# Complex number question, involving argument and modulus

• June 20th 2009, 04:35 AM
LHS
Complex 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!
• June 20th 2009, 04:54 AM
malaygoel
Let $z=2e^{i\theta_1}$

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

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

hence angle between OB and OC is $\theta_1$

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

Now, we have,
$AB^2=39$

I think AB here represents the distance between A and B.
$5^2+2^2-2.5.2.cos(\theta_2-\theta_1)=39$

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

there is only one unknown, hence it can be solved.
• June 20th 2009, 05:35 AM
LHS
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?
• June 20th 2009, 05:54 AM
running-gag
Hi

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

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

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

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

To find (OA,OB) you can use the formula AB² = OA² + OB² - 2.OA.OB.cos(AOB)
• June 20th 2009, 06:13 AM
LHS
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.
• June 20th 2009, 08:09 AM
running-gag
Quote:

Originally Posted by LHS
Ah I understand a little better now, so the angle AOB is 2pi/3?

Yes
Quote:

Originally Posted by LHS
How do you suggest finding the lines OA or OB angle with the x axis?

The argument of a complex number z is one measurement of the angle between the x axis (positive values) and the line connecting O to the point represented by the complex number z

This is why arg(w) = (ex,OB)
• June 20th 2009, 08:40 AM
LHS
Right, ok, I see, thank for you your help!