anonymous
  • anonymous
using Demoivre's theorem solve this equation .. (2+2i)^6
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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amistre64
  • amistre64
thats the z(cos sin ) stuff right?
amistre64
  • amistre64
-512i if your lucky :)
amistre64
  • amistre64
using ( ) as delimiters instead of [ ] places it inline

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amistre64
  • amistre64
if we type it out like this \(\frac{a}{b}\) it should go inline text
anonymous
  • anonymous
thank you but how can i work it out?
2bornot2b
  • 2bornot2b
work what out?
anonymous
  • anonymous
i dont understand... doesnt the 2^6 have to be a square root?
amistre64
  • amistre64
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
amistre64
  • amistre64
isnt cos(2pi) = 1?
amistre64
  • amistre64
1+i = pi/4 i believe is a better item
amistre64
  • amistre64
1 over 1i up is a 45 tri
amistre64
  • amistre64
or i might be off in my thinking there
amistre64
  • amistre64
\[(2+2i)^6=2^6(\cos(\pi/4)+i\sin(\pi/4))^6=2^6(\cos(6\pi/4)+i\sin(6\pi/4))\] 64(\cos(3\pi/2)+i\sin(3\pi/2))\] \[64(0-i)\] ugh
amistre64
  • amistre64
i got the 270^o yay!!
amistre64
  • amistre64
apparently the radius aint 2
anonymous
  • anonymous
but how do you get -512i from 64(0-i)?
amistre64
  • amistre64
the 2^6 aint correct; the (0-i) parts good
amistre64
  • amistre64
or at least thats what im having an issue with
amistre64
  • amistre64
2^9 = 512
amistre64
  • amistre64
that still gives a radius of 64 and not 512 tho
EarthCitizen
  • EarthCitizen
\[(\sqrt{8})^6 <(45*6 )\]
anonymous
  • anonymous
where do you get sqrt 8?
amistre64
  • amistre64
the original radius is 2sqrt(2); not 2 [2sqrt(2)]^6 = 64*8 = 512
amistre64
  • amistre64
we have to power up the origonal radius
amistre64
  • amistre64
i knew if i throew demovers out the window I could think of it lol
amistre64
  • amistre64
\[r^n(cos(nx)+isin(nx))\]
EarthCitizen
  • EarthCitizen
\[r ^{n}
2bornot2b
  • 2bornot2b
\[(2+2i)=2\sqrt2(\cos(\pi/4)+i\sin(\pi/4))\] \[(2\sqrt2)^6(\cos6(\pi/4)+i\sin6(\pi/4))\]
amistre64
  • amistre64
tada!! :) i knew you could do it class is starting so ciao
anonymous
  • anonymous
thanks! i kinda get it now.. but where does the 6 come from after the cos and sin
EarthCitizen
  • EarthCitizen
\[r=\sqrt{2^{2}+2^{2}}=\sqrt{4+4}=\sqrt{8}\] arg z =arctan(1)=45 \[(\sqrt{8}<45)^{6}\]
anonymous
  • anonymous
oh wow alright thank you
EarthCitizen
  • EarthCitizen
\[r ^{n}

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