## paulzord Group Title 'complex' numbers? Suppose w6 = -64i. Determine the six values of w. 3 years ago 3 years ago

1. paulzord Group Title

hm i have no idea how to even start this...

2. mathgirl Group Title

is this supposed to be w raised to the 6th power?

3. paulzord Group Title

ah yes it is sorry

4. satellite73 Group Title

is this $w^6=-64i$?

5. satellite73 Group Title

if so this is not so bad.

6. paulzord Group Title

yes it is $w ^{6} = -64i$

7. mathgirl Group Title

you just take the 6th root, right sat?

8. satellite73 Group Title

hello math girl. this site is very slow for me tonight

9. mathgirl Group Title

figure out what number you multiply by itself 6 times

10. paulzord Group Title

2

11. mathgirl Group Title

that gives you 64, and then stick on i on it

12. satellite73 Group Title

yeah just take the 6th roots. if working in degrees write $-64i=64(\cos(270)+i\sin(270))$ and then take the 6ht root of 64 which is 2 and divide 270 by 6 to get 45 so first answer is $2(\cos(45)+i\sin(45))$

13. mathgirl Group Title

where'd the trig stuff come from?

14. satellite73 Group Title

degrees means it is easier for me to divide and get whole number not write fractions in latex

15. mathgirl Group Title

i must be confused about the type of problem here

16. paulzord Group Title

how do yo know its 270?

17. satellite73 Group Title

think of where -64i is in the complex plane. it is straight down 64 units, so trig form is what i wrote. you can also use $-64i=64e^{\frac{3\pi}{2}i}$ and do it that way

18. satellite73 Group Title

oh how to i know it is 270?

19. satellite73 Group Title

draw the complex plane and imagine where -64i is. it is straight down 64 units, think of it as $0-64i$

20. mathgirl Group Title

what math class is this? i never learned this stuff

21. satellite73 Group Title

you can do this as soon as you learn trig

22. satellite73 Group Title

write $a+bi$ as $r(\cos(\theta)+i\sin(\theta)$ where $r=\sqrt{a^2+b^2}$ and $\tan(\frac{b}{a})=\theta$

23. satellite73 Group Title

in any case when you want the sixth roots, take the sixth root of the absolute value and divide the angle by 6

24. abtrehearn Group Title

If$w^{6} = -64 i,$then$|w| = \sqrt[6]{64}= 2.$The angle $\theta = 3 \pi/2,$so one of the sixth roots of -64i will be a vertex of the regular hexagon lf side length 2 centered at z = 0, and making an angle $\theta/6 = (3 \pi/2)/6 = \pi/4.$ That vertex is at$z = 2 \cos(\pi/4) + 2i \sin(\pi/4) = \sqrt{2} + i \sqrt{2}.$All the sixth roots are$2 \cos(\pi/4 + k \pi/3) + 2i \sin(\pi/4 + k \pi / 3),$where k is an integer.

25. satellite73 Group Title

so first answer is $2(\cos(45)+i\sin(45)$ $=2(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)$ $=\sqrt{2}+\sqrt{2}i$

26. satellite73 Group Title

to get second answer just divide circle up into 6 equal parts with one part at 45 degrees. or go around the circle again and write $-64i=64(\cos(630)+i\sin(630))$ and now divide 630 by 6 to get 105 so second answer is $2(\cos(105)+i\sin(105))$ etc

27. satellite73 Group Title

28. paulzord Group Title

how did you get that angle 3pi/2 when it looks like im meant to go tan 0/2

29. satellite73 Group Title

here is a picture of the unit circle in the complex plane. you can see that the angle associated with i is $\frac{3\pi }{2}$

30. satellite73 Group Title

and by the way it is $\theta =\tan(\frac{b}{a})$ which is this case is $\tan(\frac{-64}{0})$ which is undefined

31. satellite73 Group Title
32. abtrehearn Group Title

The angle$3 \pi/2$comes from plotting (0, -64) on the x-y plane and measuring counterclockwise from the positive x-axis to the point. That's three right angles, or $3 \pi/2$radians.

33. paulzord Group Title

oh... click

34. paulzord Group Title

no wonder the 6th root of it is pi/2?

35. satellite73 Group Title

if you want to do this the slow donkey way, which may not be a bad idea the first time you see it, start with 270 degrees (i know we really should work in radians but this will make my typing easier) then divide 270 by 6 for the next one add 360 to 270 to get 630, divide by 6 to get 105

36. satellite73 Group Title

then for the next one add 360 again to get 990 and divide by 6 to get 165 etc

37. satellite73 Group Title

around again you get 990+360=1350 divide by 6 to get 225 around again you get 1350+360=1710 divide by 6 get 285 around again you get 1710+360=2070 divide by 6 get 345 and finally around again you get 2070+360 = 2430 divide by 6 get 405 which is the same as 45 so you are back where you started

38. satellite73 Group Title

this is certainly not the elegant way to do it, but it gets the point across

39. abtrehearn Group Title

45 + 60 = 105 105 + 60 = 165 165 + 60 = 225 225 + 60 = 285 285 + 60 = 345. Those are the angles for the vertices in degrees.