anonymous
  • anonymous
'complex' numbers? Suppose w6 = -64i. Determine the six values of w.
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
hm i have no idea how to even start this...
anonymous
  • anonymous
is this supposed to be w raised to the 6th power?
anonymous
  • anonymous
ah yes it is sorry

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anonymous
  • anonymous
is this \[w^6=-64i\]?
anonymous
  • anonymous
if so this is not so bad.
anonymous
  • anonymous
yes it is \[w ^{6} = -64i\]
anonymous
  • anonymous
you just take the 6th root, right sat?
anonymous
  • anonymous
hello math girl. this site is very slow for me tonight
anonymous
  • anonymous
figure out what number you multiply by itself 6 times
anonymous
  • anonymous
2
anonymous
  • anonymous
that gives you 64, and then stick on i on it
anonymous
  • anonymous
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 \[2(\cos(45)+i\sin(45))\]
anonymous
  • anonymous
where'd the trig stuff come from?
anonymous
  • anonymous
degrees means it is easier for me to divide and get whole number not write fractions in latex
anonymous
  • anonymous
i must be confused about the type of problem here
anonymous
  • anonymous
how do yo know its 270?
anonymous
  • anonymous
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
anonymous
  • anonymous
oh how to i know it is 270?
anonymous
  • anonymous
draw the complex plane and imagine where -64i is. it is straight down 64 units, think of it as \[0-64i\]
anonymous
  • anonymous
what math class is this? i never learned this stuff
anonymous
  • anonymous
you can do this as soon as you learn trig
anonymous
  • anonymous
write \[a+bi\] as \[r(\cos(\theta)+i\sin(\theta)\] where \[r=\sqrt{a^2+b^2}\] and \[\tan(\frac{b}{a})=\theta\]
anonymous
  • anonymous
in any case when you want the sixth roots, take the sixth root of the absolute value and divide the angle by 6
anonymous
  • anonymous
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.
anonymous
  • anonymous
so first answer is \[2(\cos(45)+i\sin(45)\] \[=2(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)\] \[=\sqrt{2}+\sqrt{2}i\]
anonymous
  • anonymous
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
anonymous
  • anonymous
abtrehearn answer more elegant
anonymous
  • anonymous
how did you get that angle 3pi/2 when it looks like im meant to go tan 0/2
anonymous
  • anonymous
here is a picture of the unit circle in the complex plane. you can see that the angle associated with i is \[\frac{3\pi }{2}\]
anonymous
  • anonymous
and by the way it is \[\theta =\tan(\frac{b}{a})\] which is this case is \[\tan(\frac{-64}{0})\] which is undefined
anonymous
  • anonymous
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anonymous
  • anonymous
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.
anonymous
  • anonymous
oh... click
anonymous
  • anonymous
no wonder the 6th root of it is pi/2?
anonymous
  • anonymous
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
anonymous
  • anonymous
then for the next one add 360 again to get 990 and divide by 6 to get 165 etc
anonymous
  • anonymous
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
anonymous
  • anonymous
this is certainly not the elegant way to do it, but it gets the point across
anonymous
  • anonymous
45 + 60 = 105 105 + 60 = 165 165 + 60 = 225 225 + 60 = 285 285 + 60 = 345. Those are the angles for the vertices in degrees.

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