Question

Use De Moivre's Theorem to find all of the complex solutions of the equation x³-i=0. Express the answer in rectangular form.

Answer 1

havent i already done that?

Answer 2

hi I wasn't sure about the e

Answer 3

wt doubt about it?

Answer 4

I didn't know how to ask you about it without re-posting...

Answer 5

solving
\[x^3=i\] yes?

Answer 6

its done earlier satellite....he just has a doubt about the presentation

Answer 7

I learnt it a different way and ended up with

Answer 8

ooh sorry. rectangular means
\[a+bi\] yes?

Answer 9

(√3/2 + i1/2), -√3/2 + i1/2), (0-i) .... is this correct?

Answer 10

I'm not sure

Answer 11

angle for i is
\[\frac{\pi}{2}\]so one cubed root will have angle
\[\frac{\pi}{6}\]

Answer 12

no..i gave u the euler forms of the three solns..can u retype em here for me plz..

Answer 13

x = e^(ipi/6), e^(5ipi/6) and e^(7ipi/6)

Answer 14

now can u convert all these into rectangular forms for me?

Answer 15

\[cos(\frac{\pi}{6})+isin(\frac{\pi}{6})=\frac{\sqrt{3}}{2}+\frac{1}{2}i\]\]

Answer 16

no that's what I was having trouble with

Answer 17

the conversion

Answer 18

for 5pi/6
x = cos(5pi/6) + isin(5pi/6)

Answer 19

similarly for 7pi/6
x = cos(7pi/6) + isin(7pi/6)

Answer 20

can u do that nw?

Answer 21

hold on

Answer 22

I really have no idea

Answer 23

wts cos(5pi/6)? and sin 5pi/6?

Answer 24

-√3/2 & 1/2

Answer 25

righto....so
x = -√3/2 + i(1/2)
isnt it?

Answer 26

right that's one of the answers I got

Answer 27

now try for pi/6 in the same way

Answer 28

along with (√3/2 + i1/2) & (0-i)

Answer 29

how the hell did u get those two?
try the same way for pi/6 and 7pi/6

Answer 30

cos(pi/6) is √3/2 and sin(pi/6) is 1/2 right?

Answer 31

right

Answer 32

so x = √3/2 +i(1/2)

Answer 33

?

Answer 34

ys