'complex' numbers?
Suppose w6 = -64i. Determine the six values of w.

- anonymous

'complex' numbers?
Suppose w6 = -64i. Determine the six values of w.

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- anonymous

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

- anonymous

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

- anonymous

ah yes it is sorry

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## More answers

- anonymous

is this
\[w^6=-64i\]?

- anonymous

if so this is not so bad.

- anonymous

yes it is
\[w ^{6} = -64i\]

- anonymous

you just take the 6th root, right sat?

- anonymous

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

- anonymous

figure out what number you multiply by itself 6 times

- anonymous

2

- anonymous

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

- 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

where'd the trig stuff come from?

- anonymous

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

- anonymous

i must be confused about the type of problem here

- anonymous

how do yo know its 270?

- 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

oh how to i know it is 270?

- anonymous

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

- anonymous

what math class is this? i never learned this stuff

- anonymous

you can do this as soon as you learn trig

- 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

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

- 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

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

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

abtrehearn answer more elegant

- anonymous

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

- 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

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

- anonymous

http://www.google.com/imgres?imgurl=http://www.economicsnetwork.ac.uk/cheer/ch14_1/Image36.gif&imgrefurl=http://www.economicsnetwork.ac.uk/cheer/ch14_1/ch14_1p04.htm&usg=__WxOqqWc_vv0pdh4S5ZXqfog9ThQ=&h=374&w=511&sz=4&hl=en&start=0&sig2=gY2IGXqQiJMv8oTzjDrQjQ&zoom=1&tbnid=O8kGpuYwEU7LEM:&tbnh=147&tbnw=201&ei=7ZgfTqXPHoTr0gHT_7C7Aw&prev=/search%3Fq%3Dunit%2Bcircle%2Bin%2Bcomplex%2Bplane%26hl%3Den%26client%3Dubuntu%26hs%3D1DY%26sa%3DX%26channel%3Dfs%26biw%3D1179%26bih%3D803%26tbm%3Disch%26prmd%3Divns&itbs=1&iact=hc&vpx=159&vpy=99&dur=6992&hovh=192&hovw=263&tx=172&ty=90&page=1&ndsp=18&ved=1t:429,r:0,s:0

- 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

oh... click

- anonymous

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

- 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

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

- 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

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

- 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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