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- Thread starter jacks
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- Jan 26, 2012

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What have you tried?Area of Region Bounded by the locus of $z$ which satisfy the equation [tex]\displaystyle \arg \left(\frac{z+5i}{z-5i}\right) = \pm \frac{\pi}{4}[/tex] is

- Jan 26, 2012

- 66

You can take a geometric approach.Area of Region Bounded by the locus of $z$ which satisfy the equation [tex]\displaystyle \arg \left(\frac{z+5i}{z-5i}\right) = \pm \frac{\pi}{4}[/tex] is

Your relation can be written [tex]\arg(z + 5) - \arg(z - 5) = \pm \frac{\pi}{4}[/tex], that is, [tex]\alpha - \beta =\pm \frac{\pi}{4}[/tex].

Consider the line segment joining z = 5 and z = -5 as the chord on a circle and consider the rays [tex]\arg(z +5) = \alpha[/tex] and [tex]\arg(z - 5) = \beta[/tex] subject to the restriction [tex]\alpha - \beta =\pm \frac{\pi}{4}[/tex]. 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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