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

By writing a as $\vert {a}|e^{ix}$ and b as $\vert {b}|e^{iy}$ show that $$|Im(ab)|=2*Area \,OAB$$

So I'm a high schooler working on this problem. The issue I'm encountering is when we rewrite to this form and multiply our complex nos we end up with $$|a||b|(\cos(x+y)+i\sin(x+y))$$

But the area of the triangle is

$$0.5\cdot |a||b|\sin(x-y)$$

Any help would be much appreciated.

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Yes.

Consider the figure below for the geometry of the problem

enter image description here Obviously, $h=|b|\sin(y-x)$ and $$2A=|a||b|\sin(y-x).$$

While $ab=|a||b|e^{i(x+y)}=|a||b|(\cos(x+y)+i\sin(x+y))$. That is, $$\operatorname {Im}(ab)=|a||b|\sin(x+y).$$

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  • $\begingroup$ doesn't this show that 2*area OAB is in fact not |Im(ab)| $\endgroup$
    – Mojimoji
    Mar 1 '18 at 14:04
  • $\begingroup$ Yes, it does. Yes, it does. $\endgroup$
    – zoli
    Mar 1 '18 at 20:26

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