# Math Help - Complex Numbers Help - Again (Sorry!)

1. ## Complex Numbers Help again - Just 1 more question (Sorry!)

Sorry to bother you again so soon but this is another question I became stuck on.

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Consider the complex number:

$z = \frac {\left(\cos \frac {\pi}{4} - isin \frac {\pi}{4}\right)^2 \left(\cos \frac {\pi}{3} + isin \frac {\pi}{3}\right)^3}{\left(\cos \frac {\pi}{24} - isin \frac {\pi}{24}\right)^4}$

Find the $|z|$ and $arg z$

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Man, that LaTeX thing is hard to use!

Anyway this is what I have done so far:

$z = \frac {\left(\cos \frac {\pi}{4} - isin \frac {\pi}{4}\right)^2 \left(\cos \frac {\pi}{3} + isin \frac {\pi}{3}\right)^3}{\left(\cos \frac {\pi}{24} - isin \frac {\pi}{24}\right)^4}$

$z = \frac {\left(\frac {\sqrt 2}{2} - \frac {\sqrt 2}{2}i\right)^2 \left(\frac {1}{2} + \frac {\sqrt 3}{2}i\right)^3}{\left(\cos \frac {\pi}{24} - isin \frac {\pi}{24}\right)^4}$

No idea what $\frac {\pi}{24}$ is so I can't continue...

Maybe this is the wrong way - perhaps instead of tranforming them into numbers, I use cis instead:

$\left(\cos \frac {\pi}{3} + isin \frac {\pi}{3}\right)^3$ becomes $cis \frac {\pi}{3}$ and from there, use cis properties...

Anyway, this question has got me completely confused and so I'm finding it rather difficult to continue.

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PS: How do you guys manage to type up that LaTeX thing smoothly? It takes me ages to get it right! (my first time!)

2. Hi again

Originally Posted by sqleung
Sorry to bother you again so soon but this is another question I became stuck on.
No problem for that, it's enjoyable to read what you have done

Consider the complex number:

$z = \frac {\left(\cos \frac {\pi}{4} + isin \frac {\pi}{4}\right)^2 \left(\cos \frac {\pi}{3} + isin \frac {\pi}{3}\right)^3}{\left(\cos \frac {\pi}{24} + isin \frac {\pi}{24}\right)^4}$

$z = \frac {\left(\cos \frac {\pi}{4} - isin \frac {\pi}{4}\right)^2 \left(\cos \frac {\pi}{3} + isin \frac {\pi}{3}\right)^3}{\left(\cos \frac {\pi}{24} - isin \frac {\pi}{24}\right)^4}$
Which one is it ?
There are - and + sign that differ...

For the resolution, remember : $\boxed{\cos x+i \sin x=e^{ix}}$

Similarly, you can show that $\boxed{\cos x{\color{red}-}i \sin x=e^{{\color{red}-}ix}}$
This is why I ask you to clarify whether it is - or +

PS: How do you guys manage to type up that LaTeX thing smoothly? It takes me ages to get it right! (my first time!)
well, I would say that this one would have been a pain for me to write... ^^
It's pretty well written for a first time !

3. Oh, sorry! I mistyped the question! It should be:

$z = \frac {\left(\cos \frac {\pi}{4} - isin \frac {\pi}{4}\right)^2 \left(\cos \frac {\pi}{3} + isin \frac {\pi}{3}\right)^3}{\left(\cos \frac {\pi}{24} - isin \frac {\pi}{24}\right)^4}$

I edited my original post - thanks for correcting me.

4. Originally Posted by sqleung
Oh, sorry! I mistyped the question! It should be:

$z = \frac {\left(\cos \frac {\pi}{4} - isin \frac {\pi}{4}\right)^2 \left(\cos \frac {\pi}{3} + isin \frac {\pi}{3}\right)^3}{\left(\cos \frac {\pi}{24} - isin \frac {\pi}{24}\right)^4}$

I edited my original post - thanks for correcting me.
Ok, so like I said, replace with the exponential form :

$\cos \frac {\pi}{4} - isin \frac {\pi}{4}=e^{-i\frac{\pi}4}$

$\cos \frac {\pi}{24} - isin \frac {\pi}{24}=e^{-i\frac{\pi}{24}}$

$\cos \frac {\pi}{3} + isin \frac {\pi}{3}=e^{i\frac{\pi}{3}}$

Then, the rules of exponents... :

$(e^a)^b=e^{ab}$

$e^a e^b=e^{a+b}$

$\tfrac{e^a}{e^b}=e^{a-b}$

$\boxed{z=|z|e^{i arg(z)}}$

5. Originally Posted by sqleung
Anyway this is what I have done so far:

$z = \frac {\left(\cos \frac {\pi}{4} - isin \frac {\pi}{4}\right)^2 \left(\cos \frac {\pi}{3} + isin \frac {\pi}{3}\right)^3}{\left(\cos \frac {\pi}{24} - isin \frac {\pi}{24}\right)^4}$

$z = \frac {\left(\frac {\sqrt 2}{2} - \frac {\sqrt 2}{2}i\right)^2 \left(\frac {1}{2} + \frac {\sqrt 3}{2}i\right)^3}{\left(\cos \frac {\pi}{24} - isin \frac {\pi}{24}\right)^4}$

No idea what $\frac {\pi}{24}$ is so I can't continue...
$(\cos \theta + i \sin \theta)^n \equiv \cos n \theta + i \sin n\theta$

6. Originally Posted by Air
$(\cos \theta + i \sin \theta)^n \equiv \cos n \theta + i \sin {\color{red}n} \theta$
Typo

7. Originally Posted by sqleung
Oh, sorry! I mistyped the question! It should be:

$z = \frac {\left(\cos \frac {\pi}{4} - isin \frac {\pi}{4}\right)^2 \left(\cos \frac {\pi}{3} + isin \frac {\pi}{3}\right)^3}{\left(\cos \frac {\pi}{24} - isin \frac {\pi}{24}\right)^4}$

I edited my original post - thanks for correcting me.
$z = \frac{ \left[ \text{cis} \left( -\frac{\pi}{4}\right) \right]^2 \left[ \text{cis} \left(\frac{\pi}{3}\right) \right]^3}{\left[ \text{cis} \left(-\frac{\pi}{24}\right) \right]^4}$

Apply de Mooivre's Theorem:

$= \frac{\text{cis} \left[ 2\left( -\frac{\pi}{4}\right) \right]\, \text{cis} \left[3 \left(\frac{\pi}{3}\right)\right]}{\text{cis} \left[ 4 \left(- \frac{\pi}{24}\right)\right]} = \frac{\text{cis} \left( -\frac{\pi}{2}\right) \, \text{cis} (\pi)}{\text{cis} \left(- \frac{\pi}{6}\right)}$

Apply the usual theorems for multiplying and dividing complex numbers expressed in polar form:

$= \text{cis} \left( -\frac{\pi}{2} + \pi - \left[-\frac{\pi}{6} \right]\right) = \text{cis} \left( \frac{2\pi}{3} \right)$.

And of course you can easily express this in Cartesian form if desired.

8. Thank you all for your help - you guys are truly amazing!

9. There's another part to this question:

Using De Moivre's theorem, show that $z = \sqrt [3]{1}$

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Also, Mr Fantastic, thank you for your answer but there's still something I don't quite understand - how did you get:

$\left(\cos \frac {\pi}{24} - isin \frac {\pi}{24}\right)^4$ to become $\left[cis \frac {\pi}{4} \right]^6$

Your final answer is $cis \frac {\pi}{4}$ but I seem to have gotten $cis \frac {5\pi}{6}$

Is there something I'm not understanding?

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Also, how do I show $z = \sqrt [3]{1}$?

How can I apply De Moivre's theorem into this?

Thank you - just this last one!

10. Originally Posted by sqleung
[snip]
Also, Mr Fantastic, thank you for your answer but there's still something I don't quite understand - how did you get:

$\left(\cos \frac {\pi}{24} - isin \frac {\pi}{24}\right)^4$ to become $\left[cis \frac {\pi}{4} \right]^6$

Your final answer is $cis \frac {\pi}{4}$ but I seem to have gotten $cis \frac {5\pi}{6}$

Is there something I'm not understanding?

[snip]
*Sigh* I got that because I'm a blockhead. I've edited my reply.

The final answer should be $z = \text{cis} \left( \frac{2 \pi}{3} \right)$.

11. Originally Posted by sqleung
Also, Mr Fantastic, thank you for your answer but there's still something I don't quite understand - how did you get:

$\left(\cos \frac {\pi}{24} - isin \frac {\pi}{24}\right)^4$ to become $\left[cis \frac {\pi}{4} \right]^6$

Your final answer is $cis \frac {\pi}{4}$ but I seem to have gotten $cis \frac {5\pi}{6}$

Is there something I'm not understanding?
What can be not understood is why he made this mistake

$z=\frac{[cis(-\frac{\pi}{4})]^2 [cis(\frac{\pi}{3})]^3}{[cis(-\frac{\pi}{24})]^4}$

$z=\frac{cis(-\frac{\pi}{2}) \cdot cis(\pi)}{cis(-\frac{\pi}{6})}$

$z=cis\left(-\frac{\pi}{2}+\pi+\frac{\pi}6\right)=cis \left(\frac{2\pi}{3}\right)$

@Mr F : still not, because it's $-\frac{\pi}{24}$ in the denominator

12. Originally Posted by sqleung
There's another part to this question:

Using De Moivre's theorem, show that $z = \sqrt [3]{1}$

[snip] how do I show $z = \sqrt [3]{1}$?

How can I apply De Moivre's theorem into this?

Thank you - just this last one!
$z = \text{cis} \left( \frac{2 \pi}{3} \right) \Rightarrow z^3 = \left[ \text{cis} \left( \frac{2 \pi}{3} \right) \right]^3 = \text{cis} \left( 3 \left( \frac{2 \pi}{3} \right) \right) = \text{cis} (2 \pi) = 1$.

13. Originally Posted by Moo
What can be not understood is why he made this mistake

$z=\frac{[cis(-\frac{\pi}{4})]^2 [cis(\frac{\pi}{3})]^3}{[cis(-\frac{\pi}{24})]^4}$

$z=\frac{cis(-\frac{\pi}{2}) \cdot cis(\pi)}{cis(-\frac{\pi}{6})}$

$z=cis\left(-\frac{\pi}{2}+\pi+\frac{\pi}6\right)=cis \left(\frac{2\pi}{3}\right)$

@Mr F : still not, because it's $-\frac{\pi}{24}$ in the denominator
I fixed it while you were busy laughing at me

(Actually I know how much you love finding my mistakes so I left it as a gift for you (not quite a gold necklace).

)

14. Hehe, thank you very much for your help! I know I can always count on you guys