ are real numbers, and $z^2 = -5 + 12i$ , what is the value of $a^2 + b^2$ ?
Answer: $13$ We have $z^2 = (a+bi)^2 = a^2 + 2abi + b^2i^2 = a^2 - b^2 + 2abi = -5 + 12i$ . Equating real and imaginary parts: $a^2 - b^2 = -5$ and $2ab = 12$ , so $ab = 6$ . We want to find $a^2 + b^2$ . Note that $|z|^2 = a^2 + b^2$ and $|z^2| = |z|^2$ . Since $z^2 = -5 + 12i$ , we have $|z^2| = \sqrt{(-5)^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$ . But this gives us $|z|^2 = \sqrt{13}$ , which seems wrong. Actually, $|z^2| = |z|^2$ is incorrect; we should have $|z^2| = |z|^2$ . Let me reconsider: if $z^2 = -5 + 12i$ , then $|z^2| = \sqrt{25 + 144} = 13$ . Since $|z^2| = |z|^2$ , we have $|z|^2 = \sqrt{13}$ . No wait, $|z^2| = (|z|)^2$ , so $(|z|)^2 = 13$ , which means $|z| = \sqrt{13}$ . Therefore, $a^2 + b^2 = |z|^2 = 13$ . Choice B (17) might come from $(-5)^2 - 12 = 25 - 12 = 13$ , no that gives 13. Choice C (25) is $(-5)^2$ . Choice D (169) is $(13)^2$ .
Use $i^2 = -1$ to simplify the product $(3 + 2i)(1 - 4i)$ and write the result in the form $a + bi$ .
Answer: $11 - 10i$ 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) = 3*1 + 3*(-4i) + 2i*1 + 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!
What is the conjugate of the complex number $-3+8i$ ?
Answer: $-3-8i$ This question tests your understanding of complex numbers—numbers in form $a + bi$ where i is the imaginary unit with $i^2 = -1$ —and how to find the conjugate. Complex numbers extend the real number system to include square roots of negative numbers using $i = \sqrt{-1}$ , so $i^2 = -1$ . The conjugate of $a + bi$ is $a - bi$ , flipping the sign of the imaginary part only; for example, conjugate of $3 + 4i$ is $3 - 4i$ . For $-3 + 8i$ , the conjugate is $-3 - 8i$ , 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 $-3 + 8i$ to $+3 - 8i$ , 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!
Solve the quadratic equation $x^2 + 2x + 10 = 0$ and express the solutions as complex numbers in $a + bi$ form.
Answer: $x = -1 \pm 3i$ 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. For quadratic equations, when the discriminant b² - 4ac is negative, solutions involve complex numbers! For x² + 2x + 10 = 0: a = 1, b = 2, c = 10. Discriminant = 2² - 4(1)(10) = 4 - 40 = -36. Since discriminant is negative, we get complex solutions: x = (-b ± √(b² - 4ac))/(2a) = (-2 ± √(-36))/2 = (-2 ± 6i)/2 = -1 ± 3i. The √(-36) = √(36 × -1) = 6i! Choice A correctly applies the quadratic formula with complex numbers and expresses solutions in standard form a + bi as x = -1 + 3i and x = -1 - 3i. Choice B has the wrong sign for the real part: -b/(2a) = -2/2 = -1, not +1—don't drop the negative from -b in the quadratic formula! Complex number operations summary: For complex solutions to quadratics: when discriminant b² - 4ac is negative, solutions are x = (-b ± √(b² - 4ac))/(2a) = (-b ± i√|b² - 4ac|)/(2a). Write in a + bi form separating real part (-b/2a) from imaginary part (±√|discriminant|/2a × i). Complex numbers make every quadratic solvable!
Let $z_1 = 5 + 3i$ and $z_2 = 2 - 7i$ . What is $z_1 + z_2$ in standard form $a+bi$ ?
Answer: $7-4i$ 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!
Simplify using $i^2 = -1$ : $i^3 + i^4.$
Answer: $1 - i$ 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 simplify powers of i using the pattern that repeats every 4 powers. 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 powers of i follow a pattern: i¹ = i, i² = -1, i³ = i² × i = -1 × i = -i, i⁴ = (i²)² = (-1)² = 1, then the pattern repeats. So i³ = -i and i⁴ = 1, giving i³ + i⁴ = -i + 1 = 1 - i. Choice B correctly calculates i³ + i⁴ = -i + 1 = 1 - i in standard a + bi form. Choice A incorrectly calculates i³ = -1 instead of -i (confusing it with i²), and i⁴ = -i instead of 1, getting -1 - i instead of 1 - i. The key is remembering the cycle: i, -1, -i, 1, then repeat—powers of i are periodic with period 4!
Use the conjugate property to simplify. What is $(2+5i)(2-5i)?$
Answer: $29$ This question tests your understanding of complex numbers—numbers in the form $a + bi$ where i is the imaginary unit with $i^2 = -1$ —and using the conjugate property for multiplication. Complex numbers extend the real number system to include square roots of negative numbers using $i = \sqrt{-1}$ , so $i^2 = -1$ . The product $(a + bi)(a - bi) = a^2 + b^2$ , always real: here $(2 + 5i)(2 - 5i) = 4 + 25 = 29$ —outstanding! Choice B correctly applies this difference of squares pattern. Choice A might come from mistakenly subtracting instead of adding the squares, but remember it's plus b² from $- (bi)^2 = -b^2 i^2 = +b^2$ . This is key for moduli and division too— you've nailed it! Build on this for more complex problems; you're unstoppable.
Simplify using $i^2=-1$ : $i^3 + i^4.$
Answer: $1-i$ This question tests your understanding of complex numbers—numbers in the form $a + bi$ where $i$ is the imaginary unit with $i^2 = -1$ —and simplifying powers of $i$ using the cycle. Complex numbers extend the real number system to include square roots of negative numbers using $i = \sqrt{-1}$ , so $i^2 = -1$ . Powers of $i$ cycle every 4: $i^1 = i$ , $i^2 = -1$ , $i^3 = -i$ , $i^4 = 1$ , then repeat. So $i^3 + i^4 = -i + 1 = 1 - i$ —brilliant! Choice B might confuse $i^3$ with $i$ , but use the cycle to verify. Always find the exponent mod 4 for quick simplification—you're getting it! This helps in polynomials and beyond; keep the momentum!
Using $i^2=-1$ , simplify the product $(3+2i)(1-4i)$ and write the result in standard form $a+bi$ .
Answer: $11-10i$ 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 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) = 3*1 + 3*(-4i) + 2i*1 + 2i*(-4i) = 3 - 12i + 2i - $8i^2$ = 3 - 10i + 8 = 11 - 10i, since - $8i^2$ = -8*(-1) = +8. 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: leaving - $8i^2$ as is would be incorrect, but you must simplify to +8—the i squared simplification is mandatory! Complex number multiplication summary: use FOIL or distribution, then simplify i squared to -1, and remember the pattern repeats for powers of i—you're building a strong foundation here, fantastic effort!
Add and express in standard form $a+bi$ : $(5+3i)+(2-7i).$
Answer: $7-4i$ This question tests your understanding of complex numbers—numbers in the form $a + bi$ where i is the imaginary unit with $i^2 = -1$ —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 $i = \sqrt{-1}$ , so $i^2 = -1$ . 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: just like grouping like terms in algebra! Here, adding $(5 + 3i) + (2 - 7i)$ gives real parts $5 + 2 = 7$ and imaginary parts $3 - 7 = -4$ , so $7 - 4i$ —great job recognizing that! A common mistake, like in choice D, might come from subtracting instead of adding or messing up signs, but remember to treat it as combining positives and negatives carefully. Keep practicing: addition and subtraction of complex numbers are straightforward— $(a + bi) + (c + di) = (a + c) + (b + d)i$ —and you'll master it quickly! You're building a strong foundation for more advanced operations like multiplication.
When $(2 + 3i)^2$ is written in standard form $a + bi$ , what is the value of $a - b$ ?
Answer: $-17$ Expanding $(2 + 3i)^2 = (2 + 3i)(2 + 3i) = 4 + 6i + 6i + 9i^2 = 4 + 12i + 9(-1) = 4 + 12i - 9 = -5 + 12i$ . Therefore, $a = -5$ and $b = 12$ , so $a - b = -5 - 12 = -17$ . Choice A results from calculating $a + b = -5 + 12 = 7$ and then taking the negative. Choice B comes from the error $4 - 12 = -8$ . Choice D is $b - a = 12 - (-5) = 17$ with a sign error.

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