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anonymous
 4 years ago
using Demoivre's theorem solve this equation .. (2+2i)^6
anonymous
 4 years ago
using Demoivre's theorem solve this equation .. (2+2i)^6

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amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0thats the z(cos sin ) stuff right?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0512i if your lucky :)

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0using ( ) as delimiters instead of [ ] places it inline

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0if we type it out like this \(\frac{a}{b}\) it should go inline text

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0thank you but how can i work it out?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i dont understand... doesnt the 2^6 have to be a square root?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.02^6 = 1 2 4 8 16 32 64.... 64 cos(6pi) = 1 and sin(6pi)= 0 how the wolf gets 516i from that i cant tell yet

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.01+i = pi/4 i believe is a better item

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.01 over 1i up is a 45 tri

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0or i might be off in my thinking there

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0\[(2+2i)^6=2^6(\cos(\pi/4)+i\sin(\pi/4))^6=2^6(\cos(6\pi/4)+i\sin(6\pi/4))\] 64(\cos(3\pi/2)+i\sin(3\pi/2))\] \[64(0i)\] ugh

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0i got the 270^o yay!!

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0apparently the radius aint 2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but how do you get 512i from 64(0i)?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0the 2^6 aint correct; the (0i) parts good

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0or at least thats what im having an issue with

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0that still gives a radius of 64 and not 512 tho

EarthCitizen
 4 years ago
Best ResponseYou've already chosen the best response.3\[(\sqrt{8})^6 <(45*6 )\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0where do you get sqrt 8?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0the original radius is 2sqrt(2); not 2 [2sqrt(2)]^6 = 64*8 = 512

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0we have to power up the origonal radius

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0i knew if i throew demovers out the window I could think of it lol

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0\[r^n(cos(nx)+isin(nx))\]

EarthCitizen
 4 years ago
Best ResponseYou've already chosen the best response.3\[r ^{n}<n \theta\]

2bornot2b
 4 years ago
Best ResponseYou've already chosen the best response.0\[(2+2i)=2\sqrt2(\cos(\pi/4)+i\sin(\pi/4))\] \[(2\sqrt2)^6(\cos6(\pi/4)+i\sin6(\pi/4))\]

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0tada!! :) i knew you could do it class is starting so ciao

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0thanks! i kinda get it now.. but where does the 6 come from after the cos and sin

EarthCitizen
 4 years ago
Best ResponseYou've already chosen the best response.3\[r=\sqrt{2^{2}+2^{2}}=\sqrt{4+4}=\sqrt{8}\] arg z =arctan(1)=45 \[(\sqrt{8}<45)^{6}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh wow alright thank you

EarthCitizen
 4 years ago
Best ResponseYou've already chosen the best response.3\[r ^{n}<n \theta \] r=(8^1/2)^6=512, n=6 and theta=45*6=270 512<270\[r(\cos \theta+isin \theta) = 512(\cos(270)+isin(270))=i512\]
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