# Multiplication of Polar Forms

• May 17th 2010, 09:32 AM
MuhTheKuh
Multiplication of Polar Forms
Hey guys. I am struggling right now with a question where I was asked to multiply two different polar forms with each other (or take one of them to the power of something, or even both).
As you should be able to see on this picture question a) was perfectly fine and understandable for me but in question b) in during the third step he tells me to add cis(2xpi) to my existing (and with 5 multiplied) cis. Basically I would have thought that with step 2 everything is done and i could simply finish evaluating from this step on but in question b) d) e) he somehow adds or takes something to/from my cis, and I have no idea why and where he gets those values from.
http://img682.imageshack.us/img682/3753/polarforms.jpg
• May 17th 2010, 09:41 AM
tonio
Quote:

Originally Posted by MuhTheKuh
Hey guys. I am struggling right now with a question where I was asked to multiply two different polar forms with each other (or take one of them to the power of something, or even both).
As you should be able to see on this picture question a) was perfectly fine and understandable for me but in question b) in during the third step he tells me to add cis(2xpi) to my existing (and with 5 multiplied) cis. Basically I would have thought that with step 2 everything is done and i could simply finish evaluating from this step on but in question b) d) e) he somehow adds or takes something to/from my cis, and I have no idea why and where he gets those values from.
http://img682.imageshack.us/img682/3753/polarforms.jpg
This is pure trigonometry stuff: as $\cos\left(-\frac{5\pi}{4}\right)=\cos\left(-\frac{5\pi}{4}+2\pi\right)=\cos\left(\frac{3\pi}{4 }\right)$ , and the same for the sine, we get $cis\left(-\frac{5\pi}{4}\right)=cis\left(\frac{3\pi}{4}\righ t)$ , and with the other ones is the same.
In general, $cis(\phi)=cis(\theta)$ whenever $\phi-\theta=2n\pi\,,\,\,n\in\mathbb{Z}$ , because of the periodicity of cosine and sine.