Complex roots of polynomials; roots of complex numbers practice questions.

Which of the following is one of the cube roots of 8i?

• √3 − i
• √3 + i
• 2i
• 1 + √3 i

Answer: √3 + i

8i = 8 cis(π/2), so one cube root is 2 cis(π/6) = 2(√3/2 + i/2) = √3 + i. The others are −√3 + i and −2i.

A polynomial with real coefficients has 2 − 3i as a root. Which number must also be a root?

• −2 + 3i
• −2 − 3i
• 2 + 3i
• 3 − 2i

Answer: 2 + 3i

Real coefficients force non-real roots to occur in conjugate pairs, so the conjugate 2 + 3i must also be a root. −2 + 3i is the negative, not the conjugate.

Which of the following is a solution of z⁴ = −16?

• 2 + 2i
• 2i
• −2
• √2 + √2 i

Answer: √2 + √2 i

−16 = 16 cis(π), so the roots are 2 cis(π/4 + kπ/2); for k = 0 this gives 2 cis(π/4) = √2 + √2 i. Note that (2i)⁴ = +16 and (−2)⁴ = +16, not −16.

Determine the sum of all five fifth roots of unity.

• 1
• 5
• 0
• −1

Answer: 0

The roots of z⁵ − 1 = 0 sum to the negative of the z⁴ coefficient, which is 0; geometrically the five vectors are equally spaced around the unit circle, so they cancel. Answering 5 comes from adding the moduli instead of the complex numbers.

A quartic polynomial with real coefficients has roots i and 1 + i. The other two roots are

• −i and 1 − i
• −i and −1 − i
• i and 1 − i
• −1 and −1 + i

Answer: −i and 1 − i

Each non-real root brings its conjugate: the conjugate of i is −i and the conjugate of 1 + i is 1 − i.

Let ω be a cube root of unity with ω ≠ 1. Determine the value of ω² + ω.

• 1
• −1
• 0
• ω²

Answer: −1

Since ω³ = 1 and ω ≠ 1, ω satisfies ω² + ω + 1 = 0, so ω² + ω = −1.