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How do you solve a binomial equation , the answer in quadratic/polar form

Posted 10 years ago

If i put Solve[x^3==-I,x] I get the answer in a+bi form Can I got the polar solutions, and a complex graph where the solutions are?

I get that if I put the same command in Wolffram Alpha but i do not get the polar form.

POSTED BY: peter B

In Mathematica you could say

In[66]:= Graphics[{PointSize[Large], 
  Point[{Re[#], Im[#]} & /@ ComplexExpand[x /. Solve[x^3 == -I, x]]]}, 
  Frame -> True, Axes -> True, AxesLabel -> {"Re", "Im"}]

to see the roots in the Gaussian plane. And you could say this to see their polar form

In[67]:= {Abs[#], Arg[#]} & /@ ComplexExpand[x /. Solve[x^3 == -I, x]]
Out[67]= {{1, \[Pi]/2}, {1, -((5 \[Pi])/6)}, {1, -(\[Pi]/6)}}

their length is 1, of course and they trisect the unit circle, as usual.

POSTED BY: Udo Krause
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