if z=x+yi, find all possible values of z such that z^3=1

please enlighten me, no idea where to start.

Printable View

- March 15th 2010, 05:31 AMz1llchsimple complex number
if z=x+yi, find all possible values of z such that z^3=1

please enlighten me, no idea where to start. - March 15th 2010, 05:55 AMHallsofIvy
It is simplest to write z in "polar form"- [/tex]z= r(cos(\theta)+ i sin(\theta))[/tex] (Engineering notation: for "

**c**os+**i s**sin") which can also be written as where "r" is the straight line distance from 0 to x+ iy in the "complex plane" ((0,0) to (x,y)) and [tex]\theta[/itex] is the angle that line makes with the positive real axis (positive x-axis).

Then you can use De Moivres theorem: or, in exponential form, .

There, n can be any real number including 1/3: .

Of course, the solutions to are .

"1" is ON the real-axis to and its distance from 0 is 1 so r= 1: .

That's obviously true: cos(0)= 1, sin(0)= 0. It is also obvious that . Of course you knew that**one**solution of was z= 1.

Now the good part: Since sine and cosine are periodic with period , can also be written as or as but dividing by 3, so that the sine and cosine are of and gives**different**results! That's how you can get the two other roots.

(Adding**another**would give and dividing**that**by 3 gives again and so not a new solution. If you keep adding " ", you cycle though those same three solutions.)