# Thread: Another polar complex number problem

1. ## Another polar complex number problem

I've come across another question I'm having difficulty with.

The fixed points $\displaystyle A$ and $\displaystyle B$ represent the complex numbers $\displaystyle a$ and $\displaystyle b$ in an Argand diagram with origin $\displaystyle O$.

By writing $\displaystyle a = |a|e^{i\alpha}$ and $\displaystyle b = |b|e^{i\beta}$, show that $\displaystyle |Im(ab)| = 2\Delta$ where $\displaystyle \Delta$ is the area of triangle $\displaystyle OAB$.

My working takes me through these steps

$\displaystyle ab = |a||b|e^{i\alpha + i\beta}$

$\displaystyle ab = |a||b|(cos(\alpha + \beta) +isin(\alpha + \beta))$

Therefore $\displaystyle Im(ab) = |a||b|sin(\alpha + \beta)$

However from playing about with diagrams the area of the triangle $\displaystyle OAB$ looks like it should be $\displaystyle \Delta = \tfrac{1}{2}|a||b|sin(\alpha - \beta)$.

Does anyone have any thoughts on this?

2. You forgot the factor of 1/2 for the area of triangle

3. Ah so I did. I have corrected that now. What about the sine of the angles though?

4. Originally Posted by Nexus
Ah so I did. I have corrected that now. What about the sine of the angles though?
I assume that A is located in first quadrant and B is located in the fourth quadrant, then the area of the triangle will be 1/2 |a||b|sin (α + β).

Are there any restrictions to the angles? Because I think this only works if we consider case like mine or similar.

5. No other restrictions were stated in the question. I think that if the question were modified to say $\displaystyle b = |b|e^{-\beta}$ then this does lead to $\displaystyle |Im(ab)| = |a||b||sin(\alpha - \beta)| = 2\Delta$.

The book I am using is Pure Mathematics 4 by Hugh Neill and Douglas Quadling and although it is nice and accessible I have found a fair few questions which I have had to assume contain errors. I think I am going to have to add this one to that list.

Thanks for the input though!