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I am stuck on a past exam question. I don't have a clue what it's on about and would appreciate any help.

In this question, $w$ denotes the complex number $cos{\frac{2}{5}\pi} + isin{\frac{2}{5}\pi}$

i) Express $w^2$, $w^3$ and $w^*$ in polar form, with arguments in the interval $0 \le \theta < 2\pi$

My working (I'm pretty sure this is right just it's part of the same question) $|w| = 1$ $arg(w) = \frac{2}{5}\pi$ $\therefore |w^2| = 1$ $arg(w^2) = \frac{2}{5}\pi + \frac{2}{5}\pi = \frac{4}{5}\pi$ $|w^3| = 1$ $arg(w^3) = \frac{2}{5}\pi + \frac{2}{5}\pi + \frac{2}{5}\pi = \frac{6}{5}\pi$ $\therefore w^2 = cos{\frac{4}{5}\pi} + isin{\frac{4}{5}\pi}, w^3 = cos{\frac{6}{5}\pi} + isin{\frac{6}{5}\pi}, w^* = cos{\frac{2}{5}\pi} - isin{\frac{2}{5}\pi}$

ii) The points in an Argand Diagram which represent the number $ 1, 1 + w, 1 + w + w^2, 1 + w + w^2 + w^3, 1 + w + w^2 + w^3 + w^4 $ are denoted by A, B, C, D, E respectively. Sketch the Argand diagram to show these points and join them in the order stated.

I could do this by brute force just bashing numbers into my calculator to find the coordinates but surely there has to be a better way? Is there a property of complex numbers when they're added together like that?

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    Your answer for $w^\ast$ might be marked as being (technically) wrong. They specified polar form with 0 \le \theta<2\pi, so for $w^\ast$ this angle would be $8\pi/5$.2011-06-09

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Think geometrically. Do you know about Euler's formula? It states that

$e^{i\theta} = \cos\theta + i \sin\theta$

By comparing, you can see that your number is $w = e^{2\pi i/5}$, which geometrically corresponds to the point in the argand diagram a distance 1 from the origin, with argument $2\pi/5$, which is 1/5 of the way round a circle.

Squaring to get $w^2$ leaves the modulus unchanged and doubles the argument, so the point $w^2$ is 2/5 of the way round the circle, and $w^3$ and $w^4$ are 3/5 and 4/5 of the way round the circle.

Try plotting the points $1, w, w^2, w^3, w^4$ on the argand diagram and see what you notice about them. Now can you start adding them up?

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    Thanks, you've been a great help!2011-06-09