Question

list the three complex roots of -1 by rewriting -1 in trigonometric form and applying DeMoivre's theorem. You can leave the answer in trigonometric form.

Answer 1

ready?

Answer 2

\[-1=e^{i\pi}\]

Answer 3

omg someone finally answered praise the lord

Answer 4

first we can find them without any demoivre by writing
\[x^3=-1\]
\[x^3+1=0\]
\[(x+1)(x^2-x+1)=0\] so solutions are \(x=-1\) for the first factor, and quadratic equation finds the other two as \(x=\frac{1}{2}\pm \frac{\sqrt{3}}{2}i\)

Answer 5

as for demoivre we can do that as well

Answer 6

first write \(-1\) in trig form, which is kind of silly since it is a real number
but that is ok in trig form it is
\[-1=\cos(\pi)+i\sin(\pi)\]

Answer 7

hope that part is clear, the angle is obviously \(\pi\) from the picture

Answer 8

yeah its clear lol

Answer 9

now we want the cube root, and we find that by taking the real cube root of 1, which is 1, and dividing the angle by 3 (3 because it is the cube root)
so the other root is
\[\cos(\frac{\pi}{3})+i\sin(\frac{\pi}{3})\]

Answer 10

this is the one that is \(\frac{1}{2}+\frac{\sqrt{3}}{2}i\) in standard form

Answer 11

then you have a choice. easy way is to think "the other one is the conjugate, so it must be \(\frac{1}{2}-\frac{\sqrt{3}}{2}i\) "

Answer 12

so can cos(pi/2)+isin(pi/2) because it's the square root of one?

Answer 13

or you can go around the circle again since
\[\cos(\pi)+i\sin(\pi)=\cos(3\pi)+i\sin(3\pi)\] divide by 3 again and get
\[\cos(\pi)+i\sin(\pi)\] which is fairly silly because this is -1 which is what you started with

Answer 14

and finally go around the circle one more time and get
\[\cos(5\pi)+i\sin(5\pi)\] divide the angle by 3 and get
\[\cos(\frac{5\pi}{3})+i\sin(\frac{5\pi}{3})\] which gives
\[\frac{1}{2}-\frac{\sqrt{3}}{2}i\] as promised

Answer 15

now to answer you question, no the square roots of 1 are 1 and -1

Answer 16

ah true I totally forgot we are looking for -1 lol. .

Answer 17

btw the snap way to do this is to locate one root and then the others will be evenly spaced around the unit circle

Answer 18

for example to solve \(x^4=1\) the solutions are \(x=1,i,-1,-i\)

Answer 19

and to solve \(x^6=1\) the solutions are
\[1,\frac{1}{2}+\frac{\sqrt{3}}{2}i, -\frac{1}{2}+\frac{\sqrt{3}}{2}i,-1,-\frac{1}{2}+\frac{\sqrt{3}}{2}i,\frac{1}{2}-\frac{\sqrt{3}}{2}i\]