Computing a tricky complex math problem - where did I go wrong?

• B
• Mayhem
In summary: So when you take the real part, you're essentially projecting the vector onto the real axis.In summary, the conversation is discussing a problem involving complex numbers and their exponents. The goal is to compute an equation using calculator, with the expected result being 1. However, the result obtained is different, leading to confusion about where the mistake lies. The conversation goes on to discuss the use of Euler's formula and the choice of angles, eventually arriving at the correct solution of 1.
Mayhem
I stumbled upon the following problem on instagram:
$$L = \left (\frac{-1+i\sqrt{3}}{2}\right )^6+\left (\frac{-1-i\sqrt{3}}{2}\right )^6+\left (\frac{-1+i\sqrt{3}}{2}\right )^5+\left (\frac{-1-i\sqrt{3}}{2}\right )^5$$
The idea is to compute it. Using a calculator, it is supposed to equal 1. My result is something quite different, even though my work seems sound. Where did I go wrong?:

First, we recognize that the denominator of each fraction is really just a complex number in the form $$z = a+bi$$

Thus we get
$$z_1 = a_1+b_1 i = -1+\sqrt{3}i$$
$$z_2 = a_2+b_2 i = -1-\sqrt{3}i$$
by exponentiation and substitution, we may rewrite the equation as
$$\frac{z_1^6}{64}+\frac{z_2^6}{64}+\frac{z_1^5}{32}+\frac{z_2^5}{32}$$
This is still hard to compute due to the exponents, but we can rewrite z_1 and z_2 in exponential form to simplify the problem, for which we will need to determine r and θ for the respective complex numbers.

Plugging in the numbers we get:
$$r_1 = \sqrt{a_1^2+b_1^2} = \sqrt{(-1)^2+\sqrt{3}^2} = 2$$
$$r_2 = \sqrt{a_2^2+b_2^2} = \sqrt{(-1)^2+(-\sqrt{3})^2} = 2$$
and
$$\theta_1 = \arctan{\frac{b_1}{a_1}} = \arctan{\frac{\sqrt{3}}{-1}} = -\frac{\pi}{3}$$
$$\theta_2 = \arctan{\frac{b_2}{a_2}} = \arctan{\frac{\sqrt{-3}}{-1}} = \frac{\pi}{3}$$
We can now rewrite $z_1$ and $z_2$ in exponential form:
$$z_1 = r_1e^{i\theta_1} = 2e^{-i\frac{\pi}{3}}$$
$$z_2 = r_2e^{i\theta_2} = 2e^{i\frac{\pi}{3}}$$
Now we can compute:
$$z_1^6 = 64e^{-i2\pi}$$
$$z_1^5 = 32e^-i\frac{5\pi}{3}$$

$$z_2^6 = 64e^{i2\pi}$$
$$z_2^5 = 32e^i\frac{5\pi}{3}$$
We can plug this back into our equation, and cancelling out the denominators, we get
$$L = e^{-i2\pi}+e^{i2\pi} + e^{-i\frac{5\pi}{3}}+e^{i\frac{5\pi}{3}}$$
Using eiπa = (-1)a, we can compute this. It does not equal 1, but some strange expression. I have a strong intuition that I went wrong when calculating the angles, but I'm a bit rusty on this. Could someone help?

Notice that ##-\frac{1}{2} + \frac{\sqrt{3}}{2}i = e^{\frac{2\pi i}{3}}## and ##-\frac{1}{2} - \frac{\sqrt{3}}{2}i = e^{\frac{4\pi i}{3}} = e^{-\frac{2\pi i}{3}}##. You can verify this with Euler's formula. That should make the problem slightly easier.

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As for the RHS of the first L equation, the first and the second terms are conjugate complex.
The third and the fourth terms are conjugate complex.
So L = 2 Re (the first term + the third term).
$$=2 Re ( e^{6(2/3\ \pi i)}+ e^{5(2/3\ \pi i)} )= 2(1- 1/2 )= 1$$

First of all your result doesn't equal to some strange expression as you say but it is equal to ##1+1+0.5+0.5=3##. I think your mistake lies in the choice of the angles, you should take ##\theta_1=\frac{2\pi}{3}## , which still has a ##\tan\theta_1=tan\frac{2\pi}{3}=-\sqrt{3}## and ##\theta_2=-\frac{2\pi}{3}##. Why such choice? Because i think when we are looking for the primary argument of a complex number we are looking for a positive argument.

anuttarasammyak said:
As for the RHS of the first L equation, the first and the second terms are conjugate complex.
The third and the fourth terms are conjugate complex.
So L = 2 Re (the first term + the third term).
$$=2 Re ( e^{6(2/3\ \pi i)}+ e^{5(2/3\ \pi i)} )= 2(1- 1/2 )= 1$$
I am not familiar with Re() notation, although I've seen it before. Does it simply mean we only worry about the real portion of the input?

Yes if ##z = a + bi## then ##\text{Re}(z) = a## and ##\text{Im}(z) = b##, i.e. the projections onto the real and imaginary axes respectively.

Mayhem
etotheipi said:
Yes if ##z = a + bi## then ##\text{Re}(z) = a## and ##\text{Im}(z) = b##, i.e. the projections onto the real and imaginary axes respectively.
It rings a bell now. So, in order to evaluate Re(##re^{i\theta}), one simply has to convert into rectangular form and do as shown?

Mayhem said:
It rings a bell now. So, in order to evaluate Re(##re^{i\theta}##), one simply has to convert into rectangular form and do as shown?

Sure, that works. If the complex number is ##z = re^{i\theta} = r(\cos{\theta} + i\sin{\theta})## then ##\text{Re}(z) = r\cos{\theta}## and ##\text{Im}(z) = r\sin{\theta}##. The geometric interpretation is that the complex number you can think of as a vector in the complex plane and the real/imaginary components are the cosines/sines of the arguments multiplied by the magnitude.

Mayhem

1. What should I do if I get stuck on a tricky complex math problem?

If you find yourself stuck on a tricky complex math problem, the first thing you should do is take a step back and review the problem. Make sure you understand all the given information and the question being asked. Then, try breaking down the problem into smaller, more manageable parts. You can also try using different problem-solving strategies or seeking help from a teacher or tutor.

2. How do I know if I made a mistake while computing a tricky complex math problem?

If you suspect you made a mistake while computing a tricky complex math problem, the best way to check is by reworking the problem. Go through each step carefully and double-check your calculations. If you still can't find the error, ask someone else to review your work or try using a different method to solve the problem.

3. What are some common mistakes to look out for when computing a tricky complex math problem?

Some common mistakes to look out for when computing a tricky complex math problem include: misinterpreting the given information, using the wrong formula or equation, making calculation errors, and forgetting to include units or labels in your answer. It's also important to check for any algebraic or arithmetic errors, such as missing a negative sign or forgetting to carry a digit.

4. How can I avoid making mistakes while computing a tricky complex math problem?

To avoid making mistakes while computing a tricky complex math problem, it's important to practice regularly and pay attention to detail. Take your time to fully understand the problem and use a systematic approach to solve it. Double-check your work and make sure to include all necessary steps and units in your answer. It can also be helpful to work with a partner or tutor who can provide a second set of eyes on your work.

5. What should I do if I still can't solve the tricky complex math problem after multiple attempts?

If you're having trouble solving a tricky complex math problem after multiple attempts, don't get discouraged. Take a break and come back to it later with a fresh mind. You can also try approaching the problem from a different angle or seeking help from a teacher or tutor. Remember, it's okay to ask for help and it's important to keep practicing and learning from your mistakes.

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