Question Video: Locus Defined Using the Argument of a Quotient of Two Complex Numbers | Nagwa Question Video: Locus Defined Using the Argument of a Quotient of Two Complex Numbers | Nagwa

# Question Video: Locus Defined Using the Argument of a Quotient of Two Complex Numbers Mathematics

The point π§ satisfies arg ((π§ β 6)/(π§ β 6π)) = π/4. Sketch the locus of π§ on an Argand diagram.

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### Video Transcript

The point π§ satisfies the argument of π§ minus six over π§ minus six π equals π by four. Sketch the locus of π§ on an Argand diagram.

We notice that the equation in this question looks a lot like the argument of π§ minus π§ one over π§ minus π§ two equals π. Now, the locus of π§ in this case is an arc of a circle which subtends an angle of π between the points represented by π§ one and π§ two. And of course this is sketched in a counterclockwise direction from π§ one to π§ two. Comparing the general form to the equation in our question, and we find that π§ one must be equal to six and π§ two must be equal to six π. On an Argand diagram, these are points represented by the Cartesian coordinates six, zero and zero, six, respectively. We also see that π is equal to π by four. And we know if π is less than π by two, then our locus represents a major arc.

We also know that the endpoints are not included in our locus, but here we do have a bit of a problem. We know our locus is the arc of a circle. But without knowing the center of the circle, we canβt use this information to find the correct arc. In fact, it could be this major arc or this major arc. But we do know that the locus is drawn in a counterclockwise direction from π§ one, well thatβs six, zero, to π§ two, which is zero, six. For that to be the case, we have to use this arc on the right, and we see that the locus is shown. We can also choose to add the angle of π by four radians here.

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