Why does the argument change in solving z^3 = j using DeMoivre's theorem?

[aliz_khanz]

q. find all the solution of equation z^3= j

Attempt

okie now we know we have its argument as pie/2 but my friend did this and he placed the argument of it as THETA + 2*pie*k/ n

i want to ask first why the argument change ?

second , i thought in demoivers theorem we multiply n to argument but he has divided it ...

please help , its getting on my nerves :(

 

[tiny-tim]

hializ_khanz!

(have a pi: π and try using the X² icon just above the Reply box )

okie now we know we have its argument as pie/2 but my friend did this and he placed the argument of it as THETA + 2*pie*k/ n

(i'm not quite following this , but …)

remember that j = e^iπ/2 = = e^5iπ/2 = = e^9iπ/2

 

[HallsofIvy]

DeMoivre's theorem says that
z^n= (re^{j\theta})^n= r^n e^{jn\theta}

But you want to solve z^3= j so you want to calculuate
z= (j)^{1/3}, not j^3.