Timeline for How do I get the square root of a complex number?
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 30, 2018 at 1:00 | history | edited | user46234 | CC BY-SA 4.0 | added 1047 characters in body |
| Mar 28, 2017 at 11:48 | comment | added | user4414 | Better say b>=0, so that we have sqrt(-1)=i as a first solution. | |
| Jun 28, 2016 at 8:49 | comment | added | user46234 | @Wout: Thanks for spotting that error! I've now corrected, hopefully it's now good. | |
| Jun 28, 2016 at 8:48 | history | edited | user46234 | CC BY-SA 3.0 | Errors corrected (a+ib and a-ib are handled differently) |
| S Jun 27, 2016 at 15:32 | history | suggested | Wout | CC BY-SA 3.0 | Fixed incorrect sign. |
| Jun 27, 2016 at 15:15 | review | Suggested edits | |||
| S Jun 27, 2016 at 15:32 | |||||
| Jun 27, 2016 at 15:11 | comment | added | Wout | Except when b < 0 the last line isn't true because sqrt(b^2) isn't equal to b. For b < 0 replace + i... by -i... Also typo in second to last line: r + a - (r + a) should be r + a - (r - a), where r = sqrt(a^2 + b^2). | |
| Apr 26, 2016 at 3:01 | history | edited | user46234 | CC BY-SA 3.0 | deleted 9 characters in body |
| Apr 26, 2016 at 2:42 | history | answered | user46234 | CC BY-SA 3.0 |