Algebra 2 Complex Numbers: Adding, Subtracting and Multiplying Quiz: 10 Questions with Explanations

A \(9i\)B \(-9\)C \(9\)D \(81i\)

\(\sqrt{-81} = \sqrt{81}\cdot i = 9i\). The square root of a negative number is never real, so \(9\) and \(-9\) are wrong, and 81 must be square rooted.

A \(4 – 2i\)B \(2 – 4i\)C \(-2 – 4i\)D \(-2 + 4i\)

The horizontal coordinate is the real part (-2) and the vertical coordinate is the imaginary part (4). Swapping the axes gives 4 – 2i.

A \(-i\)B \(i\)C \(-1\)D \(1\)

\(43 = 4\cdot 10 + 3\), so \(i^{43} = i^{3} = -i\). A common mistake is to use the remainder as if it were the exponent of 1.

A \(5 + 4i\)B \(5 – 14i\)C \(11 + 4i\)D \(5 + 14i\)

Add the real parts: 8 + (-3) = 5. Add the imaginary parts: -5 + 9 = 4. Mixing up the signs gives the wrong imaginary part.

A \(-3 + 3i\)B \(7 + 3i\)C \(-3 – 9i\)D \(-3 + 9i\)

Distribute the minus sign: 2 + 6i – 5 + 3i. The real part is -3 and the imaginary part is 6 + 3 = 9. Forgetting to change the sign of -3i gives -3 + 3i.

A \((3 + 2i)(4 + i)\)B \((3 + 2i)(1 + 4i)\)C \((3 + 4i)(1 + 2i)\)D \((3 – 2i)(1 + 4i)\)

The side labels are 3 and 2i on the left, and 1 and 4i on top, so the product is (3 + 2i)(1 + 4i). The cell values 3, 12i, 2i, and 8i² come from multiplying each pair of labels.

A \(15\)B \(16 – i\)C \(17\)D \(16 + 2i\)

\(16 + 4i – 4i – i^{2} = 16 + 1 = 17\). The imaginary terms cancel and \(-i^{2} = +1\), so the answer is 17, not 15.

A \(4 – 9i\)B \(13 – 12i\)C \(-5 – 12i\)D \(-5 + 12i\)

\(4 – 6i – 6i + 9i^{2} = 4 – 9 – 12i = -5 – 12i\). Squaring each term separately gives 4 – 9i, which skips the middle terms.

A \(i\)B \(-1\)C \(1\)D \(-i\)

\(6 = 4\cdot 1 + 2\), so \(i^{6} = i^{2} = -1\). The second box of the cycle is \(-1\).

A \(x = 6 and x = -6\)B \(x = 6i and x = -6i\)C \(x = 36i only\)D \(There are no solutions\)

\(x^{2} = -36\) gives \(x = \pm 6i\). There are no real solutions, but there are two complex solutions.