Timeline for Does the square root function behave the same when calculating complex roots?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 9, 2021 at 17:39 | vote | accept | Deniz Gün | ||
| Mar 9, 2021 at 17:31 | history | edited | fleablood | CC BY-SA 4.0 | added 1003 characters in body |
| Mar 9, 2021 at 16:13 | history | edited | fleablood | CC BY-SA 4.0 | added 69 characters in body |
| Mar 9, 2021 at 16:04 | history | edited | fleablood | CC BY-SA 4.0 | added 1413 characters in body |
| Mar 9, 2021 at 15:42 | comment | added | fleablood | Oh... boy.... I'd say that as in the original expression it is not given that $\frac {-9}{i^2}$ will be found to be a positive real number (although it obviously will) that $\sqrt{\frac{-9}{i^2}}$ is by definition $\{w\in \mathbb C| w^2 = \frac {-9}{i^2}\}=\{w\in\mathbb C|w^2=9\}=\{3,-3\}=\pm 3$ and that there will be $8$ solutions. | |
| Mar 9, 2021 at 10:18 | comment | added | Deniz Gün | Thank you for the detailed answer. To clarify, the original question is not much more complicated it asks us to calculate the roots of: $ z^4 = \sqrt{\frac{-9}{i^2}} $ | |
| Mar 9, 2021 at 8:26 | history | edited | fleablood | CC BY-SA 4.0 | deleted 127 characters in body |
| Mar 9, 2021 at 8:18 | history | answered | fleablood | CC BY-SA 4.0 |