9231 · 2.5
Complex numbers flashcards
Revision flashcards for Cambridge 9231 Complex numbers (syllabus 2.5). Flip, recall, then mark a real past-paper question.
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What is the general form of De Moivre's theorem for finding the $n$-th roots of $w = r(\cos\theta + i\sin\theta)$?
The roots are given by $z_k = r^{1/n} \left( \cos\left(\frac{\theta + 2k\pi}{n}\right) + i\sin\left(\frac{\theta + 2k\pi}{n}\right) \right)$ for $k = 0, 1, 2, \dots, n-1$.
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What is the most common mistake when finding the $n$-th roots of a complex number?
Forgetting to add $2k\pi$ to the argument $\theta$ *before* dividing by $n$. This leads to finding only the principal root, not all $n$ distinct roots.
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What is the geometric arrangement of the $n$-th roots of a complex number $w$?
The $n$ roots are the vertices of a regular $n$-sided polygon inscribed in a circle of radius $|w|^{1/n}$ centred at the origin.
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What are the $n$-th roots of unity?
The solutions to the equation $z^n = 1$. They are given by $e^{i(2k\pi/n)}$ for $k = 0, 1, \dots, n-1$ and lie on the unit circle.
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If $z = \cos\theta + i\sin\theta$, what are the expressions for $z+z^{-1}$ and $z-z^{-1}$?
$z+z^{-1} = 2\cos\theta$ and $z-z^{-1} = 2i\sin\theta$. These are crucial for expressing powers of trig functions.
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What is the general strategy for summing a series like $\sum_{r=1}^n \cos(r\theta)$?
Consider the complex geometric series $\sum_{r=1}^n (e^{i\theta})^r$. Find its sum using the GP formula, then take the real part of the result.
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What is the formula for the sum of a finite geometric series?
$S_n = \frac{a(1-r^n)}{1-r}$, where $a$ is the first term, $r$ is the common ratio, and $n$ is the number of terms.
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How do you express $\cos^n\theta$ in terms of multiple angles?
Use the identity $2\cos\theta = z + z^{-1}$ where $z=e^{i\theta}$. Expand $(2\cos\theta)^n = (z+z^{-1})^n$ using the binomial theorem, then group terms like $(z^k + z^{-k})$ which equals $2\cos(k\theta)$.
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What is a useful trick for simplifying expressions like $\frac{1}{1-e^{i\theta}}$?
Multiply the numerator and denominator by the conjugate of the denominator, $1-e^{-i\theta}$. Alternatively, use the half-angle identity: $1-e^{i\theta} = e^{i\theta/2}(e^{-i\theta/2} - e^{i\theta/2}) = -2ie^{i\theta/2}\sin(\theta/2)$.
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When finding roots, why do we only need values of $k$ from $0$ to $n-1$?
Using $k=n$ gives an argument of $\frac{\theta + 2n\pi}{n} = \frac{\theta}{n} + 2\pi$, which is coterminal with the argument for $k=0$. The roots start repeating for $k \ge n$.