Question

can someone help me to solve this using the de moivre's formula

[(square root of 6 + square root of 2)+(square root of 6 - square root of 2) i ] raised to 7

Answer 1

Well, let's state it first, if
\[\huge z = r\left[\cos \theta + i\sin \theta\right]\]then
\[\huge z^p=r^p\left[\cos \ p\theta + i\sin \ p \theta \right]\]

Answer 2

@PeterPan : I got Z^7 = r^7[ cos 7theta +sin7theta*i. plug into the original one I have r=1
cos 7theta = sqr6 + sqr2
sin 7theta = sqr 6 -sqr2
Am I right?

Answer 3

Uhh, no :)
First thing you want to do is convert the form
\[\huge a+bi \rightarrow r\left[\cos \theta + i\sin \theta\right]\]
It's almost exactly the same as converting rectangular coordinates to Polar :)

Answer 4

yes, so a = rcos theta

Answer 5

This isn't your question, @Hoa
I wonder why you take an interest in it :>

And no, things are not that simple...

Answer 6

Well, actually, yes
\[a = r\cos \theta\]

Answer 7

hihi... just long time no see that stuff, want to review.

Answer 8

True enough
\[\huge a = r\cos \theta\]But we're not converting from polar to rectangular, in fact, the other way around :P

Answer 9

the whole thing power to the 7

Answer 10

First... convert to polar form.
Let's get r first, it's easier :P
\[\huge r = \sqrt{a^2 + b^2}\]Where
\[\huge a = \sqrt6 + \sqrt 2\]
\[\huge b = \sqrt6 - \sqrt2\]

Answer 11

So clearly,
\[\huge a^2 + b^2 = 16\]

Answer 12

ok, how about raise to 7

Answer 13

Just a sec...
It implies that
r = 4
Let's leave that be.
What's theta?

Answer 14

actually, i lost!!! just follow you

Answer 15

Well, experience might tell you that since a and b are both positive, that
\[\huge 0 \le \theta \le \frac{\pi}{2}\]

Answer 16

my answer is 4 raised to 7 cis 7pi/6