Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
paulzord
Group Title
'complex' numbers?
Suppose w6 = 64i. Determine the six values of w.
 3 years ago
 3 years ago
paulzord Group Title
'complex' numbers? Suppose w6 = 64i. Determine the six values of w.
 3 years ago
 3 years ago

This Question is Closed

paulzord Group TitleBest ResponseYou've already chosen the best response.0
hm i have no idea how to even start this...
 3 years ago

mathgirl Group TitleBest ResponseYou've already chosen the best response.0
is this supposed to be w raised to the 6th power?
 3 years ago

paulzord Group TitleBest ResponseYou've already chosen the best response.0
ah yes it is sorry
 3 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
is this \[w^6=64i\]?
 3 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
if so this is not so bad.
 3 years ago

paulzord Group TitleBest ResponseYou've already chosen the best response.0
yes it is \[w ^{6} = 64i\]
 3 years ago

mathgirl Group TitleBest ResponseYou've already chosen the best response.0
you just take the 6th root, right sat?
 3 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
hello math girl. this site is very slow for me tonight
 3 years ago

mathgirl Group TitleBest ResponseYou've already chosen the best response.0
figure out what number you multiply by itself 6 times
 3 years ago

mathgirl Group TitleBest ResponseYou've already chosen the best response.0
that gives you 64, and then stick on i on it
 3 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
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 \[2(\cos(45)+i\sin(45))\]
 3 years ago

mathgirl Group TitleBest ResponseYou've already chosen the best response.0
where'd the trig stuff come from?
 3 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
degrees means it is easier for me to divide and get whole number not write fractions in latex
 3 years ago

mathgirl Group TitleBest ResponseYou've already chosen the best response.0
i must be confused about the type of problem here
 3 years ago

paulzord Group TitleBest ResponseYou've already chosen the best response.0
how do yo know its 270?
 3 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
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
 3 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
oh how to i know it is 270?
 3 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
draw the complex plane and imagine where 64i is. it is straight down 64 units, think of it as \[064i\]
 3 years ago

mathgirl Group TitleBest ResponseYou've already chosen the best response.0
what math class is this? i never learned this stuff
 3 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
you can do this as soon as you learn trig
 3 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
write \[a+bi\] as \[r(\cos(\theta)+i\sin(\theta)\] where \[r=\sqrt{a^2+b^2}\] and \[\tan(\frac{b}{a})=\theta\]
 3 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
in any case when you want the sixth roots, take the sixth root of the absolute value and divide the angle by 6
 3 years ago

abtrehearn Group TitleBest ResponseYou've already chosen the best response.1
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.
 3 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
so first answer is \[2(\cos(45)+i\sin(45)\] \[=2(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)\] \[=\sqrt{2}+\sqrt{2}i\]
 3 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
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
 3 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
abtrehearn answer more elegant
 3 years ago

paulzord Group TitleBest ResponseYou've already chosen the best response.0
how did you get that angle 3pi/2 when it looks like im meant to go tan 0/2
 3 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
here is a picture of the unit circle in the complex plane. you can see that the angle associated with i is \[\frac{3\pi }{2}\]
 3 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
and by the way it is \[\theta =\tan(\frac{b}{a})\] which is this case is \[\tan(\frac{64}{0})\] which is undefined
 3 years ago

abtrehearn Group TitleBest ResponseYou've already chosen the best response.1
The angle\[3 \pi/2\]comes from plotting (0, 64) on the xy plane and measuring counterclockwise from the positive xaxis to the point. That's three right angles, or \[3 \pi/2\]radians.
 3 years ago

paulzord Group TitleBest ResponseYou've already chosen the best response.0
oh... click
 3 years ago

paulzord Group TitleBest ResponseYou've already chosen the best response.0
no wonder the 6th root of it is pi/2?
 3 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
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
 3 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
then for the next one add 360 again to get 990 and divide by 6 to get 165 etc
 3 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
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
 3 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
this is certainly not the elegant way to do it, but it gets the point across
 3 years ago

abtrehearn Group TitleBest ResponseYou've already chosen the best response.1
45 + 60 = 105 105 + 60 = 165 165 + 60 = 225 225 + 60 = 285 285 + 60 = 345. Those are the angles for the vertices in degrees.
 3 years ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.