Further Maths - Complex Sums

$w + wz + wz^2 + \cdots + wz^{n-1}$ What is this equal to?

\[\frac{w(z^n - 1)}{z - 1}\]

$\sum^{n - 1} _ {r = 0} wz^r$ What is this equal to?

\[\frac{w(z^n - 1)}{z - 1}\]

$w + wz + wz^2 + \cdots + wz^{n}$ What is this equal to?

Remember the plus one! Don’t you dare skip this card if you didn’t say that explicitly!

\[\frac{w(z^{n+1} - 1)}{z - 1}\]

$\sum^{n} _ {r = 0} wz^r$ What is this equal to?

Remember the plus one! Don’t you dare skip this card if you didn’t say that explicitly!

\[\frac{w(z^n - 1)}{z - 1}\]

What is the point of the complex sums of series topic?

Turning complicated series into simple expressions of $\sin$ and $\cos$.

What are the two important results when working out complex geometric sums?

$z + \frac{1}{z} = 2\cos\theta$ $z - \frac{1}{z} = 2i\sin\theta$

$\frac{-2}{e^{\frac{\pi i}{n}} - 1}$ What might your next step be here?

You’re trying to create the difference or sum of two opposite pairs, multiply the bottom by $e^{-\frac{\pi i}{2n}}$

\[-2\frac{e^{-\frac{\pi i}{2n}}}{e^{\pi i}{2n} - e^{-\frac{\pi i}{2n}}}\]

What are the two techniques for making $z + 1/z$ or $z - 1/z$ terms appear from nothing in the complex sums topic?

$1 + z + z^2 + \cdots + z^{2n-1}$ What is this equal to?

\[\frac{1(z^{2n} - 1)}{z - 1}\]

$1 + z + z^2 + \cdots + z^n$ What is this equal to?

\[\frac{1(z^{n+1} - 1)}{z - 1}\]

$\frac{-2e^{-\frac{\pi}{2n}i}}{2i\sin\left( \frac{\pi}{2n} \right)}$ How could you simplify this?

Use the fact that $-2/2i = i$

\[\frac{ie^{-\frac{\pi}{2n}}{\sin\left( \frac{\pi}{2n} \right)}\]

See Also