Practice Flashcards
If a real-coefficient polynomial has root a + bi, what else is a root?
Track your progress — Sign up free to save your progress and get smart review reminders based on spaced repetition.
All Flashcards in Topic 1.14
Below are all 24 flashcards for this topic. Sign up free to track your progress and get personalized review schedules.
1.14.18 cards
If a real-coefficient polynomial has root a + bi, what else is a root?
Its conjugate a − bi — complex roots come in conjugate pairs.
What real quadratic has roots a ± bi?
z² − 2az + (a² + b²) (middle term −sum, constant = product).
How do you finish a polynomial given one complex root?
Write the conjugate, form their real quadratic, divide it out, then solve what's left.
Why does the conjugate-pair rule need real coefficients?
The proof relies on conjugating the whole equation; with real coefficients the equation is unchanged, forcing the conjugate to be a root.
Another root if 2 + i is a root (real coefficients)?
2 − i.
Real quadratic with roots 1 ± 2i?
z² − 2z + 5 (sum 2, product 5).
Can a real cubic have exactly two real roots and one complex root?
No — complex roots come in pairs, so a cubic has either 3 real roots or 1 real + a conjugate pair.
Roots of z³ − 3z² + 7z − 5 given 1 + 2i is one?
1 + 2i, 1 − 2i, 1.
1.14.28 cards
State De Moivre's theorem.
(r cisθ)ⁿ = rⁿ cis(nθ) — power the modulus, multiply the argument by n.
How do you raise a complex number to a power?
Convert to polar form, apply De Moivre (rⁿ, nθ), then convert back if needed.
Why convert to polar before powering?
De Moivre only applies to polar/exponential form; powering a + bi directly means a messy binomial expansion.
(1 + i)⁸ = ?
(√2 cis(π/4))⁸ = (√2)⁸ cis(2π) = 16.
(√3 + i)⁶ = ?
(2 cis(π/6))⁶ = 2⁶ cis(π) = −64.
De Moivre in exponential form?
(r e^(iθ))ⁿ = rⁿ e^(inθ) — same idea via index laws.
Common De Moivre mistake?
Multiplying r by n instead of powering it (it's rⁿ), or forgetting to multiply the angle by n.
(1 − i)⁴ = ?
(√2 cis(−π/4))⁴ = 4 cis(−π) = −4.
1.14.38 cards
How many nth-roots does a non-zero complex number have?
Exactly n distinct nth-roots.
Where do the nth-roots sit on an Argand diagram?
On a circle of radius R^(1/n), equally spaced 2π/n apart (a regular n-gon).
Formula for the nth-roots of R cisφ?
z_k = R^(1/n) cis((φ + 2πk)/n) for k = 0, 1, …, n − 1.
How do you get all the roots once you have one?
Keep adding 2π/n to the argument until you have n of them.
The three cube roots of 1?
1, cis(2π/3) = −½ + (√3/2)i, cis(4π/3) = −½ − (√3/2)i.
Cube roots of 8?
2, −1 + √3 i, −1 − √3 i (modulus 2, spaced 120°).
What modulus do all the nth-roots share?
R^(1/n), where R is the modulus of the original number.
Why do the roots form a regular polygon?
They share the same modulus (so lie on a circle) and are equally spaced 2π/n apart.
Topic 1.14 study notes
Full notes & explanations for De Moivre & roots (HL only)
Math AA exam skills
Paper structures, command terms & tips
Want smart review reminders?
Sign up free to track your progress. Our spaced repetition algorithm will tell you exactly which cards to review and when.
Start Free