Question

Complex Numbers help. Will give medal and fan. *question attached below*

Answer 1

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Answer 2

Can I have a step by step approach to the solution of this problem, please?

Answer 3

can someone tell me how to tag someone plz

Answer 4

use the Moiver theorem

Answer 5

u know de moiver's theorem??

Answer 6

Yes, but I am a little confused as to how to apply it to this question :S

Answer 7

To tag someone type '@iTiaax'

Answer 8

so like this '@kmwright'

Answer 9

with quotation marks hehe

Answer 10

without*

Answer 11

then its a easy one
rcos(theta)+i r sin(theta)=r e^i(theta)
use this

Answer 12

De Moivre Theorem
\((cosx+isinx)^n=cos(nx)+isin(nx)\)

Answer 13

Just to mentioned this thm came from euler's identity
using the equation @JK24 wrote

Answer 14

mention*

Answer 15

I mean Euler's Formula^_^

Answer 16

you case you have power 4 and 5
so you apply the thm
just like with n case

Answer 17

I would simplify
\[ \frac{a^4}{a^5} = a^{-1} \]
or in this case, get

\[ \left(\cos\left(\frac{\pi}{9}\right)+ i \sin\left(\frac{\pi}{9}\right)\right)^{-1}\\
\cos\left(-\frac{\pi}{9}\right)+ i \sin\left(-\frac{\pi}{9}\right) \\
\cos\left(\frac{\pi}{9}\right)- i \sin\left(\frac{\pi}{9}\right)
\]

Answer 18

I'm lost. the pi/9 is what's confusing me :S

Answer 19

which step ?

Answer 20

do you agree we get
\[ \frac{1}{\cos(\pi/9) + i \sin(\pi/9)} \]
which we can write as
\[ \left( \cos(\pi/9) + i \sin(\pi/9)\right)^{-1} \]
?

Answer 21

@phi yes, but why is there an inverse or in other words a power to the negative one?

Answer 22

a "rule" of exponents is
\[ \frac{1}{a} = a^{-1} \]
this is one of those fundamental factoids you should learn

Answer 23

pi/9 is just an angle in radians. It is equal to 20 degrees.
It is not a standard angle such as 30, 45, 60, 60 degrees but a calculator can tell you the values. But here they want you to just simplify the expression without using tables or a calculator and so you can leave the answer as cos(pi/9) - isin(pi/9).

Answer 24

Alrighty. But don't we have to take into consideration that the numerator is raised to the power of 4 while the denominator is raised to the power of 5?

Answer 25

\[
\frac {(\cos(\pi/9) + i\sin(\pi/9))^4}{(\cos(\pi/9) + i\sin(\pi/9))^5} =
\frac {1}{(\cos(\pi/9) + i\sin(\pi/9))^1} = \\
(\cos(\pi/9) + i\sin(\pi/9))^{-1} =
\cos(\pi/9) - i\sin(\pi/9)
\]

Answer 26

as you can see, the "base" is the same "thing"
so we have something analogous to
\[ \frac{a^4}{a^5} \]

one rule for dividing is top exponent - bottom exponent
we would get -1 as the new exponent

or the long way:
a*a*a*a/ a*a*a*a*a
4 of the a's cancel, leaving 1/a

either way we get
a^(-1)

Answer 27

Oooooh! I see it now! You guys are amazing!

Answer 28

It is great when students keep asking questions until the concept is clear in their minds. Much better than those who just grab the answer and run away without trying to understand the method.

Answer 29

we use two other ideas. for angle x,we know cos(-x)= cos(x) and
sin(-x) is -sin(x)
we can use that to get the final "simplification"

Answer 30

@phi oh great..I see that.
Thank you everyone! I'll fan those who helped :)