# Thread: Complex numbers

1. ## Complex numbers

Can someone help me with this problem? I already solved a =1/2 but I'm not sure how to find the roots. 2. ## Re: Complex numbers Originally Posted by anomaly3 Can someone help me with this problem? I already solved a =1/2 but I'm not sure how to find the roots. Why do you think that $a=\dfrac{1}{2}~?$

What is $\left[2\exp\left(\dfrac{\pi i}{12}\right)\right]^4=~?$

3. ## Re: Complex numbers

$$(re^{i\theta})^n = r^ne^{i n \theta} = r^n(\cos (n\theta) + i \sin (n\theta))$$

4. ## Re: Complex numbers

In this case, how will I calculate r?

5. ## Re: Complex numbers Originally Posted by anomaly3 In this case, how will I calculate r?
@anomaly, I don't think that you understand this question.
In any case, you did not answer my question.
$\left[2\exp\left(\dfrac{\pi i}{12}\right)\right]^4=16\exp\left(\dfrac{\pi i}{3}\right)=16\left(\dfrac{1}{2}+i~\dfrac{\sqrt3} {2}\right)$

I think that you should post the original question in exact wording.

6. ## Re: Complex numbers

Sorry that I got confused by the question and to take your time.
I'll keep that in mind.
Can you please take a look at my work and see if it is correct?
Thank you very much for your help! 7. ## Re: Complex numbers

You are taking the third root, but you should be taking the fourth root of $z^4$. The first root is $z$. All four roots can be found like this:

$$\left(16e^{\tfrac{i\pi}{3}+2ni\pi}\right)^{1/4}, n=0,1,2,3$$

For $n=0$, you have $z=2e^{\tfrac{i\pi}{12}}$.
For $n=1$, you have $2e^\tfrac{7i\pi}{12}$.
For $n=2$, you have ...
For $n=3$, you have ...

Complete the last two.

9. ## Re: Complex numbers Originally Posted by anomaly3 That is correct. Really, there are an infinite number of values for $n$, but because both sine and cosine are $2\pi$-periodic, I restricted answers to where $0 \le \dfrac{2n\pi}{4} < 2\pi$, which is when $n=0,1,2,3$.

10. ## Re: Complex numbers

Thank you to both of you for taking the time to teach me! I think my understanding towards complex numbers has improved a lot. Now I can move on to the next question with a much better understanding, lol.