Number Systems: Problem Sheet 5

Here are some hints for Problem Sheet 5 given to people in office hours.

Question 2. If you are stuck, please first remind yourself of exponential form for complex numbers (there are some examples and exercises on the revision sheet) and look again at Problem 2.5.

If after doing all that you are still stuck, it might be because replacing the 8i in Problem 2.5 with 8 somehow makes it less obvious how to get started. Let w = \mathrm{e}^{i \theta}. We need to solve

r^3 \mathrm{e}^{3 i \theta} = 8.

The modulus of the left-hand side is r^3 and the modulus of the right-hand side is 8. Hence r^3 = 8 and so r=2.

The argument is the tricker bit. The argument of r^3 \mathrm{e}^{3 i \theta} is 3 \theta, and the argument of 8 is 0. But remember that we can add any multiple of 2 \pi and get another value of the argument. So the conclusion from comparing arguments is that

3 \theta = 0 + 2n \pi

for some n \in \mathbb{Z}. Now proceed as in Problem 2.5.

Finally, to convert your answers to exponential form you should use the known values of \cos (2\pi/3) and \sin (2\pi/3). You should know the values of cos and sin on the angles 0, \pi/6, \pi/3, \pi/4, \pi/2: the values needed for this problem can be deduced from these.


Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s

%d bloggers like this: