Question

If
\[Z ^{1/3} = 2 ( \cos \frac{ \pi }{ 9 } + i \sin \frac{ \pi }{ 9 }) \]
Then z = ?

Answer 1

\[\left( z ^{1/2} \right)^{2}=2^{2}\left( \cos \frac{ \pi }{ 9 }+isin \frac{ \pi }{9 } \right)^{2}\]and\[z=4\left( \cos ^{2} \frac{ \pi }{ 9 }+2icos \frac{ \pi }{ 9 }\sin \frac{ \pi }{ 9 }+\sin ^{2} \frac{ \pi }{ 9 }\right)=4\left( \cos ^{2}\frac{ \pi }{ 9 }+\sin ^{2}\frac{ \pi }{ 9 } \right)+4i\left( 2 \cos \frac{ \pi }{ 9 }\sin \frac{ \pi }{ 9 }\right)\]

Answer 2

\[(z^{1/3})^3=(2(\cos \pi/9+i \sin \pi/9))^3\]

Answer 3

\[z=4\left( 1 \right)+4i \left( \sin \frac{ 2\pi }{ 9 } \right)=4\left( 1+i \sin \frac{ 2\pi }{ 9 } \right)\]like this ??

Answer 4

perhaps it is easiest to see in polar form:

cos(x)+i*sin(x)= e^(ix)
and
(cos(x)+i*sin(x)^n= e^(i*n*x)

Answer 5

opps!!! the third term is wrong it must be \[\cos ^{2}\frac{ \pi }{ 9 }-\sin ^{2}\frac{ \pi }{ 9 }\]

Answer 6

guys I get the following:
\[8 ( \cos ^{3} \left(\begin{matrix}\pi \\ 9\end{matrix}\right) + i \sin ^{3}\left(\begin{matrix}\pi \\ 9\end{matrix}\right))\]
Is this incorrect?

Answer 7

I got \[z=4\left( \cos \frac{ 2\pi }{ 9 }+i \sin \frac{ 2\pi }{ 9 } \right)=4e ^{i \frac{ 2\pi }{ 9 }}\]for applied the polar form by @phi

Answer 8

yes, it is incorrect

use
(cos(x)+i*sin(x)^n= e^(i*n*x) = cos(n*x)+i*sin(n*x)
with n= 3

and do not forget 2^3 out front

Answer 9

pin is almost right, except he used n=2 (bad eyes?)

Answer 10

nah, I rearrange the term befor to get cos(2x)=cos^2(x)-sin^2(x) and sin(2x)=2cos(x)sin(x)
and become that answer XD

Answer 11

you want
\[ (Z^{\frac{1}{3}})^3 \]

Answer 12

If I change to polar form first I'll get (e^(i(pi)/9))^2 that's equal

Answer 13

lol, I think that z^(1/2) sry guy

Answer 14

Perhaps
\[2^{3}(\cos \frac{ \pi }{ 27 }+ isin \frac{ \pi }{ 27 })\]?

Answer 15

I am lost!!!!!!!!

Answer 16

sorry about confuseing you

Answer 17

\[ Z ^{1/3} = 2 ( \cos \frac{ \pi }{ 9 } + i \sin \frac{ \pi }{ 9 })\]
you want Z, so raise both sides to the 3rd power
\[ (Z ^{1/3})^3 = 2^3 ( \cos \frac{ \pi }{ 9 } + i \sin \frac{ \pi }{ 9 })^3\]
you get
\[ Z = 8 ( \cos \frac{ \pi }{ 9 } + i \sin \frac{ \pi }{ 9 })^3\]
now use
(cos(x)+i*sin(x))^n = cos(n*x)+i*sin(n*x)

Answer 18

What do you have as the final answer pinguine! The same as mine above?

Answer 19

\[Z=8\left( \cos \frac{ \pi }{ 9 } +isin \frac{ \pi }{ 9 }\right)^{3}=8\left( e ^{-\frac{ i \pi }{ 9 }} \right)^{3}=8e ^{-\frac{ \pi }{ 3 }}\]I got this, tochange into the sine cos form it's\[Z=8\left( \cos \frac{ \pi }{ 3 } +\sin \frac{ \pi }{ 3 }\right)\]

Answer 20

**** the power of exponential is all positive sry guy

Answer 21

look at ths Euler's formula http://fermatslasttheorem.blogspot.com/2006/02/eulers-formula.html

Answer 22

Thus also =
\[4\sqrt{3}i\]

Answer 23

How about you @phi ??

Answer 24

Chem, you have difficulty interpreting the expression...
pin is getting closer, but the "i" does not disappear. Look at my previous post and read it carefully.

Answer 25

Phi I get the answer as:
\[Z = 8( \cos \frac{ \pi }{ 3 } + i \sin \frac{ \pi }{ 3 })\]

Answer 26

yes I see, sorry sir. Pay attention of phi ^^

Answer 27

Is this not correct?

Answer 28

yes, and we know cos(pi/3) is 1/2 and sin(pi/3) is sqrt(3)/2
so you could put it in numerical form

Answer 29

My final answer in the form of a + bi:

\[4 + 4\sqrt{3}i\]
Does everyone agree???????????

Answer 30

yep

Answer 31

Thank you @ pinguine!

Answer 32

Sorry about confuseing you :P try to look at Euler's formula sir

Answer 33

I will do so. Thank you for your excellent knowledge and help!

Answer 34

Im not excellent XD