Video: Finding the Equation of a Locus

The figure shows a locus of a point 𝑧 in the complex plane. Write an equation for the locus in the form arg((𝑧 βˆ’ π‘Ž)/(𝑧 βˆ’ 𝑏)) = πœƒ, where π‘Ž, 𝑏 ∈ β„‚ and 0 < πœƒ β©½ πœ‹ are constants to be found.

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Video Transcript

The figure shows a locus of a point 𝑧 in the complex plane. Write an equation for the locus in the form the argument of 𝑧 minus π‘Ž over 𝑧 minus 𝑏 equals πœƒ, where π‘Ž and 𝑏, which are complex numbers, and πœƒ, which is greater than zero and less than or equal to πœ‹, are constants to be found.

Remember, the locus of a point 𝑧 in this form is the arc of a circle which subtends an angle of πœƒ between the points represented by 𝑧 one and 𝑧 two. We have three conditions on πœƒ. If it’s less than πœ‹ by two, the locus is a major arc. If it’s equal to πœ‹ by two, it’s a semicircle. And if it’s greater than πœ‹ 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 𝑧 is the major arc of a circle. And this makes sense because πœƒ is equal to πœ‹ by five radians.

The endpoints of our locus lie at 𝐴 and 𝐡 whose Cartesian coordinates are four, negative three and negative three, one, respectively. These represent the complex numbers four minus three 𝑖 and negative three plus 𝑖. And, remember, this locus is traced in a counterclockwise direction. Since the starting point is that represented in the complex number four minus three 𝑖, we can say that the equation of our locus is the argument of 𝑧 minus four minus three 𝑖 over 𝑧 minus negative three plus 𝑖 equals πœ‹ by five. We were actually told to find the value of the constants π‘Ž, 𝑏, and πœƒ. π‘Ž is four minus three 𝑖, 𝑏 is negative three plus 𝑖, and πœƒ is πœ‹ by five.

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