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