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  3. Algebra 2

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Algebra 2 Flashcards: Operations With Complex Numbers

Study Operations With Complex Numbers in Algebra 2 with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.

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What this deck covers

This deck focuses on Operations With Complex Numbers, giving you a quick way to review the definitions, rules, and examples that matter most for Algebra 2.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

Algebra 2 Flashcards: Operations With Complex Numbers

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QUESTION

What is (1−2i)(−4+3i)(1-2i)(-4+3i)(1−2i)(−4+3i) in standard form?

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ANSWER

2+11i2+11i2+11i. Use FOIL: (1)(−4)+(1)(3i)+(−2i)(−4)+(−2i)(3i)=−4+3i+8i+6(1)(-4) + (1)(3i) + (-2i)(-4) + (-2i)(3i) = -4 + 3i + 8i + 6(1)(−4)+(1)(3i)+(−2i)(−4)+(−2i)(3i)=−4+3i+8i+6.

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Flashcard 1: What is (1−2i)(−4+3i)(1-2i)(-4+3i)(1−2i)(−4+3i) in standard form?

Answer: 2+11i2+11i2+11i. Use FOIL: (1)(−4)+(1)(3i)+(−2i)(−4)+(−2i)(3i)=−4+3i+8i+6(1)(-4) + (1)(3i) + (-2i)(-4) + (-2i)(3i) = -4 + 3i + 8i + 6(1)(−4)+(1)(3i)+(−2i)(−4)+(−2i)(3i)=−4+3i+8i+6.

Flashcard 2: What is (10−3i)−(4+9i)(10-3i)-(4+9i)(10−3i)−(4+9i) in standard form?

Answer: 6−12i6-12i6−12i. Subtract real parts: 10−4=610-4=610−4=6, imaginary parts: −3−9=−12-3-9=-12−3−9=−12.

Flashcard 3: What is (bi)(di)(bi)(di)(bi)(di) simplified using i2=−1i^2=-1i2=−1?

Answer: −bd-bd−bd. Since (bi)(di)=bdi2=bd(−1)=−bd(bi)(di) = bdi^2 = bd(-1) = -bd(bi)(di)=bdi2=bd(−1)=−bd.

Flashcard 4: What is the value of i3i^3i3 written in simplest form?

Answer: −i-i−i. Since i3=i2⋅i=(−1)⋅i=−ii^3 = i^2 \cdot i = (-1) \cdot i = -ii3=i2⋅i=(−1)⋅i=−i.

Flashcard 5: What is the additive identity for complex numbers?

Answer: 0+0i0+0i0+0i. Adding this to any complex number yields the same number.

Flashcard 6: What is the difference (a+bi)−(c+di)(a+bi)-(c+di)(a+bi)−(c+di) in standard form?

Answer: (a−c)+(b−d)i(a-c)+(b-d)i(a−c)+(b−d)i. Subtract real parts and imaginary parts separately.

Flashcard 7: What is (3+2i)+(5−7i)(3+2i)+(5-7i)(3+2i)+(5−7i) in standard form?

Answer: 8−5i8-5i8−5i. Add real parts: 3+5=83+5=83+5=8, imaginary parts: 2+(−7)=−52+(-7)=-52+(−7)=−5.

Flashcard 8: What is the product (a+bi)(c+di)(a+bi)(c+di)(a+bi)(c+di) in standard form?

Answer: (ac−bd)+(ad+bc)i(ac-bd)+(ad+bc)i(ac−bd)+(ad+bc)i. Use FOIL and i2=−1i^2 = -1i2=−1 to simplify.

Flashcard 9: What is the value of i2i^2i2?

Answer: −1-1−1. By definition of the imaginary unit.

Flashcard 10: What is (a+bi)+(c−di)(a+bi)+(c-di)(a+bi)+(c−di) simplified in standard form?

Answer: (a+c)+(b−d)i(a+c)+(b-d)i(a+c)+(b−d)i. Add real parts and subtract imaginary parts.

Flashcard 11: What is (a+bi)−(c−di)(a+bi)-(c-di)(a+bi)−(c−di) simplified in standard form?

Answer: (a−c)+(b+d)i(a-c)+(b+d)i(a−c)+(b+d)i. Distribute the negative sign to both terms inside parentheses.

Flashcard 12: What is (a+bi)(a−bi)(a+bi)(a-bi)(a+bi)(a−bi) simplified using i2=−1i^2=-1i2=−1?

Answer: a2+b2a^2+b^2a2+b2. Conjugate multiplication yields sum of squares.

Flashcard 13: What is (4+3i)(4−3i)(4+3i)(4-3i)(4+3i)(4−3i) in standard form?

Answer: 252525. Difference of squares: (4)2−(3i)2=16−9i2=16+9=25(4)^2 - (3i)^2 = 16 - 9i^2 = 16 + 9 = 25(4)2−(3i)2=16−9i2=16+9=25.

Flashcard 14: What is (3−2i)(−1+i)(3-2i)(-1+i)(3−2i)(−1+i) in standard form?

Answer: −1+5i-1+5i−1+5i. Use FOIL: (3)(−1)+(3)(i)+(−2i)(−1)+(−2i)(i)=−3+3i+2i+2(3)(-1) + (3)(i) + (-2i)(-1) + (-2i)(i) = -3 + 3i + 2i + 2(3)(−1)+(3)(i)+(−2i)(−1)+(−2i)(i)=−3+3i+2i+2.

Flashcard 15: What is the value of i5i^5i5 written in simplest form?

Answer: iii. Since i5=i4⋅i=1⋅i=ii^5 = i^4 \cdot i = 1 \cdot i = ii5=i4⋅i=1⋅i=i.

Flashcard 16: What is (7i)(−3i)(7i)(-3i)(7i)(−3i) simplified?

Answer: 212121. Since (7i)(−3i)=−21i2=−21(−1)=21(7i)(-3i) = -21i^2 = -21(-1) = 21(7i)(−3i)=−21i2=−21(−1)=21.

Flashcard 17: What is the sum (a+bi)+(c+di)(a+bi)+(c+di)(a+bi)+(c+di) in standard form?

Answer: (a+c)+(b+d)i(a+c)+(b+d)i(a+c)+(b+d)i. Add real parts and imaginary parts separately.

Flashcard 18: What is the value of i2i^2i2?

Answer: −1-1−1. By definition of the imaginary unit.

Flashcard 19: What is (a+bi)(a−bi)(a+bi)(a-bi)(a+bi)(a−bi) simplified using i2=−1i^2=-1i2=−1?

Answer: a2+b2a^2+b^2a2+b2. Conjugate multiplication yields sum of squares.

Flashcard 20: What is (−3+4i)−(8+i)(-3+4i)-(8+i)(−3+4i)−(8+i) in standard form?

Answer: −11+3i-11+3i−11+3i. Subtract real parts: −3−8=−11-3-8=-11−3−8=−11, imaginary parts: 4−1=34-1=34−1=3.

Flashcard 21: What is the real part of the complex number a+bia+bia+bi?

Answer: aaa. The coefficient of the non-imaginary term.

Flashcard 22: What is (1+i)(1+i)(1+i)(1+i)(1+i)(1+i) in standard form?

Answer: 2i2i2i. Use FOIL: (1)(1)+(1)(i)+(i)(1)+(i)(i)=1+i+i−1=2i(1)(1) + (1)(i) + (i)(1) + (i)(i) = 1 + i + i - 1 = 2i(1)(1)+(1)(i)+(i)(1)+(i)(i)=1+i+i−1=2i.

Flashcard 23: What is the value of i6i^6i6 written in simplest form?

Answer: −1-1−1. Since i6=(i2)3=(−1)3=−1i^6 = (i^2)^3 = (-1)^3 = -1i6=(i2)3=(−1)3=−1.

Flashcard 24: What is (5−2i)(1+3i)(5-2i)(1+3i)(5−2i)(1+3i) in standard form?

Answer: 11+13i11+13i11+13i. Use FOIL: (5)(1)+(5)(3i)+(−2i)(1)+(−2i)(3i)=5+15i−2i+6(5)(1) + (5)(3i) + (-2i)(1) + (-2i)(3i) = 5 + 15i - 2i + 6(5)(1)+(5)(3i)+(−2i)(1)+(−2i)(3i)=5+15i−2i+6.

Flashcard 25: What is (2+5i)(−3−2i)(2+5i)(-3-2i)(2+5i)(−3−2i) in standard form?

Answer: 4−19i4-19i4−19i. Use FOIL: (2)(−3)+(2)(−2i)+(5i)(−3)+(5i)(−2i)=−6−4i−15i−10(2)(-3) + (2)(-2i) + (5i)(-3) + (5i)(-2i) = -6 - 4i - 15i - 10(2)(−3)+(2)(−2i)+(5i)(−3)+(5i)(−2i)=−6−4i−15i−10.

Flashcard 26: What is (8+0i)−(3+0i)(8+0i)-(3+0i)(8+0i)−(3+0i) in standard form?

Answer: 555. Subtract real parts only: 8−3=58-3=58−3=5, imaginary parts are zero.

Flashcard 27: What is (−5i)+(12i)(-5i)+(12i)(−5i)+(12i) in standard form?

Answer: 7i7i7i. Add imaginary parts: −5i+12i=7i-5i + 12i = 7i−5i+12i=7i.

Flashcard 28: What is (1−i)(1−i)(1-i)(1-i)(1−i)(1−i) in standard form?

Answer: −2i-2i−2i. Use FOIL: (1)(1)+(1)(−i)+(−i)(1)+(−i)(−i)=1−i−i−1=−2i(1)(1) + (1)(-i) + (-i)(1) + (-i)(-i) = 1 - i - i - 1 = -2i(1)(1)+(1)(−i)+(−i)(1)+(−i)(−i)=1−i−i−1=−2i.

Flashcard 29: What is (1+i)(1−i)(1+i)(1-i)(1+i)(1−i) in standard form?

Answer: 222. Difference of squares: (1)2−(i)2=1−i2=1+1=2(1)^2 - (i)^2 = 1 - i^2 = 1 + 1 = 2(1)2−(i)2=1−i2=1+1=2.

Flashcard 30: What is (2−i)(−5−i)(2- i)(-5- i)(2−i)(−5−i) in standard form?

Answer: −11+3i-11+3i−11+3i. Use FOIL: (2)(−5)+(2)(−i)+(−i)(−5)+(−i)(−i)=−10−2i+5i−1(2)(-5) + (2)(-i) + (-i)(-5) + (-i)(-i) = -10 - 2i + 5i - 1(2)(−5)+(2)(−i)+(−i)(−5)+(−i)(−i)=−10−2i+5i−1.