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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.

.

 Oct 26, 2014
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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

http://web2.0calc.com/questions/1cis60-1991-1cis-60-1991-1-1-i-need-to-prove-this-but-the-calculator-wont-give-me-the-exact-answers

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

 Oct 26, 2014

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