# Video: Finding the Equation of a Locus

The figure shows a locus of a point π§ in the complex plane. Write an equation for the locus in the form arg((π§ β π)/(π§ β π)) = π, where π, π β β and 0 < π β©½ π are constants to be found.

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### Video Transcript

The figure shows a locus of a point π§ in the complex plane. Write an equation for the locus in the form the argument of π§ minus π over π§ minus π equals π, where π and π, which are complex numbers, and π, which is greater than zero and less than or equal to π, are constants to be found.

Remember, the locus of a point π§ in this form is the arc of a circle which subtends an angle of π between the points represented by π§ one and π§ two. We have three conditions on π. If itβs less than π by two, the locus is a major arc. If itβs equal to π by two, itβs a semicircle. And if itβs greater than π by two, the locus is a minor arc. And, remember, the endpoints are not part of this locus. We can see by looking at the diagram that the locus of our π§ is the major arc of a circle. And this makes sense because π is equal to π by five radians.

The endpoints of our locus lie at π΄ and π΅ whose Cartesian coordinates are four, negative three and negative three, one, respectively. These represent the complex numbers four minus three π and negative three plus π. And, remember, this locus is traced in a counterclockwise direction. Since the starting point is that represented in the complex number four minus three π, we can say that the equation of our locus is the argument of π§ minus four minus three π over π§ minus negative three plus π equals π by five. We were actually told to find the value of the constants π, π, and π. π is four minus three π, π is negative three plus π, and π is π by five.