Another polar complex number problem

**Nexus** · September 13th 2010, 02:20 PM

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

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

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

My working takes me through these steps

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

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

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

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

Does anyone have any thoughts on this?

**songoku** · September 13th 2010, 05:21 PM

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

**Nexus** · September 13th 2010, 11:12 PM

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

**songoku** · September 15th 2010, 12:06 AM

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.

**Nexus** · September 15th 2010, 02:04 AM

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

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!

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