Thread: simple 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.

Originally Posted by z1llch

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

please enlighten me, no idea where to start.

It is simplest to write z in "polar form"- [/tex]z= r(cos(\theta)+ i sin(\theta))[/tex] (Engineering notation: for "cos+ 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.)