Algebra 2 Complex Numbers: Adding, Subtracting and Multiplying Worksheet Solutions (PDF)

Algebra 2 math worksheet solutions: Complex Numbers: Adding, Subtracting and Multiplying – Maths PDF

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This algebra 2 complex numbers answers for Algebra 2 covers complex numbers: adding, subtracting and multiplying (Common Core N-CN.A.1, N-CN.A.2), with a free PDF to print.

Every worksheet problem on adding, subtracting and multiplying complex numbers is solved here, one step at a time. Compare your method as well as your final answer, because the steps show where a forgotten sign or a missed i-squared can go wrong. Final answers appear in bold so you can find them quickly when you only need a check.

Algebra 2 · Common Core State Standards: N-CN.A.1, N-CN.A.2

Algebra 2 complex numbers answers: exercises 1 to 10

Exercise 1 – Square root of a negative

\(\sqrt{-49} = \sqrt{49}\cdot\sqrt{-1} = 7i\).

\(7i\)

Exercise 2 – Radical with a factor

\(\sqrt{-18} = i\sqrt{18}\). Since \(18 = 9 \cdot 2\), \(\sqrt{18} = 3\sqrt{2}\).

\(3i\sqrt{2}\)

Exercise 3 – Real and imaginary parts

The point is 5 units right and 3 units down.

The number \(5 – 3i\) equals \(5 + (-3)i\). The real part is the number with no \(i\), and the imaginary part is the coefficient of \(i\).

Real part \(5\), imaginary part \(-3\)

Exercise 4 – Power of i

Divide the exponent by 4: \(7 = 4 \cdot 1 + 3\), so the remainder is 3.

\(i^{7} = i^{3} = -i\).

\(-i\)

Exercise 5 – Large power of i

Divide: \(58 = 4 \cdot 14 + 2\), so the remainder is 2.

\(i^{58} = i^{2} = -1\).

\(-1\)

Exercise 6 – Add complex numbers

Real parts: \(3 + 2 = 5\). Imaginary parts: \(5 – 9 = -4\).

\(5 – 4i\)

Exercise 7 – Subtract complex numbers

Distribute the minus sign: \(7 – 2i – 4 + 6i\).

Real parts: \(7 – 4 = 3\). Imaginary parts: \(-2 + 6 = 4\).

\(3 + 4i\)

Exercise 8 – Multiply by i

Distribute: \(3i \cdot 2 – 3i \cdot 5i = 6i – 15i^{2}\).

Replace \(i^{2}\) with \(-1\): \(6i + 15\).

\(15 + 6i\)

Exercise 9 – Multiply two binomials

Partial products 8, -2i, 12i, and -3i² = 3.

\(8 – 2i + 12i – 3i^{2}\). With \(i^{2} = -1\), \(-3i^{2} = 3\).

Real: \(8 + 3 = 11\). Imaginary: \(-2 + 12 = 10\).

\(11 + 10i\)

Exercise 10 – Product with a conjugate pair

\(25 – 10i + 10i – 4i^{2} = 25 – 4(-1) = 29\).

The imaginary parts cancel, so the product is real.

\(29\)

Exercises 11 and up

Exercise 11 – Square a complex number

\((3 – 4i)(3 – 4i) = 9 – 12i – 12i + 16i^{2}\).

\(16i^{2} = -16\), so \(9 – 16 – 24i = -7 – 24i\).

\(-7 – 24i\)

Exercise 12 – Plot complex numbers

Points (2, -3) and (-4, 1).

Real part is the x-coordinate and imaginary part is the y-coordinate.

\(2 – 3i\) is at \((2, -3)\) and \(-4 + i\) is at \((-4, 1)\)

Exercise 13 – Add as vectors

The sum is the fourth corner of the parallelogram.

\((3 + 1) + (2 + 4)i = 4 + 6i\).

\(4 + 6i\), at the point \((4, 6)\)

Exercise 14 – Solve with imaginary roots

Subtract 25: \(x^{2} = -25\).

Take square roots: \(x = \pm\sqrt{-25} = \pm 5i\).

\(x = 5i\) or \(x = -5i\)

Exercise 15 – Mixed powers

\(i^{3} = -i\), \(i^{2} = -1\), \(i^{4} = 1\).

\(2(-i) + 5(-1) – 1 = -2i – 5 – 1\).

\(-6 – 2i\)

Exercise 16 – Fourth power

\((1 + i)^{2} = 1 + 2i + i^{2} = 2i\).

\((2i)^{2} = 4i^{2} = -4\).

\(-4\)

Exercise 17 – Series circuit

\(Z = (3 + 5) + (4 – 1)i\).

\(8 + 3i\) ohms

Exercise 18 – Voltage from current

\(V = (2 + i)(3 + 4i) = 6 + 8i + 3i + 4i^{2}\).

\(4i^{2} = -4\), so \(6 – 4 + 11i\).

\(2 + 11i\) volts

Exercise 19 – Rotation by i

Multiplying by i turns the point a quarter turn counterclockwise.

\(i(3 + 2i) = 3i + 2i^{2} = -2 + 3i\).

The point \((3, 2)\) moves to \((-2, 3)\): a 90 degree counterclockwise rotation about the origin, and the distance from the origin stays the same.

\(-2 + 3i\): a quarter turn counterclockwise

Exercise 20 – Make a real product

\((2 + 3i)(1 + ki) = 2 + 2ki + 3i + 3ki^{2} = (2 – 3k) + (2k + 3)i\).

The product is real when the imaginary part is 0: \(2k + 3 = 0\), so \(k = -\frac{3}{2}\).

\(k = -\frac{3}{2}\)

Exercise 21 – Find a square root

\((a + bi)^{2} = (a^{2} – b^{2}) + 2abi\).

So \(a^{2} – b^{2} = -5\) and \(2ab = 12\), giving \(ab = 6\).

Try \(a = 2\), \(b = 3\): \(4 – 9 = -5\) and \(2 \cdot 2 \cdot 3 = 12\). Both conditions hold.

\(a = 2\), \(b = 3\)

Key facts behind this algebra 2 complex numbers answers

Key facts

The imaginary unit i satisfies i² = -1, so √(-n) = i√n for n > 0.
A complex number a + bi has real part a and imaginary part b.
Add or subtract real parts and imaginary parts separately.
Multiply by distributing, then replace i² with -1.
Powers of i repeat: i, -1, -i, 1.

How to use this algebra 2 complex numbers answers

Cover each solution, solve the exercise again, then uncover it and compare every line. If your answer differs, find the first step that changed and redo the exercise from there.

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