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# Locus in the complex plane.

This is a discussion on Locus in the complex plane. within the Pre-Calculus forums, part of the Pre-University Math category; Area of Region Bounded by the locus of $z$ which satisfy the equation $\displaystyle \displaystyle \arg \left(\frac{z+5i}{z-5i}\right) = \pm ... 1. Area of Region Bounded by the locus of$z$which satisfy the equation$ \displaystyle \displaystyle \arg \left(\frac{z+5i}{z-5i}\right) = \pm \frac{\pi}{4}$is 2. Originally Posted by jacks Area of Region Bounded by the locus of$z$which satisfy the equation$ \displaystyle \displaystyle \arg \left(\frac{z+5i}{z-5i}\right) = \pm \frac{\pi}{4}$is What have you tried? 3. Originally Posted by jacks Area of Region Bounded by the locus of$z$which satisfy the equation$ \displaystyle \displaystyle \arg \left(\frac{z+5i}{z-5i}\right) = \pm \frac{\pi}{4}$is You can take a geometric approach. Your relation can be written$ \displaystyle \arg(z + 5) - \arg(z - 5) = \pm \frac{\pi}{4}$, that is,$ \displaystyle \alpha - \beta =\pm \frac{\pi}{4}$. Consider the line segment joining z = 5 and z = -5 as the chord on a circle and consider the rays$ \displaystyle \arg(z +5) = \alpha$and$ \displaystyle \arg(z - 5) = \beta$subject to the restriction$ \displaystyle \alpha - \beta =\pm \frac{\pi}{4}\$. Consider the intersection of these rays and the angle between them at their intersection point. The angle is constant .... Now think of a circle theorem involving angles subtended by the same arc at the circumference .....

It's not hard to see you that have a circle with 'holes' at z = 5 and z = -5 (why?).

Now your job is to determine the radius of this circle and use it to get the area.

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