How to simplify this complex expression?

[yamata1]

Homework Statement

Simplify $$\frac{(-1)^{2/9} + (-3/2 - \frac{i}{2} \sqrt{3})^{(1/3)}}{(-1)^{2/9}- (-3/2 - \frac{i}{2} \sqrt{3})^{(1/3)}}$$

Relevant Equations

wolfram gives us this result : $$-1 - \frac{(2 i)}{(3^{1/6} + -i)}$$

I don't know how to start with the factorization.
$$\frac{(-1)^{2/9} + (-3/2 - \frac{i}{2} \sqrt{3})^{(1/3)}}{(-1)^{2/9}- (-3/2 - \frac{i}{2} \sqrt{3})^{(1/3)}}$$
Any hints would be nice. Thank you.

 

[BvU]

I learned to multiply by 1 , like in$${a + bi\over c+di} = {a + bi\over c+di}\; {c - di\over c-di} = {(a + bi)(c-di)\over c^2+d^2} $$

But you should first read the guidelines -- they require you post your own attempt first !

 

[yamata1]

I learned to multiply by 1 , like ina+bic+di=a+bic+dic−dic−di=(a+bi)(c−di)c2+d2

But you should first read the guidelines -- they require you post your own attempt first !

My attempt : multiplying numerator and denominator by $(-1)^{2/9} + (-3/2 - \frac{i}{2} \sqrt{3})^{(1/3)} wehave \frac{(-1)^{4/9} +2(-1)^{2/9}(-3/2 - \frac{i}{2} \sqrt{3})^{(1/3)}+ (-3/2 - \frac{i}{2} \sqrt{3})^{(2/3)}}{(-1)^{4/9}+ (-3/2 - \frac{i}{2} \sqrt{3})^{(2/3)}}$

 

[BvU]

multiplying numerator and denominator

You fist want to write them in an $a+bi$ form. The $(-1)^{2\over 9}$ is a complex number.

 

[yamata1]

You fist want to write them in an $a+bi$ form. The $(-1)^{2\over 9}$ is a complex number.

How do I find the real part of $(-3/2 - \frac{i}{2} \sqrt{3})^{(1/3)}$ ? The cubic root makes it hard for me to see what to do here.

 

[Delta2]

How do I find the real part of $(-3/2 - \frac{i}{2} \sqrt{3})^{(1/3)}$ ? The cubic root makes it hard for me to see what to do here.

set $$(-3/2 - \frac{i}{2} \sqrt{3})=(a+bi)^3$$ then expand the RHS split real and imaginary parts and equate the real part of the RHS to the real part of the LHS and the imaginary part of the RHS to the imaginary part of the LHS. You then will have a system of equations with a,b as uknowns.

 

[vela]

How do I find the real part of $(-3/2 - \frac{i}{2} \sqrt{3})^{(1/3)}$ ? The cubic root makes it hard for me to see what to do here.

I'd start by expressing $-\frac 32 - i \frac{\sqrt 3}{2}$ in polar form $re^{i\theta}$.

 

[yamata1]

I'd start by expressing $-\frac 32 - i \frac{\sqrt 3}{2}$ in polar form $re^{i\theta}$.

$(-\frac 32 - i \frac{\sqrt 3}{2})^{1/3}= \sqrt[3\,] {e^{3i\pi}-e^{\frac{\pi}{3}}}$ and $(-1)^{2\over 9}=\sqrt[3\,] {e^{3i\pi}+e^{\frac{\pi}{3}}}$
When trying to solve $(-3/2 - \frac{i}{2} \sqrt{3})=(a+bi)^3$ I get $a(a^2-b-2b^2)=\frac{3}{2}$ and $b(2a^2+a-b^2)=\frac{\sqrt{3}}{2}$.
How can I sovle this 3rd order system ?

 

[Delta2]

$-\frac 32 - i \frac{\sqrt 3}{2}= \sqrt[3\,] {e^{3i\pi}-e^{\frac{\pi}{3}}}$ and $(-1)^{2\over 9}=\sqrt[3\,] {e^{3i\pi}+e^{\frac{\pi}{3}}}$
When trying to solve $(-3/2 - \frac{i}{2} \sqrt{3})=(a+bi)^3$ I get $a(a^2-b-2b^2)=\frac{3}{2}$ and $b(2a^2+a-b^2)=\frac{\sqrt{3}}{2}$.
How can I sovle this 3rd order system ?

I think my idea was not so good, we better abandon it.

 

[BvU]

Follow Vela, both for the $(-1)^{2/9}$ and for the $(-3/2 - \frac{i}{2} \sqrt{3})^{(1/3)}$

(Small disclaimer: I didn't do the exercise to the very end, just ventilated (lightheartedly ) an idea on how to attack it)

 

[vela]

$$(-\frac 32 - i \frac{\sqrt 3}{2})^{1/3}= \sqrt[3\,] {e^{3i\pi}-e^{\frac{\pi}{3}}}$$

I don't understand what you did here. For $z=-\frac 32 - i \frac{\sqrt 3}{2}$, we want you to give us two numbers, one for $r$ and one for $\theta$. Do you know how to find them?

 

[yamata1]

I don't understand what you did here. For $z=-\frac 32 - i \frac{\sqrt 3}{2}$, we want you to give us two numbers, one for $r$ and one for $\theta$. Do you know how to find them?

If I write

$e^{3i\pi}=(e^{i\pi})^3=(-1)^3=-1$

Then $z=-\frac {3}{2} - i \frac{\sqrt 3}{2}=-\sqrt 3 e^{i\pi/6}$

 

[vela]

If I write

$e^{3i\pi}=(e^{i\pi})^3=(-1)^3=-1$

Then $z=-\frac {3}{2} - i \frac{\sqrt 3}{2}=-\sqrt 3 e^{i\pi/6}$

You want $r$ to be positive and incorporate the negative sign out front into the phase to get $z=\sqrt 3 e^{i 7\pi/6}$. Now you're ready to calculate $z^{1/3}$, then you can convert the result back to rectangular form.

 

[yamata1]

If I write

$e^{3i\pi}=(e^{i\pi})^3=(-1)^3=-1$

Then $z=-\frac {3}{2} - i \frac{\sqrt 3}{2}=-\sqrt 3 e^{i\pi/6}$

I think I've clocked it.
$\sqrt[3\,] {e^{3i\pi}+e^{\frac{\pi}{3}}}= \sqrt[3\,]{e^{\frac {2i\pi}{3}}}$
$\sqrt[3\,]{e^{\frac {2i\pi}{3}}} +\sqrt[3\,]{-\sqrt 3 e^{i\pi/6}}= 3^{1/6} e^{-5i\pi/18}+e^{\frac {4i\pi}{18}}$ $=e^{\frac {4i\pi}{18}}(3^{1/6}e^{\frac {-9i\pi}{18}}+1)$ $=e^{\frac {4i\pi}{18}}(-i3^{1/6}+1)$
$\frac{(-1)^{2/9} + (-3/2 - \frac{i}{2} \sqrt{3})^{(1/3)}}{(-1)^{2/9}- (-3/2 - \frac{i}{2} \sqrt{3})^{(1/3)}}$ $= \frac{\sqrt[3\,] {e^{3i\pi}-e^{\frac{\pi}{3}}}+\sqrt[3\,] {e^{3i\pi}+e^{\frac{\pi}{3}}}}{\sqrt[3\,] {e^{3i\pi}-e^{\frac{\pi}{3}}}-\sqrt[3\,] {e^{3i\pi}+e^{\frac{\pi}{3}}}} =$

$\frac{(-i3^{1/6}+1)}{(i3^{1/6}+1)}= -1 - \frac{(2 i)}{(3^{1/6} + -i)} =\frac{-3^{1/6}-i}{3^{1/6}-i}$

 

[Delta2]

I think I've clocked it.
$\sqrt[3\,] {e^{3i\pi}+e^{\frac{\pi}{3}}}= \sqrt[3\,]{e^{\frac {2i\pi}{3}}}$

The above is clearly wrong (you are equating a complex number that has magnitude 1, with another that doesn't have magnitude 1 (correct if i am wrong but $(e^{3i\pi}+e^{\frac{\pi}{3}})$ doesn't have magnitude 1) and additionally i don't see how it relates to this problem.

$\sqrt[3\,]{e^{\frac {2i\pi}{3}}} +\sqrt[3\,]{-\sqrt 3 e^{i\pi/6}}= 3^{1/6} e^{-5i\pi/18}+e^{\frac {4i\pi}{18}}$ $=e^{\frac {4i\pi}{18}}(3^{1/6}e^{\frac {-9i\pi}{18}}+1)$ $=e^{\frac {4i\pi}{18}}(-i3^{1/6}+1)$

The above looks correct, but then again i don't see how you use it in what follows.

$\frac{(-1)^{2/9} + (-3/2 - \frac{i}{2} \sqrt{3})^{(1/3)}}{(-1)^{2/9}- (-3/2 - \frac{i}{2} \sqrt{3})^{(1/3)}}$ $= \frac{\sqrt[3\,] {e^{3i\pi}-e^{\frac{\pi}{3}}}+\sqrt[3\,] {e^{3i\pi}+e^{\frac{\pi}{3}}}}{\sqrt[3\,] {e^{3i\pi}-e^{\frac{\pi}{3}}}-\sqrt[3\,] {e^{3i\pi}+e^{\frac{\pi}{3}}}} =$

$\frac{(-i3^{1/6}+1)}{(i3^{1/6}+1)}= -1 - \frac{(2 i)}{(3^{1/6} + -i)} =\frac{-3^{1/6}-i}{3^{1/6}-i}$

 

[yamata1]

The above is clearly wrong (you are equating a complex number that has magnitude 1, with another that doesn't have magnitude 1 (correct if i am wrong but $(e^{3i\pi}+e^{\frac{\pi}{3}})$ doesn't have magnitude 1) and additionally i don't see how it relates to this problem.
The above looks correct, but then again i don't see how you use it in what follows.

Yes the mistake is in the exponent$(e^{3i\pi}+e^{\frac{\pi}{3}})$. It's $(e^{3i\pi}+e^{\frac{i\pi}{3}})$.
I just plugged $\sqrt[3\,]{e^{\frac {2i\pi}{3}}} +\sqrt[3\,]{-\sqrt 3 e^{i\pi/6}}= 3^{1/6} e^{-5i\pi/18}+e^{\frac {4i\pi}{18}}$ into $\sqrt[3\,] {e^{3i\pi}-e^{\frac{i\pi}{3}}}+\sqrt[3\,] {e^{3i\pi}+e^{\frac{i\pi}{3}}}$

 

[Delta2]

Yes the mistake is in the exponent$(e^{3i\pi}+e^{\frac{\pi}{3}})$. It's $(e^{3i\pi}+e^{\frac{i\pi}{3}})$.

It is still wrong, it is $(e^{3i\pi}+e^{\frac{i\pi}{3}})\neq e^{\frac{2i\pi}{3}}$

 

[yamata1]

It is still wrong, it is $(e^{3i\pi}+e^{\frac{i\pi}{3}})\neq e^{\frac{2i\pi}{3}}$

$(e^{3i\pi}+e^{\frac{i\pi}{3}})=-1+e^{\frac{i\pi}{3}}= -1 +1/2 + \frac{(i \sqrt(3))}{2}=-1/2 + \frac{(i \sqrt(3))}{2} =e^{\frac{2i\pi}{3}}$

 

[Delta2]

Ok I was wrong and you were right on this, but where does this expression $e^{3i\pi}+e^{\frac{i\pi}{3}}$ comes from, the original expression in post #1 doesn't contain it.

 

[yamata1]

Ok I was wrong and you were right on this, but where does this expression $e^{3i\pi}+e^{\frac{i\pi}{3}}$ comes from, the original expression in post #1 doesn't contain it.

I added it in the 8th post .

 

[Delta2]

Ok but I still don't see why we have to pass through this $e^{3i\pi}+e^{\frac{i\pi}{3}}$ as an intermediate step. It is enough to express $(-1)^\frac{2}{9}$ and $\frac{-3}{2}-\frac{\sqrt{3}}{2}$ in polar forms and then proceed.
This line you do at post #14

$\sqrt[3\,]{e^{\frac {2i\pi}{3}}} +\sqrt[3\,]{-\sqrt 3 e^{i\pi/6}}= 3^{1/6} e^{-5i\pi/18}+e^{\frac {4i\pi}{18}}$ $=e^{\frac {4i\pi}{18}}(3^{1/6}e^{\frac {-9i\pi}{18}}+1)$ $=e^{\frac {4i\pi}{18}}(-i3^{1/6}+1)$

is indeed correct and needed and you also need to find $$e^\frac{2i\pi}{9} - (-\sqrt{3}e^{i\pi/6})^{(1/3)}$$ nothing else is needed , no intermediate steps involving $e^{3i\pi}+e^{\frac{i\pi}{3}}$ are needed.

 

[archaic]

serves well as a detention exercise.