How to simplify this complex expression?

• yamata1
Sorry for the mistake. :\In summary, to solve the given expression, one should first write the complex numbers in the form of ##a+bi##, then use the property of multiplying by 1 to simplify the expression. One may also need to express the complex numbers in polar form to easily solve the problem.
yamata1
Poster has been reminded to show their work on schoolwork problems
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 : $$-1 - \frac{(2 i)}{(3^{1/6} + -i)}$$
$$\frac{(-1)^{2/9} + (-3/2 - \frac{i}{2} \sqrt{3})^{(1/3)}}{(-1)^{2/9}- (-3/2 - \frac{i}{2} \sqrt{3})^{(1/3)}}$$
Any hints would be nice. Thank you.

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 !

berkeman
BvU said:
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)}}

##

yamata1 said:
multiplying numerator and denominator
You fist want to write them in an ##a+bi## form. The ##(-1)^{2\over 9}## is a complex number.

BvU said:
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.

yamata1 said:
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.

Last edited:
yamata1 said:
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}##.

Delta2
vela said:
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 ?

Last edited:
yamata1 said:
##-\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.

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)

Delta2
yamata1 said:
$$(-\frac 32 - i \frac{\sqrt 3}{2})^{1/3}= \sqrt[3\,] {e^{3i\pi}-e^{\frac{\pi}{3}}}$$
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?

vela said:
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}
##

yamata1 said:
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
yamata1 said:
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}

##

yamata1 said:
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}

##

Delta2 said:
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}}}##

yamata1 said:
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}}##

Delta2 said:
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.

Delta2 said:
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 .

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
yamata1 said:
##\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.

serves well as a detention exercise.

1. How do I know where to start when simplifying a complex expression?

When simplifying a complex expression, it is important to follow the order of operations, also known as PEMDAS (Parentheses, Exponents, Multiplication/Division, Addition/Subtraction). Start by simplifying any operations within parentheses, then solve any exponents, followed by multiplication and division from left to right, and finally addition and subtraction from left to right.

2. Can I use any shortcuts or tricks to simplify a complex expression?

Yes, there are a few shortcuts that can make simplifying a complex expression easier. These include combining like terms, using the distributive property, and factoring out common factors. It is important to be familiar with these techniques and use them appropriately when simplifying an expression.

3. How do I handle negative numbers when simplifying an expression?

When working with negative numbers, it is important to remember the rules of operations. When multiplying or dividing two negative numbers, the result will be positive. When adding or subtracting a negative number, it is the same as subtracting or adding a positive number. It can also be helpful to rewrite the expression with positive numbers to make it easier to work with.

4. Is there a specific order to simplify different types of operations?

Yes, the order of operations should always be followed when simplifying a complex expression. This ensures that the expression is solved correctly and consistently. Remember to always start with parentheses, then solve any exponents, followed by multiplication and division, and finally addition and subtraction.

5. How do I know if I have simplified the expression correctly?

The best way to check if you have simplified an expression correctly is to substitute in values for the variables and see if the simplified expression gives the same result as the original expression. You can also use a calculator to check your work. Additionally, it is always a good idea to double check your steps and make sure you have followed the order of operations correctly.

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