Message Boards Message Boards

GROUPS:

Arg of roots of a complex number?

Posted 7 months ago
1355 Views
|
3 Replies
|
2 Total Likes
|

Consider

SolveValues[z^3 == (1 - I) Sqrt[2], z] // AbsArg

{{2^(1/3), -(\[Pi]/12)}, {2^(1/3), 
  Arg[-(-1)^(1/3) (1 - I)^(1/3)]}, {2^(1/3), 
  Arg[(-1)^(2/3) (1 - I)^(1/3)]}}

The first root is presented in the expected polar form of a complex number. But Arg has trouble with the other two roots because SolveValues (Solve is the same) doesn't seem to complete the solution (leaving ^(1/3)) behind.

This should be easy. Any suggestions to obtain the expected roots of complex numbers?

POSTED BY: Andrew Read
3 Replies

With version 12.3 I get satisfactory results with this:

SolveValues[z^3 == (1 - I) Sqrt[2], z] // FullSimplify // 
  AbsArg // FullSimplify
POSTED BY: Gianluca Gorni
Posted 7 months ago

Yes that works, thank you.

POSTED BY: Andrew Read
FullSimplify[
 AbsArg[ComplexExpand@SolveValues[z^3 == (1 - I) Sqrt[2], z]]]

Also works.

POSTED BY: Shenghui Yang
Reply to this discussion
Community posts can be styled and formatted using the Markdown syntax.
Reply Preview
Attachments
Remove
or Discard

Group Abstract Group Abstract