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

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

Mathematics
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thats the z(cos sin ) stuff right?
-512i if your lucky :)
using ( ) as delimiters instead of [ ] places it inline

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Other answers:

if we type it out like this \(\frac{a}{b}\) it should go inline text
thank you but how can i work it out?
work what out?
i dont understand... doesnt the 2^6 have to be a square root?
2^6 = 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
isnt cos(2pi) = 1?
1+i = pi/4 i believe is a better item
1 over 1i up is a 45 tri
or i might be off in my thinking there
\[(2+2i)^6=2^6(\cos(\pi/4)+i\sin(\pi/4))^6=2^6(\cos(6\pi/4)+i\sin(6\pi/4))\] 64(\cos(3\pi/2)+i\sin(3\pi/2))\] \[64(0-i)\] ugh
i got the 270^o yay!!
apparently the radius aint 2
but how do you get -512i from 64(0-i)?
the 2^6 aint correct; the (0-i) parts good
or at least thats what im having an issue with
2^9 = 512
that still gives a radius of 64 and not 512 tho
\[(\sqrt{8})^6 <(45*6 )\]
where do you get sqrt 8?
the original radius is 2sqrt(2); not 2 [2sqrt(2)]^6 = 64*8 = 512
we have to power up the origonal radius
i knew if i throew demovers out the window I could think of it lol
\[r^n(cos(nx)+isin(nx))\]
\[r ^{n}
\[(2+2i)=2\sqrt2(\cos(\pi/4)+i\sin(\pi/4))\] \[(2\sqrt2)^6(\cos6(\pi/4)+i\sin6(\pi/4))\]
tada!! :) i knew you could do it class is starting so ciao
thanks! i kinda get it now.. but where does the 6 come from after the cos and sin
\[r=\sqrt{2^{2}+2^{2}}=\sqrt{4+4}=\sqrt{8}\] arg z =arctan(1)=45 \[(\sqrt{8}<45)^{6}\]
oh wow alright thank you
\[r ^{n}

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