## iop360 3 years ago Find all complex numbers z satisfying...

1. iop360

$z^3 = -27i$ Answers are: $z = 3i$ $z = \frac{ -3\sqrt{3} }{ 2 } - \frac{ 3i }{ 2 }$ $z = \frac{ 3\sqrt{3} }{ 2 } - \frac{ 3i }{ 2 }$

2. mayankdevnani
3. mayankdevnani

ok @iop360

4. mahmit2012

|dw:1353827474036:dw|

5. iop360

how did you get pi/6?

6. iop360

i cant read the stuff at the top...

7. sirm3d

$\huge -i = e^{(3\pi/2 + 2n\pi)}$$\huge z^3 = 3^3e^{(3\pi/2 + 2n\pi)}$

8. sirm3d

$\huge z=\sqrt[3]{3^3}e^{(\large \frac{ 3\pi }{ 2 }+2n \pi)/3},\text{ } n=0,1,2$

9. iop360

how did you set up your first equation?

10. iop360

the 3pi/2

11. sirm3d

$\huge z=3e^{\frac{ 3 \pi }{ 6 }+\frac{ 2 \pi }{ 3 }n},\text { }n=0,1,2$

12. sirm3d

|dw:1353828234574:dw|

13. iop360

14. iop360

for example, the 3pi/2 in the brackets

15. sirm3d

it's in the unit circle drawn above

16. iop360

lemme try it

17. iop360

do you use eulers equation? z = r(cos(angle) + i sin(angle))

18. sirm3d

yes, just slightly different notation $re^{x i}=r(\cos x + i \sin x)$ |dw:1353828628571:dw|

19. iop360

thank you, it works!

20. iop360

only 1 question though, for your first equation -i = ... if you have a negative i, does that always mean 3pi/2