http://img34.imageshack.us/img34/6084/image2maw.jpg

If anyone could help me out here, in particular if you could explain what AB^2 is referring too. Thanks!

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- Jun 20th 2009, 03:35 AMLHSComplex number question, involving argument and modulus
http://img34.imageshack.us/img34/6084/image2maw.jpg

If anyone could help me out here, in particular if you could explain what AB^2 is referring too. Thanks! - Jun 20th 2009, 03:54 AMmalaygoel
Let $\displaystyle z=2e^{i\theta_1}$

$\displaystyle w=5e^{i\theta_2}$

we have now

$\displaystyle \frac{w}{z}=2.5e^{i(\theta_2-\theta_1)}$

hence angle between OB and OC is$\displaystyle \theta_1$

which gives$\displaystyle \theta_1=\frac{\pi}{4}$

Now, we have,

$\displaystyle AB^2=39$

I think AB here represents the distance between A and B.

$\displaystyle 5^2+2^2-2.5.2.cos(\theta_2-\theta_1)=39$

$\displaystyle 29-20cos(\theta_2-\frac{\pi}{4})=39$

there is only one unknown, hence it can be solved. - Jun 20th 2009, 04:35 AMLHS
Thank you for your help!

But unfortunately your solution is outside the level of my current course (I started complex numbers and my further course last week), I do not recognise what you are doing with e. Do you know any solutions with more basic math? - Jun 20th 2009, 04:54 AMrunning-gag
Hi

$\displaystyle arg\left(\frac{w}{z}\right) = arg(w) - arg(z)$

But

$\displaystyle arg\left(\frac{w}{z}\right) = (e_x,OC) [2\pi]$

And

$\displaystyle arg(w) - arg(z) = (e_x,OB) - (e_x,OA) = (OA,OB) [2\pi]$

Therefore

$\displaystyle (e_x,OC) = (OA,OB) [2\pi]$

To find (OA,OB) you can use the formula ABČ = OAČ + OBČ - 2.OA.OB.cos(AOB) - Jun 20th 2009, 05:13 AMLHS
Ah I understand a little better now, so the angle AOB is 2pi/3?

How do you suggest finding the lines OA or OB angle with the x axis?

Unfortunately as I said earlier, I don't really understand your lines 2-4. - Jun 20th 2009, 07:09 AMrunning-gag
- Jun 20th 2009, 07:40 AMLHS
Right, ok, I see, thank for you your help!