Melody, the question about cis(60)^1991 + cis(-60)^1991 + 1 = 1 seems to have disappeared, so I'll answer it here for you.


cis means cos + i*sin, so using De Moivre's theorem cis(60)^1991 is the same as cis(60*1991).

If we take the angles to be in degrees then 60*1991 degrees is just rotating by 360° a number of times (331 in fact) with 300° left over:

$$\left({\mathtt{60}}{\mathtt{\,\times\,}}{\mathtt{1\,991}}\right) {mod} \left({\mathtt{360}}\right) = {\mathtt{300}}$$ 

So this reduces to finding cis(300) + cis(-300).

But cis(θ) + cis(-θ) = 2*cos(θ), because sin(-θ) = -sin(θ), so the imaginary parts cancel.


Now cos(300°) = 1/2, so 2cos(300°) = 1; therefore cis(60)^1991 + cis(-60)^1991 + 1 = 2 and 2 is not equal to 1, so the statement is false.


Alan  Oct 26, 2014

Thank you very much Alan.

I whited the question because it was repeated on the next page (this page) and also I had re-thought my question.

Gino has answered here


Since your answers are different (I think) I will give it more thought.    

Melody  Oct 26, 2014

19 Online Users


New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.