Question

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

Answer 1

thats the z(cos sin ) stuff right?

Answer 2

-512i if your lucky :)

Answer 3

using ( ) as delimiters instead of [ ] places it inline

Answer 4

if we type it out like this \(\frac{a}{b}\) it should go inline text

Answer 5

thank you but how can i work it out?

Answer 6

work what out?

Answer 7

i dont understand... doesnt the 2^6 have to be a square root?

Answer 8

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

Answer 9

isnt cos(2pi) = 1?

Answer 10

1+i = pi/4 i believe is a better item

Answer 11

1 over 1i up is a 45 tri

Answer 12

or i might be off in my thinking there

Answer 13

\[(2+2i)^6=2^6(\cos(\pi/4)+i\sin(\pi/4))^6=2^6(\cos(6\pi/4)+i\sin(6\pi/4))\]
64(\cos(3\pi/2)+i\sin(3\pi/2))\]
\[64(0-i)\]
ugh

Answer 14

i got the 270^o yay!!

Answer 15

apparently the radius aint 2

Answer 16

but how do you get -512i from 64(0-i)?

Answer 17

the 2^6 aint correct; the (0-i) parts good

Answer 18

or at least thats what im having an issue with

Answer 19

2^9 = 512

Answer 20

that still gives a radius of 64 and not 512 tho

Answer 21

\[(\sqrt{8})^6 <(45*6 )\]

Answer 22

where do you get sqrt 8?

Answer 23

the original radius is 2sqrt(2); not 2

[2sqrt(2)]^6 = 64*8 = 512

Answer 24

we have to power up the origonal radius

Answer 25

i knew if i throew demovers out the window I could think of it lol

Answer 26

\[r^n(cos(nx)+isin(nx))\]

Answer 27

\[r ^{n}<n \theta\]

Answer 28

\[(2+2i)=2\sqrt2(\cos(\pi/4)+i\sin(\pi/4))\]
\[(2\sqrt2)^6(\cos6(\pi/4)+i\sin6(\pi/4))\]

Answer 29

tada!! :) i knew you could do it class is starting so ciao

Answer 30

thanks! i kinda get it now.. but where does the 6 come from after the cos and sin

Answer 31

\[r=\sqrt{2^{2}+2^{2}}=\sqrt{4+4}=\sqrt{8}\]
arg z =arctan(1)=45
\[(\sqrt{8}<45)^{6}\]

Answer 32

oh wow alright thank you

Answer 33

\[r ^{n}<n \theta \]
r=(8^1/2)^6=512, n=6 and theta=45*6=270
512<270\[r(\cos \theta+isin \theta) = 512(\cos(270)+isin(270))=-i512\]